JEE Main 2022 Physics question
The potential energy of a particle of mass 4 kg in motion along the x-axis is given by U = 4(1 − cos 4x) J. The time period of the particle for small oscillation is s. The value of K is .
Answer: 2
Solution
F = −dU/dx = −16 sin −64x for small x, so k = 64 N/m. The
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