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Question: Answered & Verified by Expert
A particle of mass 0.1 kg is executing simple harmonic motion of amplitude 0.1 m. When the particle passes through the mean position, its kinetic energy is 8×10-3 J. If the initial phase is 45°, the equation of its motion is (Assume, x t as the position of the particle at time t)
PhysicsGravitationTS EAMCETTS EAMCET 2019 (03 May Shift 2)
Options:
  • A x t=0.1 sin 4 t+π4
  • B x t=0.1 sin 16 t+π4
  • C x t=0.1 sin 2t+π4
  • D x t=0.1 sin 2 t+π4
Solution:
1348 Upvotes Verified Answer
The correct answer is: x t=0.1 sin 4 t+π4

The mass of particle is m=0.1 kg

Amplitude of particle performing SHM is A=0.1 m

The initial phase of particle is , ϕ=450=π4

The equation of particle performing SHM is given by

x=A sin ωt+ϕ     ...1

When the particle passes through the mean position its KE is maximum, which is given by

KEmax=12mω2A2=8×10-3

  12×0.1×ω2×0.12=8×10-3

  ω=4 rad s-1

Now equation (1) becomes, 

x t=0.1 sin 4 t+π4

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