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

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 (A) 12 ω2 (B) 36 ω2 (C) 133 ω2 (D) √192 ω2

Tardigrade Admin · Accepted Answer

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) ∴ maximum acceleration A max =(3 ω)2 × 4=36 ω2

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

$12 ω^{2}$

$36 ω^{2}$

$133 ω^{2}$

$192 ω^{2}$

$x = 12 sin ω t - 16 sin^{3} ω t = 4 [3 sin ω t - 4 sin^{3} ω t]$
$= 4 [sin 3 ω t] (B y$ using $sin 3 θ = 3 sin θ - 4 sin^{3} θ)$
$∴$ maximum acceleration $A_{m a x} = (3 ω)^{2} \times 4 = 36 ω^{2}$

Q. The displacement of a particle varies with time as x=12 sinωt−16sin3ωt( in cm). If its motion is S.H.M., then its maximum acceleration is

12ω2

36ω2

133ω2

192​ω2