Question

A body undergoing simple harmonic motion has a maximum acceleration of $8m/s^2$ and maximum speed of $1.6m/s$. What is the time period $T$ ?

• A. 0.1 seconds
• B. 0.2 seconds
• C. 0.3 seconds
• D. 0.4 seconds
• E. 0.5 seconds

Answer 1

Answer: (D)

In the given simple harmonic motion,
maximum acceleration $a_{\max }=8m/s^2$
and maximum speed $v_{\max }=1.6m/s$
As, we know that for simple harmonic motion
$a_{\max }={\omega }^{2}A$
and $v_{\max }=\omega A$
where, $A=$ amplitude of SHM,
$\Rightarrow a_{\max }=v_{\max }\omega$ ...(i)
So, from Eq.(i), we get
$\omega =\frac{a_{\max }}{v_{\max }}=\frac{8}{1.6}=5$
$\Rightarrow \omega =5\left(\because \omega =\frac{2\pi }{T}\right)$
$\therefore$ Time period,
$T=\frac{2\pi }{\omega }=\frac{2\pi }{5}=1.25s$