Question

A body is executing simple harmonic motion. When the displacements from the mean position is $4 \,cm$ and $5 \,cm$, the corresponding velocities of the body is $10 \,cm/sec$ and $8\, cm/\sec$. Then the time period of the body is

• A. $\frac{2\pi}{2}$ sec
• B. $ \pi / 2 $ sec
• C. $ \pi $ sec
• D. $ ( 3 \pi / 2) $ sec

Answer 1

Answer: (C)

For a body executing SHM, velocity,
$v=\sqrt{\omega^{2}\left(a^{2}-y^{2}\right)}$ we have
$10^{2}=\omega^{2}\left(a^{2}-4^{2}\right)$
and $8^{2}=\omega^{2}\left(a^{2}-5^{2}\right)$
So, $ 10^{2}-8^{2}=\omega^{2}\left(5^{2}-4^{2}\right)=(3 \omega)^{2}$
or $6=3 \omega$
or $ \omega=2$
$\because $ Time, $t=2 \pi / \omega$
$\therefore t=2 \pi / 2=\pi \,\sec$