The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t) = A sin ωt, where A is the amplitude. The maximum value of potential energy of this oscillator is found at t = T/(2β). The value of β i s ________.
Physics · Simple Harmonic Motion · kinetic and potential energy in shm
JEE Mains 2026 28 January Morning Shift

Question

The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t) = A sin ωt, where A is the amplitude. The maximum value of potential energy of this oscillator is found at t = T/(2β). The value of β i s ________.

Physics · Simple Harmonic Motion · kinetic and potential energy in shm

JEE Mains 2026 28 January Morning Shift

G1 Physics · Accepted Answer

For SHM, x(t) = A sinωt. Potential energy is U = ½kx². Therefore U is maximum at the extreme position, where |x| = A. Starting from x=0 at t=0, the particle first reaches x=A after one-quarter of a period: t = T/4. The question states t = T/(2β). Hence, T/(2β) = T/4. Therefore, 2β = 4, so β = 2. Hence the answer is 2.