For a particle performing S.H.M. the equation $\left(\frac{\mathrm{d}^2…
For a particle performing S.H.M. the equation $\left(\frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}\right)+\alpha \mathrm{x}=0$. Then the time period of the motion will be
$\frac{2 \pi}{\alpha}$
$2 \pi \alpha$
$2 \pi \sqrt{\alpha}$
$\frac{2 \pi}{\sqrt{\alpha}}$
Solution
The correct option is (D).
Concept: For SHM $\frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}=-\left(\omega^2\right) \mathrm{x}$ is the necessary condition, where $\omega$ is the angular frequency.
On comparing with the given equation $\omega=\sqrt{\alpha}$
Therefore, the time period of the SHM is:
$\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\alpha}}$
.