Question

Define second's pendulum. Derive a formula for the length of second's pendulum. A particle performing linear S.H.M. has maximum velocity 25 cm/s and maximum acceleration 100 cm/s². Find period of oscillations.

Hint:

Second's pendulum length is approximately 1 m; for SHM, \( \omega = \frac{v_{\max}}{A} = \frac{a_{\max}}{v_{\max}} \).

Answer 1

Definition: A second's pendulum is a simple pendulum with a period of exactly 2 seconds for one complete oscillation, so one swing (to and fro) takes 1 second.
Derivation: Period of simple pendulum: \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( L \) is length, \( g \) is gravitational acceleration.
For T = 2 s:
\[ 2 = 2\pi \sqrt{\frac{L}{g}} \quad \Rightarrow \quad 1 = \pi \sqrt{\frac{L}{g}} \quad \Rightarrow \quad L = \frac{g}{\pi^2}. \] Period of SHM: Maximum velocity \( v_{\max} = A \omega = 25 \) cm/s, maximum acceleration \( a_{\max} = A \omega^2 = 100 \) cm/s².
\[ \omega = \frac{a_{\max}}{v_{\max}} = \frac{100}{25} = 4 \, \text{rad/s}. \] \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{s}. \] Answer: Second's pendulum has T = 2 s; \( L = \frac{g}{\pi^2} \); period = \( \frac{\pi}{2} \) s.