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JEE Mains · Physics · STD 11 - 13. oscillations
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 \(.........\)
- A \(9\)
- B \(7\)
- C \(6\)
- D \(8\)
Answer & Solution
Correct Answer
(D) \(8\)
Step-by-step Solution
Detailed explanation
\(x = A \sin (\omega t )\) \(U _{( x )}=\frac{1}{2} kx ^2\) \(\frac{ dU }{ dt }=\frac{1}{2} k 2 x \frac{ dx }{ dt }\) \(= kA ^2 \omega \sin \omega t\,\cos \omega t \times \frac{2}{2}\) \(\left(\frac{ dU }{ dt }\right)_{\max }=\frac{ kA ^2 \omega}{2}(\sin 2 \omega t )_{\max }\)…
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