At what displacement the potential energy is equal to kinetic energy in a particle is executing SHM with amplitude A?

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4. At what displacement from equilibrium is the energy of a simple harmonic oscillator half P.E and half K.E ?

Sol: Potential energy of a SHO at any instant, P.E = ½ mω2x2
Kinetic energy at any instant, K.E = ½ mω 2(A2 − x2)

Total energy E	=	½ mω2 A2
where A is amplitude
Let x	=	pm be the displacement where the energy is half P.E and half K.E
i.e at x	=	P
P.E	=	½ E
K.E	=	½ E
We get two equations,
½ mω2P2	=	½ [½ mω2A2] and
½ mω2(A2− P2)	=	½ [½ mω2A2]

Dividing two equation we get

i.e, At a distance of 0.707 times the amplitude on either side of the equilibrium position,

Kinetic energy and potential energy are equal.

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Answer

Verified

Hint: This problem can be solved by using the direct formula for the potential energy of a body in SHM in terms of its displacement and the kinetic energy of a body in terms of its displacement and amplitude. By equating these two, we can get the required value of displacement in terms of the amplitude.

Formula used:

$KE=\dfrac{1}{2}K\left( {{A}^{2}}-{{x}^{2}} \right)$ $PE=\dfrac{1}{2}K{{x}^{2}}$

Complete answer:

Let us write the expressions for the potential and kinetic energy of a body executing simple harmonic motion (SHM).The kinetic energy $KE$ of a body in simple harmonic motion with amplitude $A$ at a point where its displacement from the mean position is $x$ is given by $KE=\dfrac{1}{2}K\left( {{A}^{2}}-{{x}^{2}} \right)$ --(1)Where $K=m{{\omega }^{2}}$, where $m$ is the mass of the body and $\omega $ is the angular frequency of the SHM.The potential energy $PE$ of a body in simple harmonic motion with amplitude $A$ at a point where its displacement from the mean position is $x$ is given by $PE=\dfrac{1}{2}K{{x}^{2}}$ --(2)Where $K=m{{\omega }^{2}}$, where $m$ is the mass of the body and $\omega $ is the angular frequency of the SHM.Now, let us analyze the question.The amplitude of the SHM is $A$.Let the displacement of the body from the mean position at a certain instant be $x$.Let the kinetic energy of the body at this instant be $KE$.Let the potential energy of the body at this instant be $PE$.Now, according to the question, the kinetic energy is equal to the potential energy. $\therefore KE=PE$ Now, using (1) and (2) in the above equation, we get$\dfrac{1}{2}K\left( {{A}^{2}}-{{x}^{2}} \right)=\dfrac{1}{2}K{{x}^{2}}$ Where $K=m{{\omega }^{2}}$, where $m$ is the mass of the body and $\omega $ is the angular frequency of the SHM.$\Rightarrow {{A}^{2}}-{{x}^{2}}={{x}^{2}}$ $\Rightarrow {{A}^{2}}={{x}^{2}}+{{x}^{2}}=2{{x}^{2}}$ $\Rightarrow {{x}^{2}}=\dfrac{{{A}^{2}}}{2}$ Square rooting both sides we get$\sqrt{{{x}^{2}}}=\sqrt{\dfrac{{{A}^{2}}}{2}}$$\Rightarrow x=\dfrac{A}{\sqrt{2}}$ Hence, the required displacement of the body from the mean position is $\dfrac{A}{\sqrt{2}}$.

So, the correct answer is “Option C”.

Note:

Students must note that the sum of the kinetic energy and the potential energy at any instance in the SHM remains the same as the total mechanical energy is conserved. This constant value is equal to $\dfrac{1}{2}K{{A}^{2}}$. The kinetic energy at the mean position has this value as the potential energy is zero at this point. Also, at the extreme positions, the potential energy is this value as the kinetic energy at the extreme points is zero (as the body has zero speed at this point).