WAEC - Physics (1999 - No. 10)
A pendulum bob executing simple harmonic motion has 2cm and 12Hz as amplitude and frequency respectively. Calculate the period of the motion.
2.00s
0.83s
0.08s
0.60s
توضیح
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:
\(T \approx 0.08 \, \text{s}\).