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← JEE Main 2022 · 28 Jul · Shift 2
PhysicsInteger+4 / -1Question 25 of 90

JEE Main 2022 · 28 Jul · Shift 2, Physics question 25

The potential energy of a particle of mass 4 kg in motion along the xx-axis is given by U=4(1−cos⁡4x)U=4(1-\cos4x) J. The time period of the particle for small oscillation (sin⁡θ≃θ)(\sin\theta\simeq\theta) is (πK)\left(\dfrac\pi K\right) s. The value of K is ______.

Answer and worked solution

Official answer: 2

Small x. U≈32x2U\approx32x^2, so k=64k=64 and ω=644=4\omega=\sqrt{\frac{64}4}=4.

T=π2T=\frac\pi2, so K=2K=2.

Answer: 2