Answer the following questions: (a) Time period of particle is S.H.M. depends on the force constant k and mass m of the particle `: T=2pisqrt(m//k).` A simple pendulum executes S.H.M. approximately. Why then is the time-period of a pendulum independent of the mass of the pendulum?
(b) The motion of simple pendulum is approximately simple harmonic for small angles of oscillation. For large angle of oscillation, a more involved analysis (beyond the scope of this book) shows that T is greater that `2pisqrt(l//g)`. Think of a quantitative argument to appreciate this result.
(c) A man with a wrist watch on his hand falls from the top of tower. Does the watch give correct time during the free fall?
(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?

Explanation

(a) In case of a spring k does not depend upon m. However, in case of a simple pendulum, k is directly proportional to m and hence the ratio `(m)/(k)` is a constant quantity.
(b). The restoring force for the bob of the pendulum is given by
`F =- mg sin theta`
If `theta` is small, then `sin theta = theta = (y)/(l) :. F =- (mg)/(l)y`
i.e, the motion is simple harmonic and time period is `T = 2pi sqrt((l)/(g))`.
Clearly, the above formula is obtained only if we apply the approximation `sin theta = theta`.
For large angles, this approximation is not valid and T is greater than `2pi sqrt((l)/(g))`.
(c) The wrist watch uses and electronic system or spring sysyem to give the time, which does not change with acceleration due to gravity. therefore, watch gives the correct time.
(d) During free fall of the cabin, the acceleration due to gravity is zero. Therefore, the frequency of oscillations is also zero i.e., the pendulum will not vibrate at all