Question

The displacement of the particle varies with time x = 12 sin $\omega$ t - 16 $si{n}^3$ $\omega$ t. If its motion is SHM, then maximum acceleration is

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

Answer 1

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

$x=12\sin \left(\omega t\right)-16{\sin }^3\left(\omega t\right)$

$x=4\left(3\sin \right(\omega t)-4{\sin }^3(\omega t)$

$x=4\sin \left(3\omega t\right)$

Maximum acceleration$=A{\omega }^2$

$A=4,\omega =3\omega$

Max acceleration$=4(3\omega {)}^2=36{\omega }^2$

Option $\text{B}$ is the correct answer.