Question

A particle executes simple harmonic motion with an angular velocity and maximum acceleration of $3.5\, rad/sec$ and $7.5\, m/s^2$ 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 $a_{\max }=7.5 \,m / s ^{2}$
We know for a $SHM$, the displacement $x=A\, \sin\, \omega t$
$\therefore \,\,\,\,\,v =\frac{d x}{d t}=A \omega \cos \omega t $
$\therefore \,\,\,\,\, a =\frac{d v}{d t}=-A \omega^{2} \sin \omega t$
$\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 $=0.6$