Oscillations

JEE Main level, numerical answer, medium

72 of 194

Question 72 of 194

The displacement of a particle, executing simple harmonic motion with time period TT, is expressed as x(t)=Asin⁡ωtx(t)=A \sin \omega t, where AA is the amplitude. The maximum value of potential energy of this oscillator is found at t=T/2βt=T / 2 \beta. The value of β\beta is ____\_\_\_\_.

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Answer: 2

For SHM with x(t)=Asin⁡(ωt)x(t) = A\sin(\omega t), the potential energy is
U=12mω2x2=12mω2A2sin⁡2(ωt)U = \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 A^2 \sin^2(\omega t).
Maximum potential energy occurs when sin⁡2(ωt)=1\sin^2(\omega t) = 1, i.e., when ωt=π2,3π2,...\omega t = \frac{\pi}{2}, \frac{3\pi}{2}, ...
The first maximum occurs at ωt=π2\omega t = \frac{\pi}{2}, giving t=π2ω=π2⋅T2π=T4t = \frac{\pi}{2\omega} = \frac{\pi}{2} \cdot \frac{T}{2\pi} = \frac{T}{4}.
Since this is given as t=T2βt = \frac{T}{2\beta}.
We have T4=T2β\frac{T}{4} = \frac{T}{2\beta}, so β=2\beta = 2.

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