Question

The instantaneous acceleration of a particle executing SHM given by $y=a\sin \omega t$ is

• A. $+{\omega }^{2}y$
• B. $+\omega y$
• C. $-\omega y^2$
• D. $-{\omega }^{2}y$

Answer 1

The correct option is D $-{\omega }^{2}y$

$y=a\sin \omega t$
Velocity, $v=\frac{dy}{dt}=a\omega \cos \omega t$
Acceleration, $A=\frac{dv}{dt}=-a{\omega }^2\sin \omega t$
$=-{\omega }^{2}y$