Question

The amplitude of a particle executing simple harmonic motion with a frequency of $60Hz$ is $0.01m$. The maximum value of acceleration of the particle is

• A. $144{\pi }^{2}m/s^2$
• B. $12m/s^2$
• C. $11m/s^2$
• D. $169m/s^2$

Answer 1

The correct option is A $144{\pi }^{2}m/s^2$

The equation of SHM can be given by:

$x=A\sin \omega t$

So, $v=\frac{dx}{dt}=A\omega \sin \omega t$

So, $a=\frac{dv}{dt}=A{\omega }^2\sin \omega t$

So, $a_{max}=A{\omega }^2=0.01\times (2\times \pi \times 60{)}^2=144{\pi }^{2}m/s^2$