Equation Of Time
高一物理力学原理英语阅读理解25题
高一物理力学原理英语阅读理解25题1<背景文章>Newton's three laws of motion are fundamental principles in physics that have had a profound impact on our understanding of the physical world.The first law, also known as the law of inertia, states that an object at rest will stay at rest, and an object in motion will stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force. For example, a book lying on a table will remain there until someone pushes or pulls it. This law was revolutionary as it challenged the then - existing ideas about motion. Newton discovered this law through his careful observations and experiments.The second law of motion is expressed as F = ma, where F is the net force acting on an object, m is the mass of the object, and a is the acceleration. This law explains how the force applied to an object is related to its mass and acceleration. In real - life applications, when you push a shopping cart, the harder you push (greater force), the faster it will accelerate, given that the mass of the cart remains the same.The third law states that for every action, there is an equal and opposite reaction. A classic example is when a rocket launches. The rocket expels gas downward (action), and in return, the gas exerts an equal andopposite force on the rocket, propelling it upward (reaction). Newton's discovery of these laws was a milestone in the history of science and has been used in various fields such as engineering, astronomy, and transportation.1. <问题1>A. According to Newton's first law, if an object is moving in a straight line at a constant speed, what will happen if no unbalanced force acts on it?A. It will gradually slow down.B. It will keep moving at the same speed and in the same direction.C. It will suddenly change direction.D. It will stop immediately.答案：B。
中考英语经典科学实验与科学理论深度剖析阅读理解20题
中考英语经典科学实验与科学理论深度剖析阅读理解20题1<背景文章>Isaac Newton is one of the most famous scientists in history. He is known for his discovery of the law of universal gravitation. Newton was sitting under an apple tree when an apple fell on his head. This event led him to think about why objects fall to the ground. He began to wonder if there was a force that acted on all objects.Newton spent many years studying and thinking about this problem. He realized that the force that causes apples to fall to the ground is the same force that keeps the moon in orbit around the earth. He called this force gravity.The discovery of the law of universal gravitation had a huge impact on science. It helped explain many phenomena that had previously been mysteries. For example, it explained why planets orbit the sun and why objects fall to the ground.1. Newton was sitting under a(n) ___ tree when he had the idea of gravity.A. orangeB. appleC. pearD. banana答案：B。
大学物理双语-练习题
Chapter 1 Particle KinematicsI) Choose one correct answer among following choices1. An object is moving along the x-axis with position as a function of time given by x=x(t). Point O is at x=0. The object is definitely moving toward O when A. 0<dt dx B. 0>dtdx C. 0)(2<dt x d D. 0)(2>dt x d 2. An object starts from rest at x=0 when t=0. The object moves in the x direction with positive velocity after t=0. The instantaneous velocity and average velocity are related by A. v v < B. v v = C. v v > D. dt dx can be larger than, smaller than, or equal to tx 3. An object is moving in the x direction with velocity )(t v x , anddt dv x is nonzero constant. With 0=x v when t=0, then for t>0 the quantity dtdv v x x is A. Negative. B. Zero. C. Positive.D. Not determined from the information given.4. An object is moving on the xy-plane with position as a function of time given by r = a t 2 i + b t 2 j (a and b are constant). Which is correct?A. The object is moving along a straight line with constant speed.B. The object is moving along a straight line with variable speed.C. The object is moving along a curved path with constant speed.D. The object is moving along a curved path with variable speed. 5. An object is thrown into the air with an initial velocity )8.99.4(0j i v +=m /s . Ignore the air resistance (空气阻力). At the highest point the magnitude of the velocity is ( )(A) 0 (B) 4.9m/s (C) 9.8m/s (D)22)8.9()9.4(+ m/s6. Two bodies are falling with negligible air resistance, side by side, above a horizontal plane. If one of the bodies is given an additional horizontal acceleration during its descent, itA. strikes the plane at the same time as the other body.B. strikes the plane earlier than the other body.C. has the vertical component of its velocity altered.D. has the vertical component of its acceleration altered.7. A toy racing car moves with constant speed around the circle shown below. When it is at point A its coordinates are x=0, y=3m and its velocity is 6m/s i . When it is at point B its velocity and acceleration are:A. -6m/s j and 12m/s2i , respectively. B. 6m/s j and -12m/s 2 i , respectively. C. 6m/s j and 12m/s 2i , respectively. D. 6m/s j and 2m/s 2 i, respectively.8. A stone is tied to a 0.50-m string and whirled at a constant speed of 4.0m/s in a vertical circle. Its acceleration at the bottom of the circle is:A. 9.8m/s 2, upB. 9.8m/s 2, downC. 8.0m/s 2, upD. 32m/s 2, up9. A boat is able to move through still water at 20m/s. It makes a round trip to a town 3.0 km upstream. If the river flows at 5m/s, the time required for this round trip is:A. 120 sB. 150 sC. 200 sD. 320 s II) Fill in the empty space with correct answer1. A particle goes from x =-2m, y =3m, z =1m to x =3m, y =-1m, z =4m. Its displacement is : .2. The x-component of the position vector of a particle is shown in the graph in Figureas a function of time. (a) The velocity component v x at the instant 3.0 s is .(b) When is the velocity component zero ? The time is .(c) Is the particle always moving in the same direction along thex-axis? .3. The angle turned through by a wheel is given by θ＝at +bt 2, where a and b areconstants. Its angular velocity ω＝ , and its angular acceleration β＝ .4. When a radio wave impinges on the antenna of your car, electrons in the antenna move back and forth along the antenna with a velocitycomponent v x as shown schematically in Figure . Roughlysketch the same graph and indicate the time instants when(a) The velocity component v x is zero;(b) The acceleration component a x is zero;(c) The acceleration has its maximum magnitude.5. A car is traveling around a banked, circular curve of radius 150 m on a test track. At the instant when t=0s, the car is moving north, and its speed is 30.0 m/s but decreasing uniformly, so that after 5.0 s its angular speed will be 3/4 that it was when t=0s. The angular speed of the car when t=0s is , the angular speed 5.0 s later is , the magnitude of the centripetal acceleration of the car when t=0s is , the magnitude of the centripetal acceleration of the car when t=5.00s is , the magnitude of the angular acceleration is , the magnitude ofthe tangential acceleration is .6. A projectile is launched at speed v 0 at an angle θ (withthe horizontal) from the bottom of a hill of constant slope βas shown in Figure. The range of the projectile up the slopeis .III) Calculate Following Problems:1. An object with mass m initially at rest is acted by a force j t k i k F 21+=, where k 1and k 2 are constants. Calculate the velocity of the object as a function of time.2. You are operating a radio-controlled model car on a vacant tennis court. Your position is the origin of coordinates, and the surface of the court lies in the xy-plane. The car, which we represent as a point, has x- and y-cooridnates that vary with time according to x=2.0m-(0.25m/s 2)t 2 , y=(1.0m/s)t+(0.025m/s 3)t 3 .a. Find the car ’s instantaneous velocity at t=2.0s.b. Find the instantaneous acceleration at t=4.0s.3. An object moves in the xy-plane. Its acceleration has components a x =2.50t 2 and a y =9.00-1.40t. At t=0 it is at the origin and has velocity j i v 00.700.10+=.Calculate the velocity and position vectors as functions of time.4. An automobile whose speed is increasing at a rate of 0.600 m/s 2 travels along a circular road of radius 20.0 m. When the instantaneous speed of the automobile is 4.00 m/s, find (a) the tangential acceleration component, (b) the radial acceleration component, and (c) the magnitude and direction of the total acceleration.5. Heather in her Corvette accelerates at the rate of (3.00i -2.00j) m/s 2, while Jill in her Jaguar accelerates at (1.00i +3.00j ) m/s 2. They both start from rest at the origin of an xy coordinate system. After 5.00 s, (a) what is Heather’s speed with respect to Jill, (b) how far apart are they, and (c) what is Heather’s acceleration relative to Jill? Chapter 2 Newton ’s laws of motionI) Choose one correct answer among following choices1. In the SI , the base units (基本单位) for length, mass, time are ( )(A) meters, grams, seconds. (B) kilometers, kilograms, seconds.(C) centimeters, kilograms, seconds. (D) meters, kilograms, seconds.2. Which one of the following has the same dimension (量纲) as time ( )(A)a x (B) a x 2 (C) xv (D) vx 3.Which of the following quantities are independent (无关) of the choice of inertial frame (惯性系)?(A)v (B)P (C)F (D) W4. Suppose the net force F on an object is a nonzero constant. Which of the following could also be constant?A. Position.B. Speed.C. Velocity.D. Acceleration. 