Oscillations

JEE Main level, numerical answer, medium

30 of 194

Question 30 of 194

A particle executes simple harmonic motion. Its amplitude is 8cm8cm and time period is 6s6s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is _____ ss

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Answer: 1

Given here, A=8cmA=8cm and T=6sT=6s
The displacement equation in SHM, when amplitude is halved is Acos⁡(2πtT)=A2A\cos \left(\dfrac{2\pi t}{T}\right)=\dfrac{A}{2}
⇒2πtT=π3\Rightarrow \dfrac{2\pi t}{T}=\dfrac{\pi }{3}
Time taken by the particle is t=t= T6=66=1sec⁡\dfrac{T}{6}=\dfrac{6}{6}=1\sec.

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