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Q.

The displacement of a particle varies with time as x=12sin⁡ωt−16sin3⁡ωt (in cm). If its motion is S.H.M., then its maximum acceleration is

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a

12ω2

b

36ω2

c

144ω2

d

192ω2

answer is B.

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Detailed Solution

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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