5. An object moves with a constant acceleration a . Which of the following expression is also constant? ( ) (A)dt v d (B) dt v d (C) dtv d )(2 (D) dt v v d )( 6. An object moving at constant velocity in an inertial frame must:A. have a net force on it.B. eventually stop due to gravity.C. not have any force of gravity on it.D. have zero net force on it.7. A heavy ball is suspended as shown. A quick jerk on the lower string will break that string but a slow pull on the lower string will break the upper string. The first result occurs because:A. the force is too small to move the ballB. action and reaction is operatingC. the ball has inertiaD. air friction holds the ball back8. A constant force of 8.0 N is exerted for 4.0 s on a16-kg object initially at rest. The change in speed of this object will be:A. 0.5m/sB. 2m/sC. 4m/sD. 8m/s9. A wedge rests on a frictionless horizontal table top. Anobject with mass m is tied to the frictionless incline of thewedge as shown in figure. The string is parallel to theincline. If the wedge accelerates to the left, when theobject leaves the incline, the magnitude of its acceleration isA. gsin θB. gcos θC. gtan θD. gcot θ10. A crane operator lowers a 16,000-N steel ball with a downward acceleration of 3m/s 2. The tension force of the cable is:A. 4900NB. 11, 000NC. 16, 000ND. 21, 000N11. A 1-N pendulum bob is held at an angle θ from the verticalby a 2-N horizontal force F as shown. The tension in the stringsupporting the pendulum bob (in newtons) is:A. cos θB. 2/ cos θC. 5D. 112. A car moves horizontally with a constant acceleration of 3m/s 2. A ball is suspended by a string from the ceiling of the car. The ball does not swing, being at rest with respect to the car. What angle does the string make with the vertical?A. 17◦B. 35◦C. 52◦D. 73◦13. A 32-N force, parallel to the incline, is required to push a certain crate at constant velocity up a frictionless incline that is 30◦ above the horizontal. The mass of the crateis:A. 3.3kgB. 3.8kgC. 5.7kgD. 6.5kg II) Fill in the empty space with correct answer 1. A 2.5 kg system has an acceleration 2)4(s m i a =. There are two forces acting onthe system, and One of the forces is N j i F )63(1 -=. The other force is .2. Two masses, m 1 and m 2, hang over an ideal pulley and the system is free to move. The magnitude of the acceleration aof the system of two masses is . The magnitude of the tension in the cord is .3. You are swinging a mass m at speed v around on astring in circle of radius r whose plane is 1.00 m abovethe ground as shown in Figure. The string makes anangle θ with the vertical direction.(a) Apply Newton’s second law to the horizontal andvertical direction to calculate the angle θ is .(b) If the angle θ = 47.4° and the radius of the circle is1.50 m, the speed of the mass is .(c) If the mass is 1.50 kg, the magnitude of the tensionin the string is .(d) The string breaks unexpectedly when the mass is movingexactly eastward. The location the mass will hit the groundis . III) Calculate Following Problems:1. A wedge with mass M rests on a frictionless horizontal tabletop. A block with mass m is placed on the wedge, and ahorizontal force F is applied to the wedge. What must be the magnitude of F if the block is to remain at a constant height above the table top?2. The mass of blocks A and B in Figure are 20.0kg and 10.0kg, respectively. The blocks are initially at rest on the floor and are connected by a massless string passing over a massless and frictionless pulley. An upward force F is applied to the pulley. Find the accelerations a 1 ofblock A and a 2 of block B when F is(a) 124N ; (b) 294N ; (c) 424N.3. An object is drop from rest. Find the function of speed withrespect to time and the terminal speed. Assuming that the drag forceis given by D = bv 2.4. A small bead can slide without friction on a circular hoop that is ina vertical plane and has a radius of 0.100m. The hoop rotates at aconstant rate of 4.00rev/s about a vertical diameter.(a) Find the angle β at which the bead is in vertical equilibrium.(b) Is it possible for the bead to “ride ” at the same elevation as thecenter of the hoop?(c) What will happen if the hoop rotates at 1.00rev/s.Chapter 3 Linear momentum, Conservation of momentum I) Choose one correct answer among following choices1. An object is moving in a circle at constant speed v . The magnitude of the rate of change of momentum of the objectA. is zero.B. is proportional to v .C. is proportional to v 2.D. is proportional to v 3.2. If the net force acting on a body is constant, what can we conclude about its momentum? A. The magnitude and/or the direction of P may change. B. The magnitude of P r remains fixed, but its direction may change. C. The direction of P remains fixed, but its magnitude may change. D. P remains fixed in both magnitude and direction.3. If I is the impulse of a particular force, what is dt I d / ?A. The momentumB. The change in momentumC. The forceD. The change in the force4. A variable force acts on an object from 0=i t to f t . The impulse of the force is zero. One can conclude that A. 0=∆r and 0=∆P . B. 0=∆r but possibly 0≠∆P . C. possibly 0≠∆r but 0=∆P . D. possibly 0≠∆r and possibly 0≠∆P .5. A system of N particles is free from any external forces. Which of the following is true for the magnitude of the total momentum of the system?A. It must be zero.B. It could be non-zero, but it must be constant.C. It could be non-zero, and it might not be constant.D. The answer depends on the nature of the internal forces in thesystem.6. The x and y coordinates of the center of mass of thethree-particle system shown below are:A. 0, 0B. 1.3m, 1.7mC. 1.4m, 1.9mD. 1.9m, 2.5m7. Block A, with a mass of 4 kg, is moving with a speed of2.0m/s while block B, with a mass of 8 kg, is moving in theopposite direction with a speed of 3m/s. The center of mass ofthe two block-system is moving with a velocity of:A. 1.3m/s in the same direction as A.B. 1.3m/s in the same direction as B.C. 2.7m/s in the same direction as A.D. 1.0m/s in the same direction as B.8. A large wedge with mass of 10kg rests on a horizontal frictionless surface, as shown. A block with a mass of 5.0kg starts from rest and slides down the inclinedsurface of the wedge, which is rough. At one instantthe vertical component of the block ’s velocity is3.0m/s and the horizontal component is 6.0m/s. Atthat instant the velocity of the wedge is:A. 3.0m/s to the leftB. 3.0m/s to the rightC. 6.0m/s to the rightD. 6.0m/s to the left9. A 1.0-kg ball moving at 2.0m/s perpendicular to a wall rebounds from the wall at1.5m/s. The change in the momentum of the ball is:A. zeroB. 0.5N · s away from wallC. 0.5N · s toward wallD. 3.5N · s away from wallII) Fill in the empty space with correct answer1. Two objects, A and B , collide (碰撞). A has a mass of kg m A 2=, and B has a mass of kg m B 4=. The velocities before the collision are )32(j i v A +=m /s and )24(j i v B +=m /s . After the collision, )23(j i v A +='m /s . The final velocity of B ='B v m /s .2. A stream of water impinges on(撞击) a stationary “dished ”turbine blade, as shown in Fig.8. The speed of the water is v ,both before and after it strikes the curved surface of the blade,and the mass of water striking the blade per unit time is constantat the value dt dm /=μ. The force exerted by the water on theblade is ___________.3. A 320g ball with a speed v of 6.22m/s strikes a wall at angle θ of 30.0o and then rebounds with the same speed and angle. It is in contact with the wall for 10.4 ms.(a) The impulse was experienced by the wall is .(b) The average force exerted by the ball on the wall is .4. The muzzle speed of a bullet can be determined using a devicecalled a ballistic pendulum, shown in Figure. A bullet of mass mmoving at speed v encounters a large mass M hanging vertically asa pendulum at rest. The mass M absorbs the bullet. The hangingmass (now consisting of M + m) then swings to some height habove the initial position of the pendulum as shown. The initialspeed v ′of the pendulum (with the embedded bullet) after impactis . The muzzle speed v of the bullet is .III) Calculate Following Problems:1. A block of mass m 1=1.60kg initially moving to the right with a speed of 4.00 m/s on a frictionless horizontal track collides with a spring attached to a second block of mass m 2=2.10kg initially moving to the left with a speed of 2.50 m/s, as shown in Figure. The spring constant is 600 N/m.(a) At the instant block 1 is moving to the right with a speed of 3.00 m/s, as in Figure, determine the velocity of block 2.