Question

If the differential equation for a simple harmonic motion is $\frac{d^2 y}{dt^2} + 2 y = 0$, the time-period of the motion is

• A. $\pi \sqrt{2} s $
• B. $ \frac{\sqrt{2}}{\pi} s $
• C. $ \frac{\pi}{\sqrt{2}} s $
• D. $2 \pi s$
• E. $ \frac{\sqrt{\pi}}{2} s $

Answer 1

Answer: (A)

Given,
$\frac{d^{2} y}{d t^{2}}+2 y=0\,...(i)$
But, we know that
$\frac{d^{2} y}{d t^{2}}+\omega^{2} Y=0\,...(ii)$
Comparing both equation, we get
$\omega^{2} =2 $
$\Rightarrow \omega =\sqrt{2} $
The periodic time
$T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{2}}$
$=\sqrt{2} \pi=\pi \sqrt{2} \,s$