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PhysicsSHMCalculation of time periodMedium2 minQB

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

Options:

Answer:

Solution:

From the graph, Umax = 1 J at amplitude A = 0.4 m.

$U_{ma x} = \frac{1}{2} m ω^{2} A^{2} \Rightarrow 1 = \frac{1}{2} (2) (0.4)^{2} ω^{2} \Rightarrow ω = \frac{5}{2} rad/s$

$T = \frac{2 π}{ω} = \frac{4 π}{5} s$

Answer: D

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Calculation of time period

⏱ 2mℹ️ Source: QB

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