(b) Determine the distance the spring is compressed at that instant.2. A3.00-kg steel ball strikes a wall with a speed of 10.0m/s at an angle of 60.0° with the surface. It bounces offwith the same speed and angle. If the ball is in contactwith the wall for 0.200 s, what is the average forceexerted on the ball by the wall?3. A small ball with mass m is released from rest at thetop of a container which inside wall is semicircle-shapedand frictionless. The container with mass M and radius R rests on a frictionless horizontal surface, as shown. When the ball slides to point B at the bottom of the container, find the normal force exerted by the container on the ball.Chapter 4 Work and EnergyI) Choose one correct answer among following choices1. The work done by gravity during the descent of a projectile:A. is positiveB. is negativeC. is zeroD. depends for its sign on the direction of the y axis2. A particle has a constant kinetic energy E k . Which of the following quantities must also be constant? ( )(A)r (B) v (C) v (D) P3. A 0.2kg block slides (滑行) across a frictionless floor with a speed of 10 m /s . The net work done on the block is ( )(A) -20J (B) -10J (C) 0J (D) 20J4. A 0.50kg object moves in a horizontal circular track with a radius of 2.5m. An external force of 3.0N, always tangent to the track, causes the object to speed up as it goes around. The work done by the external force as the mass makes one revolution is:A. 24 JB. 47 JC. 59 JD. 94 J5. A man pushes an 80-N crate a distance of 5.0m upward along a frictionless slope that makes an angle of 30◦ with the horizontal. His force is parallel to the slope. If the speed of the crate decreases at a rate of 1.5m/s 2, then the work done by the man is:A. −200 JB. 61 JC. 140 JD. 200 J6. When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude F = a x+b x 2, where a and b are constants. The work done in stretching this rubber band from x = 0 to x = L is:A. a L 2 + b Lx 3B. a L + 2b L 2C. a + 2b LD. a L 2/2 +b L 3/37. An ideal spring is hung vertically from the ceiling. When a 2.0-kg mass hangs at rest from it the spring is extended 6.0cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10cm. While the spring is being extended by the force, the work done by the spring is:A. −3.6JB. −3.3JC. 3.6 JD. 3.3J8. Two objects with masses of m 1 and m 2 have the same kinetic energy and are both moving to the right. The same constant force F is applied to the left to both masses. If m 1 = 4m 2, the ratio of the stopping distance of m 1 to that of m 2 is:A. 1:4B. 4:1C. 1:2D. 1:19. At time t = 0 a 2-kg particle has a velocity of (4m/s)i − (3m/s)j. At t = 3s its velocity is (2m/s)i + (3m/s)j . During this time the work done on it was:A. 4 JB. −4JC. −12 JD. −40 J10. A 2-kg block starts from rest on a rough inclined plane that makes an angle of 60o with the horizontal. The coefficient of kinetic friction is 0.25. As the block goes 2.0m down the plane, the mechanical energy of the Earth-block system changes by:A. 0B. −9.8JC. 9.8JD. −4.9 J II) Fill in the empty space with correct answer1. A chain (链条) is held on a frictionless table withone-fourth of its length hanging over the edge, as shown infigure. If the chain has a length L and a mass m , the work required to pull the hanging part back on the table isJ .2. A 0.1kg block is dropped from a height of 2m onto a spring offorce constant k = 2N/m, as shown. The maximum distance thespring will be compressed is _________m . (g=10m/s 2)3. A single constant force j i F 53+=N acts on a4.00-kg particle.(a) If the particle moves from the origin to the point having the vector positionj i r 32-=m, the work down by this force is .(b) If its speed at the origin is 4.00 m/s, the speed of the particle at r is . 4l(c) The change in the potential energy of the system is .III) Calculate Following Problems:1. A 3.00-kg mass starts from rest and slides a distance ddown a frictionless 30.0° incline. While sliding, it comes intocontact with an unstressed spring of negligible mass, asshown. The mass slides an additional 0.200 m as it is broughtmomentarily to rest by compression of the spring (k=400N/m). Find the initial separation d between the mass and thespring.2. Two masses are connected by a light string passing over alight frictionless pulley as shown. The 5.00-kg mass is released fromrest.(a) Determine the speed of the 3.00-kg mass just as the 5.00-kg masshits the ground.(b) Find the maximum height to which the 3.00-kg mass rises.Chapter 5 Angular momentum and Rigid bodyI) Choose one correct answer among following choices1. A particle moves with position given by j i t r 43+=, where r is measured in meters when t is measured in seconds. For each of the following, consider only t > 0. The magnitude of the angular momentum of this particle about the origin isA. increasing in time.B. constant in time.C. decreasing in time.D. undefined2. A solid object is rotating freely without experiencing any external torques. In this caseA. Both the angular momentum and angular velocity have constant direction.B. The direction of angular momentum is constant but the direction of the angular velocity might not be constant.C. The direction of angular velocity is constant but the direction of the angular momentum might not be constant.D. Neither the angular momentum nor the angular velocity necessarily has a constant direction.3. A 2.0-kg block travels around a 0.50-m radius circle with an angular velocity of 12 rad/s. The magnitude of its angular momentum about the center of the circle is:A. 6.0kg · m 2/sB. 12 kg · m 2/sC. 48 kg/m 2 · sD. 72 kg · m 2/s 24. A 6.0-kg particle moves to the right at 4.0m/s as shown. Themagnitude of its angular momentum about the point O is:A. zeroB. 288 kg·m 2/sC. 144 kg·m 2/sD.24kg·m 2/s5. Two objects are moving in the x, y plane as shown.The magnitude of their total angular momentum (aboutthe origin O) is:A. zeroB. 6kg · m2/sC. 12kg · m2/sD. 30kg · m2/s6. A 2.0-kg block starts from rest on the positive x axis 3.0m from the origin and thereafter has a constant acceleration given by )/(342s m j i a-=. At the end of 2s its angular momentum about the origin is:A. 0B. (−36 kg·m 2/s)kC. (+48 kg·m 2/s)kD. (−96 kg·m 2/s)k7. As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of 12rad/s. The circle is parallel to the xy plane and is centered on the z axis, a distance of 0.75m from the origin. The z component of the angular momentum around the origin is: A. 6.0kg · m 2/s B. 9.0kg · m 2/s C. 11kg · m 2/s D. 14kg · m 2/sII) Fill in the empty space with correct answer1. A particle located at the position vector )2(j i r-=m is acted by a force )3(j i F+=N . The torque about the origin should be_______m N ⋅.2. The velocity of a m =2kg body moving in the xy plane is given by )2(j i v-=m /s .Its position vector is )2(j i r+=m . Its angular momentum L about the origin should be___________s m kg 2⋅.3. Two particles each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. The total angular momentum of the system about any origin is .4. A particle is located at r= (0.5m)i+ (−0.3m)j+ (0.8m)k. A constant force of magnitude 2N acts on the particle. When the force acts in the positive x direction, the components of the torque about the origin is , and when the force acts in the negative x direction, the components of the torque about the origin is .5. A uniform beam of length l is in a vertical position with its lower end on a rough surface that prevents this end from slipping. The beam topples. At the instant before impact with the floor, the angular speed of the beam about its fixed end is .6. A disk of mass m and radius R is free to turn about a fixed, horizontal axle. The disk has an ideal string wrapped around its periphery from which another mass m (equal to the mass of the disk) is suspended, as indicated in Figure. The magnitude of the acceleration of the falling mass is , the magnitude of the angular acceleration of the disk is .III) Calculate Following Problems:1. The pulley has radius 0.160m and moment of inertia 0.480kg ·m2. The rope does not slip on the pulley rim. Use energy methods to calculate the speed of the 4.00-kg block just before it strikes the floor.2. A block with mass m slides down a surface inclined 30to the horizontal . The coefficient of kinetic friction is μ. A string attached to the block is wrapped around a wheel on a fixed axis. The wheel has mass m and radius R with respect to the axis of rotation. The string pulls without slipping .a) What is the acceleration of the block down the plane?b) What is the tension in the string?3. A wooden block of mass M resting on a frictionless horizontal surface is attached to a rigid rod of length l and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and normal to the rod with speed v hits the block and becomes embedded in it. What is the angular momentum of the bullet –block system?Chapter 9 Mechanic oscillationI) Choose one correct answer among following choices1. A particle on a spring executes simple harmonic motion. If the mass of the particle and the amplitude are both doubled then the period of oscillation will change by a factor ofA. 4.B. 8.C. 2.D. 2 2. A particle is in simple harmonic motion with amplitude A. At time t=0 it is at x=-A/2 and is moving in the negative direction, then the initial phase is:A. 2π/3 radB. 4π/3 radC. π radD. 3π/2 rad3. A particle is in simple harmonic motion with period T. At time t = 0 it is at the equilibrium point. Of the following times, at which time is it furthest from the equilibrium point?A. 0.5TB. 0.7TC. TD. 1.4T4. A weight suspended from an ideal spring oscillates up and down with a period T. If the amplitude of the oscillation is doubled, the period will be:A. TB. 2TC. T/2D. 4T5. The displacement of an object oscillating on a spring is given by x(t) = A cos(ωt + φ). If the initial displacement is zero and the initial velocity is in the negative x direction, then the phase constant φ is:A. 0B. π/2 radC. π radD. 3π/2 rad6. An object is undergoing simple harmonic motion with period T, amplitude A and initial phase πϕ31-=. Its graph of x versus t is:7. An object of mass m, oscillating on the end of a spring with spring constant k , has amplitude A. Its maximum speed is:A. m k A /B.m k A /2C. k m A /D. k Am /II) Fill in the empty space with correct answer1. The total energy of a simple harmonic oscillator (谐振子) with amplitude A and force constant k is_________.2. Find the initial phases (初相) of the simple harmonic motion as shown in figure.1ϕ= 2ϕ=III) Calculate Following Problems:1. An object oscillates with simple harmonic motion along the x axis. Its displacement from the origin varies with time according to the equation: )4cos()00.4(ππ+=t m x .where t is in seconds and the angles in the parentheses are in radians.－－1/21/2A －1/21/2A A －1/21/2A A －1/21/2A A。
