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Questions  

The displacement of a particle varies with time as x=12sinωt16sin3ωt (in cm). If its motion is S.H.M., then its maximum acceleration is 

a
12ω2
b
36ω2
c
144ω2
d
192ω2

detailed solution

Correct option is B

x=12sin⁡ωt−16sin3⁡ωt=43sin⁡ωt−4sin3⁡ωt =4[sin⁡3ωt] By using sin⁡3θ=3sin⁡θ−4sin3⁡θ∴A max=(3ω)2×4=36ω2

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