Periodic Motion

1. How does the frequency of vibration of a simple pendulum related with its length? Hence estimate the frequency of second pendulum.

The time period of simple pendulum is, $$T=2\pi \sqrt{\frac{l}{g}}$$

Then the frequency of vibration of a simple pendulum is given by $$f=\frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$$ Hence, the frequency is inversely proportional to the square root of the effective length of the simple pendulum.

Now, the time period of second’s pendulum is T = 2 seconds, $$\therefore frequency (f) = \frac{1}{T} = \frac{1}{2} = 0.5Hz$$

2. In usual notation, a simple harmonic motion is given as $$y=a sin (\omega t - \phi).$$ Find its acceleration.

The displacement of the SHM is given as, $$y=a sin (\omega t - \phi) ... (i)$$

Then the acceleration is given by $$a=\frac{d^2y}{dt^2} = \frac{d^2}{dt^2} [a sin (\omega t - \phi)] $$

$$= \frac{d}{dt} [a \omega cos (\omega t - \phi)] $$

$$ = -a \omega^2 sin (\omega t - \phi) $$

$$ = - \omega^2 y \hspace{2cm} [from (i)] $$

Therefore, acceleration , $$a = -\omega^2 y$$

3. A pendulum clock is taken to moon, will it gain or loss the time? Why?

The time period of simple pendulum is, $$T=2\pi \sqrt{\frac{l}{g}}$$

where is the length and g is the acceleration due to gravity. On the moon surface, the value of g is less than that on the earth surface. So, the time period will be more. Hence, the clock pendulum will be late and thus it will lose time.

4. Explain why soldiers are ordered to break steps while crossing a bridge.

The bridge vibrates with its own natural frequency. When soldiers walk on the bridge with periodic steps, the bridges is set to in vibration by the external periodic force. If the frequency of this periodic force is equal to the natural frequency of the bridges, the amplitudes of vibration increases at each step and become very large. Then, the bridge might break. Hence, soldiers are ordered to break the steps.

5. Why are bells made of metal and not of wood? Explain

When bell is rang, it starts vibrating with gradual decrease in amplitudes. This is damped oscillation. In case of wood bell, the damping is more than that with metal bell. So, the metal bell vibrates for long time but wood bell does not vibrate for long time. Hence, bells are made of metal.

6. A pendulum clock is in an elevator that descends at constant velocity. Does it keep correct time? If the same clock is in an elevator in free fall, does it keep correct time?

The time period of simple pendulum is, $$T=2\pi \sqrt{\frac{l}{g}}$$

where is the length and g is the acceleration due to gravity. When the pendulum clock is in an elevator that descends at constant velocity, g = 9.8 ms^-2. So, the time period does not change and hence it keep correct time. When, the clock is in an elevator in free fall, effective value of g = 0. So, $$T=\infty$$ Thus, the pendulum will be at rest and does not keep time.

7. A body is moving in a circular path with constant speed. Is this motion a simple harmonic? Why?

When a body is moving in a circular path with constant speed, it is said to be in uniform circular motion. The radial acceleration is $$a=\omega^2 r$$ and tangential acceleration is $$\alpha = \frac{a}{r}$$ then, the acceleration is not directly proportional to the displacement. Hence, this motion is not a simple harmonic. But the projection of such uniform motion along the diameter of cicular path is SHM.

8. What are the drawbacks of simple pendulum?

The main drawbacks of simple pendulum are:

a. The assumption of mass-less and inextensible string for simple pendulum is difficult to achieve.

b. The range of vibration has to be less.

c. The same plane for vibration is difficult to achieve.

d. No heavy point mass object is found.

9. On what factors does the period of a simple pendulum depends?

The time period of simple pendulum is, $$T=2\pi \sqrt{\frac{l}{g}}$$ Thus, $$T\propto \sqrt{l}$$ and $$T\propto \frac{1}{\sqrt{g}}$$

Hence, the time period depends on effective length and acceleration due to gravity as well as the temperature of the pendulum.

10. If length of a simple pendulum increased by 4 times its original length, will its time period change? If yes, by how much?

The time period of simple pendulum is, $$T=2\pi \sqrt{\frac{l}{g}}$$ Let the increased length be $$l'=4l,$$ then, the time period becomes

$$T'=2\pi \sqrt{\frac{l'}{g}}=2\pi \sqrt{\frac{4l}{g}}=2(2\pi \sqrt{\frac{l}{g}})$$ $$T'=2T$$ Thus, the time period is increased to 2 times its initial time period.

11. If a pendulum clock is taken to a mountain top, does it gain or loss time?

The time period of simple pendulum is, $$T=2\pi \sqrt{\frac{l}{g}}$$

where is the length and g is the acceleration due to gravity. At the top of the mountain, g is less than at the sea level. So, the time period of the clock will be more. Hence, such clock will lose time.

12. The position of a certain object in SHM is given as x = 0.05 cos (290t + 2.5); where x is in meter and t is in sec. What are the amplitude, period and initial phase angle for this motion?

The displacement of the SHM is given as, $$y=a sin (\omega t + \phi)$$ comparing with the given equation of SHM as x = 0.05 cos (290t + 2.5), we get

amplitude, a=0.05 m

Angular frequency, ω=290

or, 2πf=290

f=290/2π=46.155 Hz

T=1/f=0.022 sec

And phase angle, φ=2.5 rad