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Question: Answered & Verified by Expert
A particle executes simple harmonic motion with a time period of 16 s along the x - axis. At time t=2 s, the particle crosses the mean position while at t=4 s, the velocity of the particle is  + 4 m s-1. The amplitude of motion (in m) is
PhysicsOscillationsJEE Main
Options:
  • A 2π
  • B 162π
  • C 242π
  • D 322π
Solution:
2917 Upvotes Verified Answer
The correct answer is: 322π

The general equation of SHM is

x=Asinωt+ϕ, here

ω=2π16=π8 rad s-1

At t=2 s, the particle is at the mean position. Hence,

0=Asinπ8×2+ϕ

ϕ=-π4

So the equation of SHM will become

x=Asinπ8t-π4

v=πA8cosπ8t-π4

At t= 4 s

4=πA8cosπ8×4-π4

4=πA8cosπ4

A=322π m

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