First slide

Simple hormonic motion

Question

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

Moderate

A

$12 ω^{2}$
B

$36 ω^{2}$
C

$133 ω^{2}$
D

$\sqrt{192} ω^{2}$

Solution

$x = 12 \sin ωt - 16 \sin^{3} ωt = 4 [3 \sin ωt - 4 \sin^{3} ωt]$
$= 4 [\sin 3 ωt] (By using \sin 3 θ = 3 \sin θ - 4 \sin^{3} θ)$
$∴ maximum acceleration A_{\max} = {(3 ω)}^{2} \times 4 = 36 ω^{2}$

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