Simple Harmonic Motion Ques 8
- A particle undergoing simple harmonic motion has time dependent displacement given by $x(t)=A \sin \frac{\pi t}{90}$. The ratio of kinetic to potential energy of this particle at $t=210 s$ will be
(2019 Main, 11 Jan I)
(a) $2$
(b) $1$
(c) $\frac{1}{9}$
(d) $3$
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Answer:
Correct Answer: 8.(*)
Solution:
Formula:
- Here given, displacement, $x(t)=A \sin \frac{\pi t}{90}$
where $A$ is amplitude of S.H.M., $t$ is time taken by particle to reach a point where its potential energy $U=\frac{1}{2} k x^{2}$ and kinetic energy $=\frac{1}{2} k\left(A^{2}-x^{2}\right)$ here $k$ is force constant and $x$ is position of the particle.
Potential energy $(U)$ at $t=210 s$ is
$ \begin{aligned} U & =\frac{1}{2} k x^{2}=\frac{1}{2} k A^{2} \sin ^{2} (\frac{210}{90} \pi )\\ & =\frac{1}{2} k A^{2} \sin ^{2} (2 \pi+\frac{3}{9} \pi)=\frac{1}{2} k A^{2} \sin ^{2} (\frac{\pi}{3}) \end{aligned} $
Kinetic energy at $t=210 s$, is
$ \begin{aligned} K & =\frac{1}{2} k\left(A^{2}-x^{2}\right) \\ & =\frac{1}{2} k A^{2} [1-\sin ^{2} (\frac{210 \pi}{90} )]\\ & =\frac{1}{2} k A^{2} \cos ^{2}(210 \pi / 90) \end{aligned} $
$\Rightarrow \quad K=\frac{1}{2} k A^{2} \cos ^{2}(\pi / 3)$
So, ratio of kinetic energy to potential energy is
$ \frac{K}{U}=\frac{\frac{1}{2} k A^{2} \cos ^{2}(\pi / 3)}{\frac{1}{2} k A^{2} \sin ^{2}(\pi / 3)}=\cot ^{2}(\pi / 3)=\frac{1}{3} $
$\therefore$ No option given is correct.
BETA
