A pendulum bob executing simple harmonic motion has 2cm and 12Hz as amplitude and frequency respectively. Calculate the period of the motion.
A
2.00s
B
0.83s
C
0.08s
D
0.60s
Verified Correct Answer: C.
Physical Principles & Numerical Solution
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:
\(T \approx 0.08 \, \text{s}\).
Source: Myschool Past Questions Archive (Record #25476)