Question

A particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta .$ Its time period of vibration will be

• A. $\frac{2\pi \beta }{\alpha }$
• B. $\frac{{\beta }^2}{{\alpha }^2}$
• C. $\frac{\alpha }{\beta }$
• D. $\frac{{\beta }^2}{\alpha }$

Answer 1

The correct option is A $\frac{2\pi \beta }{\alpha }$

We know that,
Magnitude of maximum acceleration $a_{max}=A{\omega }^2$
Magnitude of maximum velocity$v_{max}=A\omega$

Given,
$a_{max}=A{\omega }^2=\alpha$ --- (1)
$v_{max}=A\omega =\beta$ --- (2)

Dividing (1) by (2),
$\omega =\frac{\alpha }{\beta }$
Then time period
$T=\frac{2\pi }{\omega }=\frac{2\pi \beta }{\alpha }$