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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2023
  • Hard
  • Numerical

The general displacement of a simple harmonic oscillator is \(x = A \sin \omega t\). Let \(T\) be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when \(t=\frac{T}{\beta}\). The value of \(\beta\) is \(.........\)

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Correct Answer

(D) \(8\)

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Detailed explanation

\(x = A \sin (\omega t )\)

\(U _{( x )}=\frac{1}{2} kx ^2\)

\(\frac{ dU }{ dt }=\frac{1}{2} k 2 x \frac{ dx }{ dt }\)

\(= kA ^2 \omega \sin \omega t\,\cos \omega t \times \frac{2}{2}\)

\(\left(\frac{ dU }{ dt }\right)_{\max }=\frac{ kA ^2 \omega}{2}(\sin 2 \omega t )_{\max }\)

\(2 \omega t =\frac{\pi}{2} \Rightarrow t =\frac{\pi}{4} \omega=\frac{ T }{8} \Rightarrow \beta=8\)

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