A simple harmonic motion is represented by $\alpha \frac{d^2 x}{d t^2}+\beta x=0$. Its period is
A simple harmonic motion is represented by $\alpha \frac{d^2 x}{d t^2}+\beta x=0$. Its period is
$\frac{2 \pi \beta}{\alpha}$
$2 \pi \sqrt{\frac{\alpha}{\beta}}$
$2 \pi \sqrt{\frac{\beta}{\alpha}}$
$\frac{2 \pi \alpha}{\beta}$
Solution
The standard equation of SHM is: $\frac{d^2 x}{d t^2}=-\omega^2 x$
On comparing with equation: $\alpha \frac{d^2 x}{d t^2}+\beta x=0$
The angular frequency is given by: $\omega=\sqrt{\frac{\beta}{\alpha}}$.
Therefore, the time period of SHM is:
$T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\alpha}{\beta}}$