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QJEEM22Phab08: JEE Main 2022 Physics - Class 11 JEEM22Phab08 Solution: The potential energy of a particle of mass in motion along the…

The potential energy of a particle of mass 4 kg4 \mathrm{~kg} in motion along the x-axis is given by U=4(1−cos⁡4x)\mathrm{U}=4(1-\cos 4 x) J. The time period of the particle for small oscillation (sin⁡θ≃θ)(\sin \theta \simeq \theta) is (πK)s\left(\frac{\pi}{K}\right) s . The value of K\mathrm{K} is _________.

Step-by-Step Verified Solution

U=4(1−cos⁡4x)U = 4(1 - \cos 4x)

⇒F=−dUdx=−(4)(4sin⁡4x)\Rightarrow F = - {{dU} \over {dx}} = - (4)(4\sin 4x)

=−16sin⁡4x= - 16\sin 4x

as small x

F=−16(4x)=−64x≡−kxF = - 16(4x) = - 64x \equiv - kx

T=2πmk=2π464=π2T = 2\pi \sqrt {{m \over k}} = 2\pi \sqrt {{4 \over {64}}} = {\pi \over 2}

⇒K=2\Rightarrow K = 2

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