Simple Harmonic Motion 1 Question 3

3. 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: 3. (d)

Solution:

  1. 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

$$ \begin{aligned} \therefore & \frac{1}{2} k x^{2} & =\frac{1}{2} k A^{2}-\frac{1}{2} k x^{2} \\ \text { or } & k x^{2} & =\frac{k A^{2}}{2} \text { or } x= \pm \frac{A}{\sqrt{2}} \end{aligned} $$



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