Question

A particle executes simple harmonic motion with an angular velocity and maximum acceleration of $3.5rad/sec$ and $7.5m/s^2$ respectively. Amplitude of the oscillation is

• A. 0.36
• B. 0.28
• C. 0.61
• D. 0.53

Answer 1

Answer: (C)

The angular velocity $\omega =3.5$ red/sec
maximum acceleration $a_{\max }=7.5m/s^2$
We know for a $SHM$, the displacement $x=A\sin \omega t$
$\therefore v=\frac{dx}{dt}=A\omega \cos \omega t$
$\therefore a=\frac{dv}{dt}=-A{\omega }^2\sin \omega t$
$\therefore$ Maximum acceleration
$\left|a_{\max }\right|=A{\omega }^2$
Now $A{\omega }^2=7.5$
$\Rightarrow A=\frac{7.5}{{\omega }^2}$
$=\frac{7.5}{(3.5{)}^2}=0.6$
$\therefore$ Amplitude $=0.6$