麦克斯韦方程 讲义说明书
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Sources of electromagnetic fields
Differential form of Maxwell’s equation
Stokes’ and Gauss’ law to derive integral form of Maxwell’s equation
Some clarifications on all four equations
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Moving charge creates magnetic field
－ Typically charge and current densities are utilized in Maxwell’s equations
to quantify the effects of fields:
1. Uniform field
Gauss: Integration over closed surface
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- angle between field and normal vector
to surface matters
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ρ = electric charge density – total electric charge per unit volume V
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lim electric
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current density – total electric current per unit area S
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µ0 =permeability of free space
常用机电英语词汇(上下)
cooling water pump 冷却水泵 cooling 冷却 copy 拷贝 core 铁心、核心、磁心 correct 正确的,改正 correction 修正、改正 corrosion 腐蚀 cost of fuel 燃料费用 counting 计数器 couple CPL 联轴器 crane 起重机 curdle 凝固 currency 流动、流通 current n. 电流，水流、当前、气流 current transformer CT 电流互感器 cursor 光标 curve 曲线 cutter 切削工具 ccycle 循环、周期、周波 cylinder CYL 汽缸，圆柱体 cymometer 频率表，频率计 damage 损坏、破坏 danger zone 危险区 danger 危险、危险物 dangerous 危险的 dank 潮湿 data base 数据库 data pool 数据库 data 数据 deactivate 使无效 dead band 死区 debugging .调试 deceleration n. 减速，减速度 decrease DEC 减少 deep 深度、深的、深 default 默认(值),缺省(值) degree 度、等级 delay time 延时 delay 延迟 delete 删除 defective 有缺陷的 description 说明、描述 detect 发现、检定 detector 检测器，探测器 deviate 偏离、偏差
clockwise 顺时针、右旋的 clogged 障碍，塞满，粘注 close 关闭 closed-loop 闭环 coarse 粗的、不精确的 code 代号、密码 coder 编码器 coil 线圈 cold 冷 collect 收集 colour 颜色 command 命令、指挥 communication 通信、通讯 compensation 补偿，矫正 component 元件 compress air 压缩空气 compress 压缩 compressor 压缩机 computer 计算机 condensate 冷凝、使凝结 condition 条件、状况、环境 conduct 传导 conductivity 导电率 conductor 导体，导线 configure 组态 congealer 冷却器、冷冻器 connect 连接 connection 联接 connector 联接器、接线盒 console 控制台 constant 恒定的 contact 接触,触点,vt.接触,联系 contact to earth 接地、触地、碰地 contact 触点 contactor （电流）接触器、触头 continuous 连续的 control CNTR/CNTPL 控制 control panel 控制盘 control valve 调节阀 controller 控制器 convert 转换 vt.使转变，转换…. conveyor 传送带，输送机 cooktop 炉灶 cool 冷的 cooler 冷却器 cooling fan 冷却风机 cooling tower 冷却塔
The empirical mode decomposition and the Hilbert spectrum for nonlinear and non-stationary time seri
904N.E.Huang and others10.Discussion98711.Conclusions991References993 A new method for analysing has been devel-oped.The key part of the methodany complicated data set can be decomposed intoof‘intrinsic mode functions’Hilbert trans-This decomposition method is adaptive,and,highly efficient.Sinceapplicable to nonlinear and non-stationary processes.With the Hilbert transform,Examplesthe classical nonlinear equation systems and dataare given to demonstrate the power new method.data are especially interesting,for serve to illustrate the roles thenonlinear and non-stationary effects in the energy–frequency–time distribution.Keywords:non-stationary time series;nonlinear differential equations;frequency–time spectrum;Hilbert spectral analysis;intrinsic time scale;empirical mode decomposition1.Introductionsensed by us;data analysis serves two purposes:determine the parameters needed to construct the necessary model,and to confirm the model we constructed to represent the phe-nomenon.Unfortunately,the data,whether from physical measurements or numerical modelling,most likely will have one or more of the following problems:(a)the total data span is too short;(b)the data are non-stationary;and(c)the data represent nonlinear processes.Although each of the above problems can be real by itself,the first two are related,for a data section shorter than the longest time scale of a sta-tionary process can appear to be non-stationary.Facing such data,we have limited options to use in the analysis.Historically,Fourier spectral analysis has provided a general method for examin-the data analysis has been applied to all kinds of data.Although the Fourier transform is valid under extremely general conditions(see,for example,Titchmarsh1948),there are some crucial restrictions of Proc.R.Soc.Lond.A(1998)Nonlinear and non-stationary time series analysis905the Fourier spectral analysis:the system must be linear;and the data must be strict-ly periodic or stationary;otherwise,the resulting spectrum will make little physicalsense.to the Fourier spectral analysis methods.Therefore,behoves us review the definitions of stationarity here.According to the traditional definition,a time series,X (t ),is stationary in the wide sense,if,for all t ,E (|X (t )2|)<∞,E (X (t))=m,C (X (t 1),X (t 2))=C (X (t 1+τ),X (t 2+τ))=C (t 1−t 2),(1.1)in whichE (·)is the expected value defined as the ensemble average of the quantity,and C (·)is the covariance function.Stationarity in the wide sense is also known as weak stationarity,covariance stationarity or second-order stationarity (see,forexample,Brockwell &Davis 1991).A time series,X (t ),is strictly stationary,if the joint distribution of [X (t 1),X (t 2),...,X (t n )]and [X (t 1+τ),X (t 2+τ),...,X (t n +τ)](1.2)are the same for all t i and τ.Thus,a strictly stationaryprocess with finite second moments is alsoweakly stationary,but the inverse is not true.Both definitions arerigorous but idealized.Other less rigorous definitions have also beenused;for example,that is stationary within a limited timespan,asymptotically stationary is for any random variableis stationary when τin equations (1.1)or (1.2)approaches infinity.In practice,we can only have data for finite time spans;these defini-tions,we haveto makeapproximations.Few of the data sets,from either natural phenomena or artificial sources,can satisfy these definitions.It may be argued thatthe difficulty of invoking stationarity as well as ergodicity is not on principlebut on practicality:we just cannot have enough data to cover all possible points in thephase plane;therefore,most of the cases facing us are transient in nature.This is the reality;we are forced to face it.Fourier spectral analysis also requires linearity.can be approximated by linear systems,the tendency tobe nonlinear whenever their variations become finite Compounding these complications is the imperfection of or numerical schemes;theinteractionsof the imperfect probes even with a perfect linear systemcan make the final data nonlinear.For the above the available data are ally of finite duration,non-stationary and from systems that are frequently nonlinear,either intrinsicallyor through interactions with the imperfect probes or numerical schemes.Under these conditions,Fourier spectral analysis is of limited use.For lack of alternatives,however,Fourier spectral analysis is still used to process such data.The uncritical use of Fourier spectral analysis the insouciant adoption of the stationary and linear assumptions may give cy range.a delta function will giveProc.R.Soc.Lond.A (1998)906N.E.Huang and othersa phase-locked wide white Fourier spectrum.Here,added to the data in the time domain,Constrained bythese spurious harmonics the wide frequency spectrum cannot faithfully represent the true energy density in the frequency space.More seri-ously,the Fourier representation also requires the existence of negative light intensity so that the components can cancel out one another to give thefinal delta function. Thus,the Fourier components might make mathematical sense,but do not really make physical sense at all.Although no physical process can be represented exactly by a delta function,some data such as the near-field strong earthquake records areFourier spectra.Second,tions;wave-profiles.Such deformations,later,are the direct consequence of nonlinear effects.Whenever the form of the data deviates from a pure sine or cosine function,the Fourier spectrum will contain harmonics.As explained above, both non-stationarity and nonlinearity can induce spurious harmonic components that cause energy spreading.The consequence is the misleading energy–frequency distribution forIn this paper,modemode functions The decomposition is based on the direct extraction of theevent on the time the frequency The decomposition be viewed as an expansion of the data in terms of the IMFs.Then,based on and derived from the data,can serve as the basis of that expansion linear or nonlinear as dictated by the data,Most important of all,it is adaptive.As will locality and adaptivity are the necessary conditions for the basis for expanding