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A particle starts oscillating simple harmonically from its equilibrium position with time period ' $\mathrm{T}$ '. At time $\mathrm{t}=\frac{\mathrm{T}}{12}$, the ratio of its kinetic energy to potential energy is
$\left[\sin \frac{\pi}{3} \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \sin \frac{\pi}{6}=\cos \frac{\pi}{3}=\frac{1}{2}\right]$
Options:
$\left[\sin \frac{\pi}{3} \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \sin \frac{\pi}{6}=\cos \frac{\pi}{3}=\frac{1}{2}\right]$
Solution:
2563 Upvotes
Verified Answer
The correct answer is:
$3: 1$
Take equation of motion as $\mathrm{x}=\mathrm{a} \sin \omega \mathrm{t}$,
we have $\mathrm{t}=\frac{\mathrm{T}}{12}$
$x\left(\frac{T}{12}\right)=\sin \left(\frac{2 \pi}{T} \times \frac{T}{12}\right)=\operatorname{asin}\left(\frac{\pi}{6}\right)=\frac{a}{2}$
Then the ratio of the kinetic energy to the potential energy at $x=\frac{a}{2}$ will be given by
$\frac{\text { K.E. }}{\text { P.E. }}=\frac{\frac{1}{2} k\left(a^2-x^2\right)}{\frac{1}{2} k x^2}=\frac{a^2-\left(\frac{a}{2}\right)^2}{\left(\frac{a}{2}\right)^2}=\frac{3}{1}$
we have $\mathrm{t}=\frac{\mathrm{T}}{12}$
$x\left(\frac{T}{12}\right)=\sin \left(\frac{2 \pi}{T} \times \frac{T}{12}\right)=\operatorname{asin}\left(\frac{\pi}{6}\right)=\frac{a}{2}$
Then the ratio of the kinetic energy to the potential energy at $x=\frac{a}{2}$ will be given by
$\frac{\text { K.E. }}{\text { P.E. }}=\frac{\frac{1}{2} k\left(a^2-x^2\right)}{\frac{1}{2} k x^2}=\frac{a^2-\left(\frac{a}{2}\right)^2}{\left(\frac{a}{2}\right)^2}=\frac{3}{1}$
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