Oscillations

JEE Main level, numerical answer, medium

84 of 194

Question 84 of 194

The general displacement of a simple harmonic oscillator is x=Asin⁡ωtx=A\sin \omega t. Let TT be its time period. The slope of its potential energy (UU) – time (tt) curve will be maximum when t=Tβt=\dfrac{T}{\beta }. The value of β\beta is ______.

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Answer: 8

Given here, displacement x=Asin⁡(ωt)x=A\sin \left(\omega t\right)
Now, potential energy of a simple harmonic oscillator is U=12mω2x2=12mω2A2sin⁡2(ωt)U=\dfrac{1}{2}m{\omega }^{2}{x}^{2}=\dfrac{1}{2}m{\omega }^{2}{A}^{2}{\sin }^{2}\left(\omega t\right),
So, slope of potential energy and time curve is dUdt=12mω3A22sin⁡2ωt\dfrac{dU}{dt}=\dfrac{1}{2}\dfrac{m{\omega }^{3}{A}^{2}}{2}\sin 2\omega t
Now, (dUdt)max{\left(\dfrac{dU}{dt}\right)}_{max} will be maximum when sin⁡2ωt=1\sin 2\omega t=1
⇒2ωt=π2⇒2(2πT)t=π2⇒t=T8⇒β=8\Rightarrow 2\omega t=\dfrac{\pi }{2} \Rightarrow 2\left(\dfrac{2\pi }{T}\right)t=\dfrac{\pi }{2} \Rightarrow t=\dfrac{T}{8} \Rightarrow \beta =8

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