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) For a simple pendulum, k itself is proportional to m, therefore, m//k is constant, Hence, the time period of a pendulum is independent of the mass of the pendulum.
`F = -mg sin theta`
For small values of `theta, sin theta = theta`
`therefore` We ger `T = 2pi sqrt((l)/(g))`
For larger value of `theta, sin theta` will be greater than `theta`.
This will decrease the value of g and hence, time period will increase.
(c ) As the working of a wristwatch depends on the spring action, it is not affected by acceleration due to gravity during free fall. Therefore, the watch gives correct time.
(d) In a freely falling pendulum, gravity disappears, so frequency of oscillation of simple pendulum will be zero.