Question

A particle executes SHM between x = - A and x = + A. The time taken by it to go from 0 to $\frac{A}{2}$ is $T_1$ and to go from $\frac{A}{2}$ to A is $T_2$. Then

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

When $y=\frac{A}{2},t=T_{1}A=A$
and $y=\frac{A}{2}+\frac{A}{2}=A,t=T_1+T_2,A=A.$ Now, $y=A\sin \omega t$
$\therefore \frac{A}{2}=A\sin \omega T_{1}or{T}_1=\frac{1}{2}=\sin \frac{\pi }{6}$
$\therefore \omega T_1=\frac{\pi }{6}\Rightarrow T_1=\frac{\pi }{6\omega }...\left(1\right)$
and $A=A\sin \omega (T_1+T_2)$
$\therefore \omega (T_1+T_2)=\frac{\pi }{2}\therefore T_1+T_2=\frac{\pi }{2\omega }\therefore T_2=\frac{\pi }{2\omega }-\frac{\pi }{6\omega }$ from (1)
$T_2=\frac{2\pi }{6\omega }\therefore T_2>T_1$