Question

A body is executing simple harmonic motion. When the displacements from the mean position is $4cm$ and $5cm$, the corresponding velocities of the body is $10cm/sec$ and $8cm/\sec$. 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, $t=2\pi /\omega$
$\therefore t=2\pi /2=\pi \sec$