nonlinear and non-stationary time orthogonality is not a necessary criterionselection for a nonlinearon the physical time scaleslocal energy and the instantaneous frequencyHilbert transform can give us a full energy–frequency–time distribution of the data. Such a representation is designated as the Hilbert spectrum;it would be ideal for nonlinear and non-stationary data analysis.We have obtained good results and new insights by applying the combination of the EMD and Hilbert spectral analysis methods to various data:from the numerical results of the classical nonlinear equation systems to data representing natural phe-nomena.The classical nonlinear systems serve to illustrate the roles played by the nonlinear effects in the energy–frequency–time distribution.With the low degrees of freedom,they can train our eyes for more complicated cases.Some limitations of this method will also be discussed and the conclusions presented.Before introducing the new method,we willfirst review the present available data analysis methods for non-stationary processes.Proc.R.Soc.Lond.A(1998)Nonlinear and non-stationary time series analysis9072.Review of non-stationary data processing methodsWe willfirstgivea brief survey of themethodsstationary data.are limited to linear systems any method is almost strictly determined according to the special field in which the application is made.The available methods are reviewed as follows.(a )The spectrogramnothing but a limited time window-width Fourier spectral analysis.the a distribution.Since it relies on the tradition-al Fourier spectral analysis,one has to assume the data to be piecewise stationary.This assumption is not always justified in non-stationary data.Even if the data are piecewise stationary how can we guarantee that the window size adopted always coincides with the stationary time scales?What can we learn about the variations longer than the local stationary time scale?Will the collection of the locally station-ary pieces constitute some longer period phenomena?Furthermore,there are also practical difficulties in applying the method:in order to localize an event in time,the window width must be narrow,but,on the other hand,the frequency resolu-tion requires longer time series.These conflicting requirements render this method of limited usage.It is,however,extremely easy to implement with the fast Fourier transform;thus,ithas attracted a wide following.Most applications of this methodare for qualitative display of speech pattern analysis (see,for example,Oppenheim &Schafer 1989).(b )The wavelet analysisThe wavelet approach is essentially an adjustable window Fourier spectral analysiswith the following general definition:W (a,b ;X,ψ)=|a |−1/2∞−∞X (t )ψ∗ t −b ad t,(2.1)in whichψ∗(·)is the basic wavelet function that satisfies certain very general condi-tions,a is the dilation factor and b is the translationof theorigin.Although time andfrequency do not appear explicitly in the transformed result,the variable 1/a givesthe frequency scale and b ,the temporal location of an event.An intuitive physical explanation of equation (2.1)is very simple:W (a,b ;X,ψ)is the ‘energy’of X ofscale a at t =b .Because of this basic form of at +b involvedin thetransformation,it is also knownas affinewavelet analysis.For specific applications,the basic wavelet function,ψ∗(·),can be modified according to special needs,but the form has to be given before the analysis.In most common applications,however,the Morlet wavelet is defined as Gaussian enveloped sine and cosine wave groups with 5.5waves (see,for example,Chan 1995).Generally,ψ∗(·)is not orthogonalfordifferent a for continuous wavelets.Although one can make the wavelet orthogonal by selecting a discrete set of a ,thisdiscrete wavelet analysis will miss physical signals having scale different from theselected discrete set of a .Continuous or discrete,the wavelet analysis is basically a linear analysis.A very appealing feature of the wavelet analysis is that it provides aProc.R.Soc.Lond.A (1998)908N.E.Huang and othersuniform resolution for all the scales.Limited by the size of thebasic wavelet function,the downside of the uniform resolution is uniformly poor resolution.Although wavelet analysis has been available only in the last ten years or so,it hasbecome extremelypopular.Indeed,it is very useful in analysing data with gradualfrequency changes.Since it has an analytic form for the result,it has attracted extensive attention of the applied mathematicians.Most of its applications have been in edge detection and image compression.Limited applications have also been made to the time–frequency distribution in time series (see,for example,Farge 1992;Long et al .1993)andtwo-dimensionalimages (Spedding et al .1993).Versatile as the wavelet analysis is,the problem with the most commonly usedMorlet wavelet is its leakage generated by the limited length of the basic wavelet function,whichmakesthe quantitativedefinitionof the energy–frequency–time dis-tribution difficult.Sometimes,the interpretation of the wavelet can also be counter-intuitive.For example,to define a change occurring locally,one must look for theresult in the high-frequencyrange,for the higher the frequency the more localized thebasic wavelet will be.If a local event occurs only in the low-frequency range,one willstill be forced to look for its effects inthe high-frequencyrange.Such interpretationwill be difficultif it is possible at all (see,for example,Huang et al .1996).Another difficulty of the wavelet analysis is its non-adaptive nature.Once the basic waveletis selected,one will have to use it to analyse all the data.Since the most commonlyused Morlet wavelet is Fourier based,it also suffers the many shortcomings of Fouri-er spectral analysis:it can only give a physically meaningful interpretation to linear phenomena;it can resolve the interwave frequency modulation provided the frequen-cy variationis gradual,but it cannot resolve the intrawave frequency modulation because the basic wavelet has a length of 5.5waves.Inspite of all these problems,wavelet analysisisstillthe bestavailable non-stationary data analysis method so far;therefore,we will use it in this paper as a reference to establish the validity and thecalibration of the Hilbert spectrum.(c )The Wigner–Ville distributionThe Wigner–Ville distribution is sometimes alsoreferred toas the Heisenberg wavelet.By definition,it is the Fourier transform of the central covariance function.For any time series,X (t ),we can define the central variance as C c (τ,t )=X (t −12τ)X ∗(t +12τ).(2.2)Then the Wigner–Ville distribution is V (ω,t )=∞−∞C c (τ,t )e −i ωτd τ.(2.3)This transform has been treated extensively by Claasen &Mecklenbr¨a uker (1980a ,b,c )and by Cohen (1995).It has been extremely popular with the electrical engi-neering community.The difficulty with this method is the severe cross terms as indicated by the exis-tence of negativepowerfor some frequency ranges.Although this shortcoming canbe eliminated by using the Kernel method (see,for example,Cohen 1995),the resultis,then,basically that of a windowed Fourier analysis;therefore,itsuffers all thelim-itations of the Fourier analysis.An extension of this method has been made by Yen(1994),who used the Wigner–Ville distribution to define wave packets that reduce Proc.R.Soc.Lond.A (1998)Nonlinear and non-stationary time series analysis909 a complicated data set to afinite number of simple components.This extension is very powerful and can be applied to a variety of problems.The applications to complicated data,however,require a great amount of judgement.(d)Evolutionary spectrumThe evolutionary spectrum wasfirst proposed by Priestley(1965).The basic idea is to extend the classic Fourier spectral analysis to a more generalized basis:from sine or cosine to a family of orthogonal functions{φ(ω,t)}indexed by time,t,and defined for all realω,the frequency.Then,any real random variable,X(t),can beexpressed asX(t)= ∞−∞φ(ω,t)d A(ω,t),(2.4)in which d A(ω,t),the Stieltjes function for the amplitude,is related to the spectrum asE(|d A(ω,t)|2)=dµ(ω,t)=S(ω,t)dω,(2.5) whereµ(ω,t)is the spectrum,and S(ω,t)is the spectral density at a specific time t,also designated as the evolutionary spectrum.If for eachfixedω,φ(ω,t)has a Fourier transformφ(ω,t)=a(ω,t)e iΩ(ω)t,(2.6) then the function of a(ω,t)is the envelope ofφ(ω,t),andΩ(ω)is the frequency.If, further,we can treatΩ(ω)as a single valued function ofω,thenφ(ω,t)=α(ω,t)e iωt.(2.7) Thus,the original data can be expanded in a family of amplitude modulated trigono-metric functions.The evolutionary spectral analysis is very popular in the earthquake communi-ty(see,for example,Liu1970,1971,1973;Lin&Cai1995).The difficulty of its application is tofind a method to define the basis,{φ(ω,t)}.In principle,for this method to work,the basis has to be defined a posteriori.So far,no systematic way has been offered;therefore,constructing an evolutionary spectrum from the given data is impossible.As a result,in the earthquake community,the applications of this method have changed the problem from data analysis to data simulation:an evo-lutionary spectrum will be assumed,then the signal will be reconstituted based on the assumed spectrum.Although there is some general resemblance to the simulated earthquake signal with the real data,it is not the data that generated the spectrum. Consequently,evolutionary spectrum analysis has never been very useful.As will be shown,the EMD can replace the evolutionary spectrum with a truly adaptive representation for the non-stationary processes.