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Difficulty: MediumSimple Harmonic Motion

A particle executes simple harmonic motion with an amplitude of 0.10 m0.10\text{ m}. At what displacement from its equilibrium position, in meters, is the kinetic energy of the particle equal to three times its potential energy?

Answer: 0.05 m

Answer

The displacement from the equilibrium position is 0.05 m0.05\text{ m}.
In simple harmonic motion, potential energy is Ep=12kx2E_p = \frac{1}{2}kx^2 and kinetic energy is Ek=12k(A2−x2)E_k = \frac{1}{2}k(A^2 - x^2). Setting Ek=3EpE_k = 3E_p gives A2−x2=3x2A^2 - x^2 = 3x^2, which simplifies to A2=4x2A^2 = 4x^2, or x=A2x = \frac{A}{2}. Given an amplitude A=0.10 mA = 0.10\text{ m}, the displacement is x=0.10 m2=0.05 mx = \frac{0.10\text{ m}}{2} = 0.05\text{ m}.

Step-by-Step Solution

1
Set up the relation between kinetic energy and potential energy using the given condition.
Ek=3Ep  ⟹  12k(A2−x2)=3(12kx2)E_k = 3E_p \implies \frac{1}{2}k(A^2 - x^2) = 3\left(\frac{1}{2}kx^2\right)
In simple harmonic motion, energy is partitioned between kinetic and potential forms based on displacement xx.
2
Solve the algebraic equation for displacement xx in terms of amplitude AA.
A2−x2=3x2  ⟹  A2=4x2  ⟹  x=A2A^2 - x^2 = 3x^2 \implies A^2 = 4x^2 \implies x = \frac{A}{2}
Canceling the common factor 12k\frac{1}{2}k isolates the geometric parameters AA and xx.
3
Substitute the known amplitude value into the expression for xx.
x=0.10 m2=0.05 mx = \frac{0.10\text{ m}}{2} = 0.05\text{ m}
Plugging in A=0.10 mA = 0.10\text{ m} yields the required displacement.

Key Concept

Energy Conservation in Simple Harmonic Motion
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Topics

Simple Harmonic Motion

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