First slide

Simple harmonic motion

Question

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

Moderate

A

$12 ω^{2}$
B

$36 ω^{2}$
C

$144 ω^{2}$
D

$\sqrt{192} ω^{2}$

Solution

$\begin{aligned} \begin{aligned} x & = 12 \sin ω t - 16 \sin^{3} ω t = 4 [3 \sin ω t - 4 \sin^{3} ω t] \\ = 4 [\sin 3 ω t] (By using \sin 3 θ = 3 \sin θ - 4 \sin^{3} θ) \\ ∴ A & _{max} = (3 ω)^{2} \times 4 = 36 ω^{2} \end{aligned} \end{aligned}$

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