First slide

Simple harmonic motion

Question

A particle is undergoing SHM along a straight line so that its period is 12 s. The time it takes in traversing a distance equal to half its amplitude from the equilibrium position is

Moderate

A

3 s
B

2 s
C

1 s
D

0.5 s

Solution

$\begin{array}{l} y = A \sin ω t \\ \frac{A}{2} = A \sin \frac{2 π}{12} t \\ \frac{1}{2} = \sin \frac{π}{6} t ∴ \frac{π t}{6} = \frac{π}{6} \Rightarrow t = 1 s \end{array}$

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