Question

A particle of mass 4 kg moves in simple harmonic motion and its potential energy U varies with position x as shown in the figure. The time period of oscillation of the particle is 2πα second. Find the value of α.___

Solution

In SHM, U-x graph is a parabola.

∴U=C1 x2, where C1 is a constant.

From figure whe, x = 0.2m, U = 2J

∴2=C1(0.2)2

⇒C1=20.04=50

∴U=50x2

Also, in SHM U=12mω2 x2

∴12mω2=50

⇒ω2=1004=25

⇒ω=5s−1

∴T=2πω=2π5

T=2π5s

∴U=C1 x2, where C1 is a constant.

From figure whe, x = 0.2m, U = 2J

∴2=C1(0.2)2

⇒C1=20.04=50

∴U=50x2

Also, in SHM U=12mω2 x2

∴12mω2=50

⇒ω2=1004=25

⇒ω=5s−1

∴T=2πω=2π5

T=2π5s

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