Simple Harmonic Motion Ques 9
- A particle is executing simple harmonic motion (SHM) of amplitude $A$, along the $X$-axis, about $x=0$. when its potential energy (PE) equals kinetic energy (KE), the position of the particle will be
(2019 Main, 9 Jan II)
(a) $A$
(b) $\frac{A}{2}$
(c) $\frac{A}{2 \sqrt{2}}$
(d) $\frac{A}{\sqrt{2}}$
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Answer:
Correct Answer: 9.(d)
Solution:
Formula:
- Here, $A$ =amplitude of particle in SHM
We know that in SHM potential energy $(U)$ of a particle is given by the relation at distance $x$ from the mean position is
$$ U=\frac{1}{2} k x^{2} $$
and at the same point kinetic energy $(K)$
$$ \begin{aligned} & =\text { Total energy }- \text { Potential energy } \\ & =\frac{1}{2} k A^{2}-\frac{1}{2} k x^{2} \end{aligned} $$
According to the question,
potential energy $=$ kinetic energy
$$ \begin{aligned} \therefore & \frac{1}{2} k x^{2} & =\frac{1}{2} k A^{2}-\frac{1}{2} k x^{2} \\ \text { or } & k x^{2} & =\frac{k A^{2}}{2} \text { or } x= \pm \frac{A}{\sqrt{2}} \end{aligned} $$
BETA
