Question

A particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$. Then, 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 $\frac{2\pi \beta }{\alpha }$

For SHM, maximum acceleration, $a_{max}={\omega }^{2}A=\alpha$, where $A$ is the amplitude of SHM.
Maximum velocity, $v_{max}=\omega A=\beta$
So, $\frac{\alpha }{\beta }=\frac{{\omega }^{2}A}{\omega A}=\omega$
or $\frac{\alpha }{\beta }=\frac{2\pi }{T}$
or $T=\frac{2\pi \beta }{\alpha }$