First slide

Simple harmonic motion

Question

The acceleration of a particle performing S.H.M. is $12 cm / \sec^{2}$ at a distance of 3 cm from the mean position. Its time period is

Moderate

A

0.5 sec
B

1.0 sec
C

2.0 sec
D

3.14 sec

Solution

$T = 2 π \sqrt{\frac{Displacement}{Acceleration}}$ $= 2 π \sqrt{\frac{3}{12}} = π = 3 .14 \sec$

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