(e)The empirical orthogonal function expansion(EOF)The empirical orthogonal function expansion(EOF)is also known as the principal component analysis,or singular value decomposition method.The essence of EOF is briefly summarized as follows:for any real z(x,t),the EOF will reduce it toz(x,t)=n1a k(t)f k(x),(2.8)Proc.R.Soc.Lond.A(1998)910N.E.Huang and othersin whichf j·f k=δjk.(2.9)The orthonormal basis,{f k},is the collection of the empirical eigenfunctions defined byC·f k=λk f k,(2.10)where C is the sum of the inner products of the variable.EOF represents a radical departure from all the above methods,for the expansion basis is derived from the data;therefore,it is a posteriori,and highly efficient.The criticalflaw of EOF is that it only gives a distribution of the variance in the modes defined by{f k},but this distribution by itself does not suggest scales or frequency content of the signal.Although it is tempting to interpret each mode as indepen-dent variations,this interpretation should be viewed with great care,for the EOF decomposition is not unique.A single component out of a non-unique decomposition, even if the basis is orthogonal,does not usually contain physical meaning.Recently, Vautard&Ghil(1989)proposed the singular spectral analysis method,which is the Fourier transform of the EOF.Here again,we have to be sure that each EOF com-ponent is stationary,otherwise the Fourier spectral analysis will make little sense on the EOF components.Unfortunately,there is no guarantee that EOF compo-nents from a nonlinear and non-stationary data set will all be linear and stationary. Consequently,singular spectral analysis is not a real improvement.Because of its adaptive nature,however,the EOF method has been very popular,especially in the oceanography and meteorology communities(see,for example,Simpson1991).(f)Other miscellaneous methodsOther than the above methods,there are also some miscellaneous methods such as least square estimation of the trend,smoothing by moving averaging,and differencing to generate stationary data.Methods like these,though useful,are too specialized to be of general use.They will not be discussed any further here.Additional details can be found in many standard data processing books(see,for example,Brockwell &Davis1991).All the above methods are designed to modify the global representation of the Fourier analysis,but they all failed in one way or the other.Having reviewed the methods,we can summarize the necessary conditions for the basis to represent a nonlinear and non-stationary time series:(a)complete;(b)orthogonal;(c)local;and (d)adaptive.Thefirst condition guarantees the degree of precision of the expansion;the second condition guarantees positivity of energy and avoids leakage.They are the standard requirements for all the linear expansion methods.For nonlinear expansions,the orthogonality condition needs to be modified.The details will be discussed later.But even these basic conditions are not satisfied by some of the above mentioned meth-ods.The additional conditions are particular to the nonlinear and non-stationary data.The requirement for locality is the most crucial for non-stationarity,for in such data there is no time scale;therefore,all events have to be identified by the time of their occurences.Consequently,we require both the amplitude(or energy) and the frequency to be functions of time.The requirement for adaptivity is also crucial for both nonlinear and non-stationary data,for only by adapting to the local variations of the data can the decomposition fully account for the underlying physics Proc.R.Soc.Lond.A(1998)Nonlinear and non-stationary time series analysis911of the processes and not just to fulfil the mathematical requirements for fitting the data.This is especially important for the nonlinear phenomena,for a manifestation of nonlinearity is the ‘harmonic distortion’in the Fourier analysis.The degree of distortion depends on the severity of nonlinearity;therefore,one cannot expect a predetermined basis to fit all the phenomena.An easy way to generate the necessary adaptive basis is to derive the basis from the data.In this paper,we will introduce a general method which requires two steps in analysing the data as follows.The first step is to preprocess the data by the empirical mode decomposition method,with which the data are decomposed into a number of intrinsic mode function components.Thus,we will expand the data in a basis derived from the data.The second step is to apply the Hilbert transform to the decomposed IMFs and construct the energy–frequency–time distribution,designated as the Hilbert spectrum,from which the time localities of events will be preserved.In other words,weneed the instantaneous frequency and energy rather than the global frequency and energy defined by the Fourier spectral analysis.Therefore,before goingany further,we have to clarify the definition of the instantaneous frequency.3.Instantaneous frequencyis to accepting it only for special ‘monocomponent’signals 1992;Cohen 1995).Thereare two basicdifficulties with accepting the idea of an instantaneous fre-quency as follows.The first one arises from the influence of theFourier spectral analysis.In the traditional Fourier analysis,the frequency is defined for thesineor cosine function spanning the whole data length with constant ampli-tude.As an extension of this definition,the instantaneous frequencies also have torelate to either a sine or a cosine function.Thus,we need at least one full oscillationof a sineor a cosine wave to define the local frequency value.According to this logic,nothing full wave will do.Such a definition would not make sense forThe secondarises from the non-unique way in defining the instantaneousfrequency.Nevertheless,this difficulty is no longer serious since the introduction ofthe meanstomakethedata analyticalthrough the Hilbert transform.Difficulties,however,still exist as ‘paradoxes’discussed by Cohen (1995).For an arbitrary timeseries,X (t ),we can always have its Hilbert Transform,Y (t ),as Y (t )=1πP∞−∞X (t )t −t d t,(3.1)where P indicates the Cauchy principal value.This transformexists forallfunctionsof class L p(see,for example,Titchmarsh 1948).With this definition,X (t )and Y (t )form the complex conjugate pair,so we can have an analytic signal,Z (t ),as Z (t )=X (t )+i Y (t )=a (t )e i θ(t ),(3.2)in which a (t )=[X 2(t )+Y 2(t )]1/2,θ(t )=arctanY (t )X (t ).(3.3)Proc.R.Soc.Lond.A (1998)912N.E.Huang andothers Theoretically,there are infinitely many ways of defining the imaginary part,but the Hilbert transform provides a unique way of defining the imaginary part so that the result is ananalyticfunction.A brief tutorial on the Hilbert transform with theemphasis on its physical interpretation can be found in Bendat &Piersol is the bestlocal fitan amplitude and phase varying trigonometric function to X (t ).Even with the Hilbert transform,there is still controversy in defining the instantaneous frequency as ω=d θ(t )d t .(3.4)This leads Cohen (1995)to introduce the term,‘monocomponent function’.In prin-ciple,some limitations on the data are necessary,forthe instantaneous frequencygiven in equation (3.4)is a single value function of time.At any given time,thereis only one frequency value;therefore,it can only represent one component,hence ‘monocomponent’.Unfortunately,no cleardefinition of the ‘monocomponent’signalwas given to judge whether a function is or is not ‘monocomponent’.For lack ofa precise definition,‘narrow band’was adopted a on the data for the instantaneous frequency to make sense (Schwartz et al .1966).There are two definitions for bandwidth.The first one is used in the study of the probability properties of the signalsand waves,wherethe processes are assumed tobe stationary and Gaussian.Then,the bandwidth can be defined in spectral moments The expected number of zero crossings per unit time is given byN 0=1π m 2m 0 1/2,(3.5)while the expected number of extrema per unit time is given byN 1=1π m 4m 2 1/2,(3.6)in which m i is the i th moment of the spectrum.Therefore,the parameter,ν,definedas N 21−N 20=1π2m 4m 0−m 22m 2m 0=1π2ν2,(3.7)offers a standard bandwidth measure (see,for example,Rice 1944a,b ,1945a,b ;Longuet-Higgins 1957).For a narrow band signal ν=0,the expected numbers extrema and zero crossings have to equal.the spectrum,but in a different way.coordinates as z (t )=a (t )e i θ(t ),(3.8)with both a (t )and θ(t )being functions of time.If this function has a spectrum,S (ω),then the mean frequency is given byω = ω|S (ω)|2d ω,(3.9)Proc.R.Soc.Lond.A (1998)Nonlinear and non-stationary time series analysis913which can be expressed in another way asω =z ∗(t )1i dd tz (t )d t=˙θ(t )−i ˙a (t )a (t )a 2(t )d t =˙θ(t )a 2(t )d t.(3.10)Based on this expression,Cohen (1995)suggested that ˙θbe treated as the instanta-neous frequency.With these notations,the bandwidth can be defined asν2=(ω− ω )2 ω 2=1 ω 2(ω− ω )2|S (ω)|2d ω=1 ω 2z ∗(t ) 1i d d t− ω 2z (t )d t =1 ω 2 ˙a 2(t )d t +(˙θ(t )− ω )2a 2(t )d t .