Question

A particle performing simple harmonic motion has a frequency of $1.2Hz$ and a maximum acceleration of $4m/s^2$. Find
The maximum velocity.

Answer 1

Now, $f=1.2Hz$

Now, Angular frequency $\omega =2\pi f=2\pi \times 1.2=2.4\pi$

If A is the amplitude then maximum acceleration, $a_{max}=A{\omega }^2=4$

Then maximam velocity $v_{max}=A\omega =\frac{a_{max}}{\omega }=\frac{4}{2.4\pi }=0.53m/s^2$