Mechanics Exercise Class Ⅳ

Review Ⅰ Simple Harmonic Motion Velocity Acceleration Ⅱ Energy Ⅲ Pendulums Torsion pendulum Simple pendulum Physical pendulum

Review Ⅳ Damped Harmonic Motion Ⅴ Forced Oscillations and Resonance The greatest Ⅵ Transverse and Longitudinal Waves Sinusoidal Waves

Review Ⅶ Wave Speed on Stretched String The average power Ⅷ Interference of Waves resonance Ⅸ Standing Waves

Review Ⅶ Sound Waves The longitudinal displacement Ⅷ Interference Constructive interference Destructive interference (m=0.1.2…) Ⅸ The Doppler Effect The speed of the detector The speed of the source

The block –springs system forms a linear simple harmonic oscillator, [例1] Suppose that the two springs in figure have different spring constants k1and k2. Show that the frequency f of oscillation of the block is then given by where f1 and f2 are the frequencies at which the block would oscillate if connected only to spring 1or only to spring 2. p367 25 Solution Key idea: The block –springs system forms a linear simple harmonic oscillator, with the block undergoing SHM. o x Assuming there is a small displacement x , then the spring 1 is Stretched for x and the spring 2 is compressed for x at the same time. From the Hook’s law we can write The coefficient of a simple spring

Using the Newton’s Second Low , we can obtain Thus the angular frequency is And the frequency f of oscillation of the block is

So the x-component of the force that the spring exerted on the mass is [例2] A spring with spring constant k is attached to a mass m that is confined to move along a frictionless rail oriented perpendicular to the axis of the spring as indicated in the figure. The spring is initially unstretched and of length l0 when the mass is at the position x = 0 m in the indicated coordinate system. Show that when the mass is released from the point x along the rail, the oscillations occur but their oscillations are not simple harmonic oscillations. Solution: The mass is pulled out a distance x along the rail, the new total length of the spring is So the x-component of the force that the spring exerted on the mass is

From the diagram , so the x-component of the force is For the x-component of the force is Thus when the mass is released from the point x along the rail, the oscillations occur but their oscillations are not simple harmonic oscillations.

[例3]质量为m的物体自倾角为 的光滑斜面顶点处由静止 而滑下，滑行了l 远后与一质量为 的物体发生完全非弹性 碰撞. 与劲度系数为k的弹簧相连.碰撞前 静止于斜面上， 如图所示.问两物体碰撞后作何种运动，并解出其运动方程.已知: 漆安慎p330 9.2.10 [解 答] a为弹簧自由伸展位置,b为加 后平衡位置，O为 发生完全非弹性碰撞后的平衡位置，以O为原点建立坐标系O-x如图： 故

以物体m为隔离体，物体m由斜面顶滑下，做匀加速运动滑行l 远后速度为 再以 为系统。以完全非弹性碰撞前后为过程始末，且近 似认为碰撞过程中 位置不变。 当 发生完全非弹性碰撞之后沿ox方向的动力学方程为

受线性恢复力，做简谐运动。根据定义 的运动方程为 若以碰撞后弹簧压缩最甚时为计时起点，设此时坐标为x0 则 以弹簧自由伸长位置a为重力势能、弹性势能零点。在由碰撞后到达压缩最甚的过程中机械能守恒，有 运动方程

[例4](1)一简谐振动的运动规律为 若计时起点 提前0.5s，其运动学方程如何表示？欲使其初相为零，计时起 点应提前或推迟若干？漆安慎 P329 9.2.8 解答： 已知 （1） 计时起点提前0.5， 则 代入（1）式，运动方程为： 秒，可使初相为零，即 设计时起点提前 代入（1）式得： 则 即提前 秒时计 时可使其初相为零。

点推迟1s，它的初相是多少？欲使其初相为零，应怎样调整 计时起点？ （2）一简谐振动的运动学方程为 若计时起 点推迟1s，它的初相是多少？欲使其初相为零，应怎样调整 计时起点？ (2) 解答： 计时起点提前 秒时将 代入(2)式 . 若计时起点推迟一秒，则 ，此时初相为 若要 即推迟 秒计时时，可使初相为零。

（3）画出上面两种简谐振动在 计时起点改变前后时旋转矢量的位置. （3）画出上面两种简谐振动在 计时起点改变前后时旋转矢量的位置. 见图(a), (b)

[例5] Two sinusoidal waves , identical except for phase, travel in the same direction along a string and interfere to produce a resultant wave given by with x in meters and t in seconds. What are (a) the wavelength of the two waves, (b) the phase difference between them , and (c) their amplitude ? P395. 28 solution From the equation , we can get the wavelength of the two waves: (b) From the equation we can obtain the phase difference (c) Because we know , their amplitude is

[例6]对于平面简谐波 画出 处体元的位移-时间曲线。画出 时的波形图。 解答： （1） 可求得T=12st=0时；

（2）t=3s、6s时波形图 其相位与t=3s时相差 ，所以将t=3s的波形图向右移 ，即得t=6s时的波形图。

解 ： 写出波动方程的标准式 [例7]一平面简谐波沿 O x 轴正方向传播， 已知振幅 在 时坐标原点处的质点位于平衡位置沿 O y 轴正方向运动 . 求（1）波动方程 解 ： 写出波动方程的标准式 O

2）求 波形图. 波形方程 o 2.0 1.0 -1.0 时刻波形图

3） 处质点的振动规律并做图 . 处质点的振动方程 1.0 -1.0 2.0 O 1 2 3 4 * 处质点的振动曲线

[例8] A bat is flitting about in a cave , navigating via ultrasonic bleeps. Assume that the sound emission frequency of the bat is 39000Hz. During one fast swoop directly toward a flat wall surface , the bat is moving at 0.025 times the speed of sound in air. What frequency does the bat hear reflected off the wall? P424 . 54 Solution: There are two Doppler shifts in this situation. First, the emitted wave strikes the wall, so the sound wave of frequency is (source moving toward the stationary wall)

(observer moving toward the Second , the wall reflects the wave of frequency and reflects It , so the frequency detected , will be given : (observer moving toward the stationary source)