(3.11)For a narrow band signal,this value has to be small,then both a and θhave to begradually varying functions.Unfortunately,both equations (3.7)and (3.11)defined the bandwidth in the global sense;they are both overly restrictive and lack preci-sion at the same time.Consequently,the bandwidth limitation on the Hilbert trans-form to give a meaningful instantaneous frequency has never been firmly established.For example,Melville (1983)had faithfully filtered the data within the bandwidth requirement,but he still obtained many non-physical negative frequency values.It should be mentioned here that using filtering to obtain a narrow band signal is unsat-isfactory for another reason:the filtered data have already been contaminated by the spurious harmonics caused by the nonlinearity and non-stationarity as discussed in the introduction.In order to obtain meaningful instantaneous frequency,restrictive conditions have to be imposed on the data as discussed by Gabor (1946),Bedrosian (1963)and,more recently,Boashash (1992):for any function to have a meaningful instantaneous frequency,the real part of its Fourier transform has to have only positive frequency.This restriction can be proven mathematically as shown in Titchmarsh (1948)but it is still global.For data analysis,we have to translate this requirement into physically implementable steps to develop a simple method for applications.For this purpose,we have to modify the restriction condition from a global one to a local one,and the basis has to satisfy the necessary conditions listed in the last section.Let us consider some simple examples to illustrate these restrictions physically,by examining the function,x (t )=sin t.(3.12)Its Hilbert transform is simply cos t .The phase plot of x –y is a simple circle of unit radius as in figure 1a .The phase function is a straight line as shown in figure 1b and the instantaneous frequency,shown in figure 1c ,is a constant as expected.If we move the mean offby an amount α,say,then,x (t )=α+sin t.(3.13)Proc.R.Soc.Lond.A (1998)。
Simultaneous Equation Models (1)
Simultaneous Equations Models (1)The Nature of Simultaneous Equations Models● In the previous study, we considered single equation models – one Y and one or more X ’s.● The cause-and-effect relationship is from X ’s to Y .● There may be a two-way, or simultaneous, relationship between Y and X ’s.● It is difficult to distinguish dependent and explanatory variables.● Set up simultaneous equations model where variables are jointly dependent or endogenous.● Can not estimate the parameters of a single equation without taking into account of theinformation provided by other equations.● OLS estimator for single equation in simultaneous model is biased and inconsistent.● i i i i ii i i u X Y Y u X Y Y 21211222021111212101+++=+++=γββγββY 1i and u 2i are correlated, Y 2i and u 1i are correlated, so OLS leads to inconsistent estimates.Examples of Simultaneous Equations Models● Example1: Demand –and-supply modelDemand function: 0 1110<++=αααt t d t u P Q Supply function:0 1210>++=βββt t s t u P QEquilibrium condition: s t d t Q Q = Wheretime t supplied, Q demanded,quantity s t ===quantityQ d t● Price P and quantity Q are determined by the intersection of the demand and supply curves.Demand and supply curves are linear.● P and Q are jointly dependent.● the demand curve will shift upward if u 1t is positive and downward if u 1t is negative. ● A shift in the demand curve changes both P and Q● A change in u 2t will shift supply curve then change both P and Q● So u 1t and P, u 2t and P are correlated – violate the important assumption of CLRM. ● Example 2: Keynesian model of income determination● Consumption function: 10 1 10<<++=βββt t t u Y CIncome identity:)(t t t t S I C Y =+=Where C = consumption expenditure, Y = income, I = investment (assumed exogenous), S = savings, t = time, u = stochastic disturbance term● β1 is marginal propensity to consume(MPC) lying between 0 and 1.● C and Y are interdependent and Y is not expected to be independent of the disturbance term. ● Because U i shifts, then the consumption function also shifts, which, in turn, affects Y.The simultaneous equation bias: Inconsistency of OLS estimator● Use simple Keynesian model of income determination to show OLS estimator is inconsistentin simultaneous model● We want to estimate consumption function10t t t u Y C ++=ββ● First show that Y t and u t are correlated.Substituting consumption function into income identity:t t t u I Y 111011111ββββ-+-+-=t t I Y E 11011)(ββββ-+-=so12t 1]E[u)]()][([),cov(β-=--=t t t t t t u E u Y E Y E u Y● Second show that the OLS estimator 1ˆβis an inconsistent estimator of 1β, because of the correlation of Y t and u t∑∑∑∑∑∑∑∑+=++==---=21210221)( )())((ˆtttt ttt t t t t t tyy u y y u Y yy C Y Y Y Y C C ββββSo)(11 )/lim()/lim()ˆlim(2211211Yt t t N y p N u y p p σσββββ-+=+=∑∑● Plim(1ˆβ) will be always be greater than 1βThe Identification Problem● Recall the demand and supply model, if we have time-series data on P and Q only and noadditional information, can we estimate the demand function?● Need to solve the identification problem.Notations and Definitions● Take income determination model as example:Consumption function : 10 1 10<<++=βββt t t u Y CIncome identity : )(t t t t S I C Y =+=● - Endogenous variables: determined within the model- Predetermined variables: determined outside the model.- Predetermined variables include current and lagged exogenous variables and lagged endogenous variables.- Lagged endogenous variable is nonstochastic, hence a predetermined variable. - Be careful to defend the classification.●β’s are known as the structural parameters or coefficients.● Solve for endogenous variables to derive the reduced-form equations.● Reduced-form equation is the one which expresses an endogenous variable solely in terms ofthe predetermined variables and the stochastic disturbances.● Substitute consumption function into income identity:t t t w I Y +∏+∏=10Where 1t 111001 w ,11,1ββββ-=-=∏-=∏t uSubstitute income identity into consumption functiont t t w I C +∏+∏=32Where 1t 1131021 w ,1 ,1βββββ-=-=∏-=∏t u● 31 and ∏∏ are impact multipliers● Reduced form equations give the equilibrium values of the relevant endogenous variables.● The OLS method can be applied to estimate the coefficients of the reduced –form equations● Structural coefficients can be “retrieved ” from the reduced form coefficients.The identification problem● The identification problem is whether numerical estimates of the parameters of a structuralequation can be obtained from the estimated reduced form coefficients.● Identified, underidentified, exactly identified and overidentified.Underidentified● Consider the demand and supply model, together with market clearing condition.(Insert equations)● There are four structural coefficients corresponding two reduced form coefficients. – modelcan not be solved.● What does “underidentified ” mean? See figures● An alternative way to looking at the identification problem. – “mongrel ” equations. Ifmongrel equation is observational indistinguishable with demand function, then demand function is underidentified.Just , or exact, identification● Demand function: 0 ,0211210><+++=αααααt t t t u I P Q Supply function: 0 1210>++=βββt t t u P Q● There is an additional variable in the demand equation● Derive reduced form equations● Five structural coefficients corresponding with four reduced form coefficients – remainunderidentified.● Demand curve is underidentified, but supply curve is identified.● “mongrel ” equation is distinguishable from supply function but not from demand function.● The presence of an additional variable in the demand function enables us to identify thesupply function.● ConsiderDemand function : 0,0 211210><+++=αααααt t t t u I P QSupply function:0 ,0 21 21210>>+++=-βββββt t t t u P P QExactly identified!Overidentified● Demand function : 13210t t t t t u R I P Q ++++=ααααSupply function: 21210t t t t u P P Q +++=-βββ● Solvefor the structural equations and get reduced form:t t t t t v P R I P +∏++∏+∏+∏=-13210t t t t t w P R I Q +∏+∏+∏+∏=-17654● 7 coefficients corresponding 8 equations. Will have multiple solutions. For example:151201 ∏∏=∏∏=ββ● The reason for the multiple solution is that we have “too much ” information to identify thesupply curve.● “too much ” reflects by the exclusion of two variables in the supply function. One should beenough.Rules for identification● Solve structural equations, then get reduced form, check how many structural coefficientsand how many reduced form coefficients – no need for this time-consuming process● Order conditions of identificationM – number of endogenous variables in the modelm – number of endogenous variables in a given equation K – number of predetermined variables in the modelk – number of predetermined variables in a given equationA equivalent explanation is, in a model of M simultaneous equations, in order for an equationto be identified, the number of predetermined variables excluded from the equation must not be less than the number of endogenous variables included in that equation less 1. that is: K-k>= m-1● Check the previous examples.。
《时间英文表达法》课件
Preliminary expression method
Summary word
The prepositional expression method uses prepositions to describe temporal relationships, usually used in conjunction with time nouns.
