The general displacement of a simple harmonic | vedclass

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Question

(D) The displacement is given by $x = A \sin \omega t$.The potential energy $U$ is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2 \omega t

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) The displacement is given by $x = A \sin \omega t$.The potential energy $U$ is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2 \omega t$.The slope of the potential energy-time curve is $\frac{dU}{dt}$.$\frac{dU}{dt} = \frac{d}{dt} (\frac{1}{2} k A^2 \sin^2 \omega t) = \frac{1}{2} k A^2 (2 \sin \omega t \cos \omega t) \cdot \omega = \frac{1}{2} k A^2 \omega \sin(2 \omega t)$.For the slope to be maximum,$\sin(2 \omega t)$ must be maximum,i.e.,$\sin(2 \omega t) = 1$.This occurs when $2 \omega t = \frac{\pi}{2}$.Substituting $\omega = \frac{2 \pi}{T}$,we get $2 (\frac{2 \pi}{T}) t = \frac{\pi}{2}$.$\frac{4 \pi t}{T} = \frac{\pi}{2} \implies t = \frac{T}{8}$.Comparing this with $t = \frac{T}{\beta}$,we get $\beta = 8$.

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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