A particle executes SHM of amplitude A. The distance from the mean position when its kinetic energy is equal to its pote

Question

A particle executes SHM of amplitude A. The distance from the mean position when its kinetic energy is equal to its potential energy is:

Accepted Answer

Given: A particle executes SHM with amplitude $A$ .

Find: The distance from the mean position when K.E. = P.E.

For SHM, the total energy is

E = \frac{1}{2}M \omega^2 A^2

At displacement $x$ , the kinetic energy is

\text{K.E.} = \frac{1}{2}M \omega^2 (A^2 - x^2)

and the potential energy is

\text{P.E.} = \frac{1}{2}M \omega^2 x^2

When kinetic energy equals potential energy,

\frac{1}{2}M \omega^2 (A^2 - x^2) = \frac{1}{2}M \omega^2 x^2

So,

A^2 - x^2 = x^2

2x^2 = A^2

x^2 = \frac{A^2}{2}

x = \pm \frac{A}{\sqrt{2}}

Hence, the distance from the mean position is $\frac{A}{\sqrt{2}}$ . The solution working gives $\frac{1}{\sqrt{2}}A$ , which corresponds to option C. The solution's label says B, but that conflicts with the working, so the working is taken as authoritative.

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