The general displacement of a simple harmonic oscillator is x = A sin ω t. Let T be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when t = T/β. The value of β is ______________.

Solution

Physics sir · 4 steps

Write U as a sine-squared, differentiate it, and you get a sine of twice the angle.

1. Potential energy. With x = A sin ω t and k = mω², the potential energy is U = 1/2kx² = 1/2kA²sin² ω t.
2. Slope of U-t curve. Differentiate with respect to time. Using 2 sin ω t cos ω t = sin 2ω t, the slope becomes a single sine of 2ω t.

   dU/dt = 1/2kA² · 2 sin ω t cos ω t · ω = 1/2kA²ω sin 2ω t

3. Maximise the slope. The slope is largest when sin 2ω t = 1, that is 2ω t = π/2. So t = π/4ω = π/4 × T/2π = T/8.
4. So. t = T/β with β = 8.

The common slip: 4. Finding where U itself is maximum (at the extreme position, t = T/4) instead of where its slope is maximum gives 4, not 8.

The idea to keep. Energy in SHM goes as sin² ω t, so it oscillates at twice the frequency. Its rate of change is a sin 2ω t term, peaking at T/8, not T/4.

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