Question

The displacement of a particle varies with time as $x=12\sin \omega t-16{\sin }^3\omega t\left(\text{in cm}\right)$. If the particle is executing S.H.M, then its maximum acceleration is

• A. $12{\omega }^2$
• B. $36{\omega }^2$
• C. $144\omega$
• D. $\sqrt{192}{\omega }^2$

Answer 1

The correct option is B $36{\omega }^2$

Given ,
$x=12\sin \omega t-16{\sin }^3\omega t$
From this, we can write that,
$x=4(3\sin \omega t-4{\sin }^3\omega t)$
By using the formula, $\sin 3\theta =3\sin \theta -4{\sin }^3\theta$
we get ,
$x=4\sin \left(3\omega t\right)$
Comparing this with $x=A\sin {\omega }_{1}t$
${\omega }_1=3\omega ;A=4\text{m}$
Maximim acceleration of the particle $a_{max}=A{\omega }_1^2=4\times (3\omega {)}^2=36{\omega }^2$
Thus, option (b) is the correct answer.