Oscillations

JEE Main level, single correct, medium

120 of 194

Question 120 of 194

The displacement of simple harmonic oscillator after 33 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is

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Answer: (C) 36s36s

Displacement equation is given by x=Asin⁡ωtx=A\sin \omega t
Now, when amplitude is halved, A2=Asin⁡ωt\dfrac{A}{2}=A\sin \omega t
Or ωt=π6\omega t=\dfrac{\pi }{6}
At, t=3st=3s, ω(3)=π6\omega \left(3\right)=\dfrac{\pi }{6}⇒ω=π18\Rightarrow \omega =\dfrac{\pi }{18}
Time period, T=2πω=2ππ18=36sT=\dfrac{2\pi }{\omega }=\dfrac{2\pi }{\dfrac{\pi }{18}}=36s

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