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. $2 \pi s$

Answer 1

Answer: (A)

$\pi \sqrt{2} s$

Answer 2

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
$T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{2}}$
$=\sqrt{2} \pi=\pi \sqrt{2} \,s$