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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 \(.........\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(8\)
Step-by-step Solution
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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