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Transcript
CHAPTER 8 VIBRATIONS AND WAVES 8.1 HOOKE’S LAW -The motion that repeat itself over and over is referred as periodic motion. -The motion of an oscillating object depends on the restoring forces that make it go back and forth. Eg: the beating of your heart, the ticking of a clock, the movement of a child on a swing. -If the spring is stretched or compressed a small distance, x, from its unstretched (equilibrium) position, and then released, it exerts a force on the mass. The spring force ( or restoring force), FS kx Hooke's law (spring-mass system) …(8.1) where k = spring constant, 1 x = displacement of the mass from its unstretched (not too large) (-) sign: the force exerted by the spring is always directed opposite the displacement of the mass. -The value of k is a measure of the stiffness of the spring. -Stiff spring large k -Soft spring small k -A particular type of periodic motion is Simple Harmonic Motion (SHM). -Simple harmonic motion occurs when the net force along the direction of motion is a Hooke’s law type of force-that is, when the net force is linearly proportional to the magnitude of the system's displacement from equilibrium and in the opposite direction. Eg: A clock pendulum moving back & forth SHM is described by the following parameters: Amplitude (A) : the magnitude of the maximum displacement (+A or-A) of a mass from its equilibrium position, Period (T) : the time needed to complete one cycle of oscillation. SI unit: second (s) Frequency (f) : the number of cycles (oscillations) per second SI unit : Hertz (Hz) or 1/s or cycle per second (cycle/s) -How to find acceleration??? -By combining Newton's 2nd law with Hooke's law: 2 -kx = ma a 8.2 F k x m m …(8.2) ELASTIC POTENTIAL ENERGY -Mechanical energy – 4 types: 1. 2. 3. 4. Gravitational Potential Energy Translational Kinetic Energy Rotational Kinetic Energy Elastic Potential Energy -An object has potential energy by virtue of its shape or position. -Energy is stored in a spring only when it is either stretched or compressed. -The energy stored in a stretched or compressed spring or other elastic material is called elastic potential energy (PE). PE of a spring of the system: KE of a spring: PEs 1 2 kx 2 KE 1 m 2 2 …(8.3) …(8.4) -The elastic potential energy is stored in a spring is zero when the spring is unstretched or uncompressed (x=0) -The elastic potential energy is a maximum when a spring has reached its maximum compression or extension. -PE is always positive because it proportional to x2. 3 Energy E=1/2 kA2 PE=1/2kx2 KE KEmax=E KE=1/2 mv2 -A x=0 x1 +A Figure shows the energy transfers as the spring-mass system oscillates -In SHM, the total energy is proportional to the square of the amplitude of the motion. -From conservation of mechanical energy Ei E f …(8.5) KE PE g PEs i KE PEg PEs f …(8.6) -If nonconservative forces (eg friction) are present, the final mechanical energy dos not equal the initial mechanical energy. From work-energy theorem: Wnc KE PEg PEs f KE PEg PEs i …(8.7) Because the spring force is conservative and since there are no external forces that can do work on the system, the total mechanical energy of the system consisting the block and spring remains constant. 4 Example: A 0.50-kg object is attached to a spring of spring constant 20N/m along a horizontal, frictionless surface. The object oscillates in SHM and has a speed of 1.5m/s at the equilibrium position. a. What is the total energy of the system? b. What is the amplitude c. At what locations are the valued for the PE and KE the same? Solution: a. At x=0, PE=0, So total E=KE EK b. 1 1 2 m max (0.50kg)(1.5m / s) 2 0.56 J 2 2 At x= A, v=0, so total E =PE E PE 1 2 1 2 kx kA 2 2 5 A c. 2E k 2(0.56 J ) 0.24m 20 N / m 1 2 When KE = PE, E = KE+PE = PE + PE = 2 kx2 = kx2 So x E k 0.56 J 0.17m 20 N / m 8.3 VELOCITY AS A FUNCTION OF POSITION -Refer to Fig, as the mass moves toward the origin to some new position, x, part of this energy is transformed into KE, and the PE stored in the spring is reduced to 1/2kx2. Figure (a): A mass attached to a spring on a frictionless surface is released from rest with the spring extended a distance of A. Just before released, E=1/2kx2. Figure (b) : when the mass reaches position x, it has KE and PE (decreased) Total energy, E=KE + PE 6 1 2 1 1 kA m 2 kx2 2 2 2 Solving for v, k 2 A x2 m At x=0, v=vmax At v= A , v=0 8.4 COMPARING SIMPLE HARMONIC MOTION WITH UNIFORM CIRCULAR MOTION -As the turntable rotates with constant angular speed, the shadow of the ball moves back and forth with SHM. 7 -Figure shows the ball moving with constant speed, v0, in a direction tangent to circular path. 0 sin sin 0 …(8.8) A2 x 2 A -Equating the right-hand sides of the last two expressions. 0 A A2 x 2 C A 2 x 2 …(8.9) Period and Frequency -Period, T=time required for one complete trip back and forth and T=the time it takes the ball to make one complete circular trip on the turntable. -For the ball moves through a distance 2 A, the speed of the ball around the circular path is 0 Period T 2A T 2A 0 …(8.10) For a mass-spring system moving with SHM 8 Period : T 2 m k …(8.11) Unit : second (s) Frequency f : 1 1 T 2 k m …(8.12) Unit : s-1 or hertz (Hz) Angular frequency 8.5 : 2f k m …(8.13) POSITION, VELOCITY AND ACCELERATION AS A FUNCTION OF TIME -See back the relationship between SHM and uniform circular motion. x A cos A cos(t ) …(8.14) 9 x Acost Displacement: Velocity : Acceleration : …(8.15) Asin t ..(8.16) a 2 A cos t 2 x …(8.17) Where A = amplitude = angular frequency of the motion f = frequency of revolution/rotation The maximum speed : The maximum acceleration: max A amax A 2 10 8.6 MOTION OF A PENDULUM Restoring force Ft mg sin …(8.18) For motion with a small angles Ft mg So where s L mg Ft s L T 2 L g ..(8.19) Eg : Pendulum clocks Comparison : 11 For a simple pendulum T 2 L g 1 f 2 g L g L where L = length of the pendulum, g = acc due to gravity. -The period is independent of the suspended mass. For a mass-spring system: m T 2 k f 1 2 k m k m m = mass of the object, k = spring constant -The period is dependent of the suspended mass Example: An object with a mass of 1.0kg is attached to a spring with a spring constant of 10N/m. The object is displaced by 3.0cm from the equilibrium position and let go. a. What is the amplitude A? b. What is the period T? c. What is the frequency f? Solution: a. b. c. The initial stretch is the max displacement So A=3.0cm T 2 f m 1.0kg 2 2.0s k 10 N / m 1 1 0.50 Hz T 2.0s 12