NVAEasyJEE Main 2026 · 28 January, Shift 1Simple Harmonic Motion (SHM)

Physics Question from JEE Main 2026 · 28 January, Shift 1

The displacement of a particle executing simple harmonic motion with time period TT is expressed as x(t)=Asinωt,x(t)=A\sin\omega t, where AA is the amplitude of oscillation. If the maximum value of the potential energy of the oscillator is found at t=T2β,t=\frac{T}{2\beta}, then the value of β\beta is _____.

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