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Question

A particle of mass 2 kg executes simple harmonic motion and its potential energy U varies with position x as shown below. The period of oscillation of the particle is



A
2π5 s
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B
22π5 s
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C
2π5 s
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D
4π5 s
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Solution

The correct option is D 4π5 s
Given , mass of the particle m=2 kg
Potential energy as a function of the position of particle is given by
U=12mω2x2
At extreme positions, x=±A. From the data given in the diagram, we can say that
U=12mω2A212mω2(0.4)2=1.0 J
ω=2×1.0(0.4)2×2=2.5 rad/s
Time period of oscillation of a particle executing SHM is given by,
T=2πω
T=2π2.5=4π5 s
Thus, option (d) is the correct answer.

Physics

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