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 motino of simple pendulum is approximatley 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 quanlitative 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 teh frequency of oscillation of a simple pendulum mounted iin a cabin that is freely falling under gravity?

Explanation

(a) For a simple pendulum, force constant or spring factor k is proportional to mass m, therefore, m cancels out in denominator as well as in numerator. That is why the time period of simple pendulum is independent of the mass of the bob.
(b) The effective restoring force acting on the bob of simple pendulum in displaced position is `F=-mg sin theta`. When `theta` is small, `sintheta~~theta`. Then the expression for time period of simple pendulum is give by `T=2pisqrt(l//g)`
When `theta` is large, `sinthetalttheta`. If the restoring force `mg sin theta` is replaced by `mg theta`, this amounts to effective reduction in the value of 'g' for large angles and hence an increase in the value of time period T.
(c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do with gravity.
(d) We know that gravity disappears for a man under free fall, so frequency is zero.