Format of time
12 hour system
Divide the morning and afternoon into 12 hours per day using AM and PM.
24-hour system
Using 24 hours as a day, from 0:00 to 23:59, time is expressed in hours and minutes.
Context • Notes on Time Expression
01
Basic knowledge of time
Definition of time units
Second
One of the fundamental units of time, used to measure extremely short time periods.
02
The expression of time
Direct expression method
Summary
The direct expression method is the most common way of expressing time, which directly uses time nouns to represent time.
时间是什么？物理学家探究其终极理论(附英文)
时间是什么？物理学家探究其终极理论(2010-03-01 11:14:08)时间是什么？物理学家探究其终极理论艾琳·彼芭（Erin Biba）2010-2-26 译者：kiwi艺术家的多元宇宙想像图/贾森·托尔钦斯基（JasonTorchinsky）美国圣地亚哥（SAN DIEGO）——作为科学家使自己被别人所知的一个方法就是去解决一个真正困难的问题。
物理学家肖恩·卡罗尔（Sean Carroll）就是其中之一。
他因为尝试去回答一个目前还没有科学家能够真正解释的古老问题，已经有了一点极客圈摇滚明星的感觉；这个问题就是：时间是什么？这里，在美国科学促进会（American Association forthe Advancement of Science）的年会上，他作了关于时间之箭（the arrow of time）的报告；走廊中的科学家们拦下他，告诉肖恩他们对他的工作有多么感兴趣。
2月19日，在美国科学促进会（AAAS）的会场，卡罗尔坐下来向《连线》杂志记者解释他的理论，以及为什么马蒂麦福莱（Marty McFly，经典时光旅行电影《回到未来》中的角色；驾驶时间机器从1985年回到三十年前的1955年）的冒险在现实世界中永远不会存在，因为时间只能向前走不可能回到过去。
肖恩·卡罗尔（Sean Carroll）是加州理工大学的一位理论物理学家，他的研究兴趣是宇宙学的理论、场论（field theory）和地心引力（gravitation），他主要研究宇宙的进化。
卡罗尔的最新著作《从永恒到现在：探求时间的终极理论》（From Eternity to Here: The Quest for theUltimate Theory of Time），试图将他的时间和宇宙理论带给物理学家，同样也带给普通大众。
连线：你能从一个外行人的角度解释一下你的时间理论吗？卡罗尔：我正在试着去理解时间是如何工作的。
含常数的微分方程求时间常数
含常数的微分方程求时间常数英文版Solving Time Constants in Differential Equations with ConstantsDifferential equations are a crucial tool in various fields of science and engineering, particularly when modeling phenomena that involve rates of change. One common scenario encountered in such equations is the presence of a constant, often referred to as the "time constant." This article aims to explain how to approach and solve for the time constant in differential equations.Introduction to Time ConstantsTime constants are fundamental in understanding how systems respond to changes. They represent the time it takes for a system to reach a particular state, such as equilibrium, after a disturbance. In differential equations, time constantsoften appear as coefficients alongside the derivative terms, governing the rate at which the system evolves.Solving Differential Equations with Time ConstantsWhen dealing with differential equations containing time constants, the first step is to identify the equation's order and type. This classification helps determine the appropriate method for solving it. For example, first-order linear differential equations can be solved using integration or by separation of variables.Once the equation type is identified, the next step is to rewrite the equation in a form that facilitates solving for the time constant. This often involves rearranging terms and manipulating the equation algebraically.Techniques for SolvingTechniques such as Laplace transforms or integrating factors can be employed to solve differential equations with time constants. Laplace transforms convert the differential equation into an algebraic equation, which can be solved more easily.Integrating factors, on the other hand, are used to turn certain types of differential equations into exact equations, which can then be integrated directly.Case Studies and ApplicationsReal-world applications of differential equations with time constants abound in fields like electronics, physics, and biology. For instance, in electronics, time constants describe how capacitors and inductors respond to changes in voltage or current. In biology, they model how populations grow or decay over time.ConclusionIn summary, understanding and solving for time constants in differential equations is crucial for analyzing systems' responses to changing conditions. By applying appropriate techniques and methods, researchers and engineers can gain valuable insights into system behavior and make informed decisions based on quantitative models.中文版含常数的微分方程求时间常数微分方程是科学和工程领域中各种学科的重要工具，特别是当需要建模涉及变化率的现象时。
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THE EQUATION OF TIME
The time on a sundial is not the same as the time you read from a watch. There are two reasons for this difference. One is due to our use of zonal times that are taken from fixed longitudes 15o apart. Australian Eastern Standard Time (AEST) is referred to a longitude of 150o east of the prime meridian, and this gives a time which is 10 hours ahead of Universal Time (excluding the effect of change in local time due to day light saving). Sydney (longitude 151.18o) is quite close to longitude 150o, and so when the Sun in Sydney reaches it’s highest point (e.g. on 10 July 2010) in the sky quite close to noon on a clock, in Melbourne the Sun would reach it’s highest point around 12.25 pm, Melbourne being about 6.2 o west of Sydney in longitude.
There is a second effect which displaces the actual time of solar central meridian passage or CMP from the times listed above. This effect is called the "Equation of Time", and it is the time in minutes by which the time of CMP is displaced from mean solar noon (clock time).
The equation of time results from: 1) the 23o tilt of the Earth's rotational axis, 2) the fact that the Earth moves around the Sun in an elliptical orbit. When a planet moves in an elliptical orbit, it moves faster when it is at perihelion than when it is at aphelion. This appears to surface inhabitants as if the Sun moves around the Earth faster at perihelion (January) and slower at aphelion (July). The tilt of the Earth's axis makes the Sun appear to cross one's central meridian at different times than if the axis were perpendicular to the orbital plane. These two effects add to produce the equation of time as shown above.
When the sun is “fast”, CMP will occur earlier. When it is “slow”, CMP will occur later. So in Melbourne (excluding the effect of change in local time due to day light saving), apparent solar noon will occur around 12:04 AEST in the first week of November, and around 12:34 in the middle of February.
The actual equation of time, which is mathematically defined as apparent solar time - mean solar time, varies slightly from year to year due to variations in the Earth's eccentricity and obliquity and in the time of the solstices and equinoxes. But, for a century either side of the year 2000 it may be approximated (to an accuracy of better than one percent) by the formula:
EoT(minutes) = 9.87 sin (2B o) - 7.67 sin (B o + 78.7o)
where B o = 360 (N-81) / 365
and N is the day of the year (eg N=1 on Jan 1 and N=33 on Feb 2)
The equation of time affects any thing to do with the sun, such as sunrise and sunset times as well as the CMP time.
Material Prepared by John Kennewell and Andrew McDonald and later updated by Murray Parkinson and Rakesh Panwar. © Copyright IPS - Radio and Space Services. Comments or suggestions can be directed to education@.au。