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

Answer: (A)

$\frac{2\pi \beta }{\alpha }$

For SHM, maximum acceleration, $a_{max}={\omega }^{2}a=\alpha$ where a is the amplitude of SHM.
and 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 $T=\frac{2\pi \beta }{\alpha }$