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MCQEasyJEE 2023Energy in SHM

JEE Physics 2023 Question with Solution

A particle is executing Simple Harmonic Motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be:

  • A

    1:11 : 1

  • B

    2:12 : 1

  • C

    1:41 : 4

  • D

    1:31 : 3

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A particle executes SHM and its displacement is x=A2x = \frac{A}{2}.

Find: The ratio of potential energy to kinetic energy at this displacement.

In SHM,

P.E.=12kx2P.E. = \frac{1}{2}kx^2

and

K.E.=12kA2−12kx2K.E. = \frac{1}{2}kA^2 - \frac{1}{2}kx^2

Substitute x=A2x = \frac{A}{2}:

P.E.=12k(A2)2=A2k8P.E. = \frac{1}{2}k\left(\frac{A}{2}\right)^2 = \frac{A^2k}{8} K.E.=12kA2−A2k8=3A2k8K.E. = \frac{1}{2}kA^2 - \frac{A^2k}{8} = \frac{3A^2k}{8}

Therefore,

P.E.K.E.=A2k83A2k8=13\frac{P.E.}{K.E.} = \frac{\frac{A^2k}{8}}{\frac{3A^2k}{8}} = \frac{1}{3}

Thus, the ratio of potential energy to kinetic energy is 1:31:3. The correct option is D.

Common mistakes

  • Using x=Ax = A instead of x=A2x = \frac{A}{2}. This gives the energy ratio at the extreme position, not at half amplitude. Always substitute the given displacement carefully before calculating energies.

  • Assuming potential energy and kinetic energy are equal at x=A2x = \frac{A}{2}. In SHM, equal energies occur at x=A2x = \frac{A}{\sqrt{2}}, not at half amplitude. Use the standard energy expressions to avoid this misconception.

  • Taking total energy as only 12kx2\frac{1}{2}kx^2. That expression is only the potential energy at displacement xx. The total energy in SHM is 12kA2\frac{1}{2}kA^2 because it depends on amplitude, not instantaneous displacement.

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