A pendulum bob executing simple harmonic motion has 2cm and 12Hz as amplitude and frequency...

A pendulum bob executing simple harmonic motion has 2cm and 12Hz as amplitude and frequency respectively. Calculate the period of the motion.

Given:
- Frequency \( f = 12 \, \text{Hz} \)

The period \( T \) is the reciprocal of the frequency:
\(T = \frac{1}{f}\)

Substituting the given frequency:
\(T = \frac{1}{12 \, \text{Hz}} \approx 0.0833 \, \text{s}\)

Thus, the period of the motion is approximately:
\(T \approx 0.08 \, \text{s}\).

Previous Next

{{ settings.no_comment_msg ? settings.no_comment_msg : 'There are no comments' }}