Question

The equation of a particle executing SHM is $\frac{2d^2x}{dt^2}+32x=0$Then the time period of the body is

• A. zero
• B. $\frac{\pi}{2}$
• C. $\pi$
• D. $2\pi$

Answer 1

Answer: (B)

$\frac{\pi}{2}$

Answer 2

$\frac{d^2 x}{dt^2}+16x=0$ $\therefore \, \, \, \, \, \omega^2 =16 \Rightarrow \, \, \, \omega=4$ $\, \, \, \, \, \, \, \, \, \, T=\frac{2\pi}{\omega} =\frac{2\pi}{4}=\frac{\pi}{2}$