Question

$y=2\left(cm\right)\sin [\frac{\pi t}{2}+\varphi ]$ what is the maximum acceleration of the particle doing the S.H.M ?

• A. $\frac{\pi }{2}cm/s^2$
• B. $\frac{{\pi }^2}{2}cm/s^2$
• C. $\frac{{\pi }^2}{4}cm/s^2$
• D. $\frac{\pi }{4}cm/s^2$

Answer 1

The correct option is B $\frac{{\pi }^2}{2}cm/s^2$

$y=2\sin (\frac{\pi t}{2}+\varphi )$
Velocity of particle $\frac{dy}{dt}=2\times \frac{\pi }{2}\cos (\frac{\pi t}{2}+\varphi )$
Acceleration $\frac{d^{2}y}{dt}=-\frac{{\pi }^2}{4}2\sin (\frac{\pi t}{2}+\varphi )$
Thus $a_{max}=\frac{{\pi }^2}{2}$