Question

The maximum speed of a particle executing S.H.M. is 1m/s and its maximum acceleration is 1.57m/$se{c}^2$ . The time period of the particle will be

[DPMT 2002]

• A.
• B. 1.57 sec
• C. 2 sec
• D. 4 sec

Answer 1

The correct option is D

4 sec

Given max velocity $\omega a=1$ and maximum

acceleration ${\omega }^{2}a=1.57$

$\therefore \frac{{\omega }^{2}a}{\omega a}=1.57\Rightarrow \omega =1.57\Rightarrow \frac{2\pi }{T}=1.57\Rightarrow T=4$