The potential energy of a particle of mass 4 kg in motion along the x-axis is given by U=4(1-cos 4 x) J. The time period of the particle for small oscillation (sin θ ≃ θ) is (π/K) s. The value of K is _________.

Solution

Physics sir · 3 steps

Find the force from U, make it look like -kx for small x, and read off the spring constant.

1. Force from potential. Use F = -dU/dx. Differentiating 4(1-cos 4x) gives F = -16 sin 4x, where the chain rule brings out a factor of 4.
2. Small oscillation. For small x, sin 4x ≈ 4x, so F ≈ -64x. This is the form F = -kx, so the effective spring constant is k = 64 N m⁻¹.

   T = 2π√(m/k) = 2π√(4/64) = 2π × 1/4 = π/2 s

3. So. Comparing T = π/2 with π/K gives K = 2.

The common slip: 1. Dropping the chain-rule factor of 4 when differentiating cos 4x gives k = 16, T = π and so K = 1.

The idea to keep. For any potential with a minimum, the effective spring constant is k = d²U/dx² at that minimum. Then T = 2π√(m/k), just like a spring.

Answer from the NTA final answer key. Solution by PrepWiser.