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 $t = {T \over \beta }$. The value of $\beta$ is ______________.
Answer (integer)
8
Solution
<p>$U = {1 \over 2}m{\omega ^2}{A^2}{\sin ^2}\omega t$</p>
<p>So, ${{dU} \over {dt}} = {{m{\omega ^3}{A^2}} \over 2}\sin 2\omega t$</p>
<p>This value will be maximum when $\sin 2\omega t = 1$</p>
<p>or $2\omega t = {\pi \over 2}$</p>
<p>$2 \times {{2\pi } \over T}t = {\pi \over 2}$</p>
<p>$\Rightarrow t = {T \over 8}$</p>
<p>So $\beta = 8$</p>
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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