Oscillations Question 6
Question 6 - 30 January - Shift 1
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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Answer: (8)
Solution:
Formula: Equation of Simple harmonic motion, Potential energy of Simple harmonic motion
$x=A \sin (\omega t)$
$U _{(x)}^{math}=\frac{1}{2} kx^{2}$,
$\frac{dU}{dt}=\frac{1}{2} k 2 x \frac{dx}{dt}$
$=kA^{2} \omega \sin \omega t \cos \omega t \times \frac{2}{2}$
2
$(\frac{dU}{dt}) _{\max }=\frac{kA^{2} \omega}{2}(\sin 2 \omega t) _{\max }$
$2 \omega t=\frac{\pi}{2} \Rightarrow t=\frac{\pi}{4} \omega=\frac{T}{8} \Rightarrow \beta=8$
