First slide

Simple hormonic motion

Question

A particle executes S.H.M between x = -A to x = +A. The time taken for it in going from 0 to $\frac{A}{2}$ is $T_{1}$ and from $\frac{A}{2}$ to A is $T_{2}$ . Then

Moderate

A

$T_{1} < T_{2}$
B

$T_{1} > T_{2}$
C

$T_{1} = T_{2}$
D

$T_{1} = 2 T_{2}$

Solution

The speed at mean position is maximum and at extreme position, speed is zero, so $T_{1} < T_{2}$ .
$T_{1} = T / 12 a n d T_{2} = T / 6$

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