Q130 / 170 · Manipal (MET) 2020 · Physics

A particle, doing simple harmonic motion, at a distance 3 cm from mean position has acceleration $12 \mathrm{~cm} / \mathrm{s}^2$. What is its time period?

A
0.5 s
B
1s
C
2 s
Correct answer
D
3.14 s

Solution

The time period of a particle executing SHM is $T=2 \pi \sqrt{\frac{x}{a}}=2 \pi \sqrt{\frac{\text { Displacement }}{\text { Acceleration }}}$ $=2 \pi \sqrt{\frac{3 \times 10^{-2}}{12 \times 10^{-2}}}=2 \pi \times \frac{1}{2}$ $=\pi=3.14 \mathrm{~s}$

Hence, the correct answer is option: 'D'.

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