Question:
A particle starts oscillating simple harmonically from its equilibrium position with time period T. The ratio of kinetic energy to potential energy of the particle at time \(t=T/12\) is
A particle starts oscillating simple harmonically from its equilibrium position with time period T. The ratio of kinetic energy to potential energy of the particle at time \(t=T/12\) is
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Start from equilibrium so displacement is A sin(omegat); at t = T/12 the phase is 30 degrees.
Updated On: Nov 12, 2026
- \(1:4\)
- \(2:1\)
- \(3:1\)
- \(4:1\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Starting from the mean position: \(x=A\sin\omega t\), \(\omega=\dfrac{2\pi}{T}\).
Step 2: Find the Displacement
At \(t=\dfrac{T}{12}\), \(\omega t=\dfrac{\pi}{6}\), so \(x=A\sin30^{\circ}=\dfrac{A}{2}\).
Step 3: Energies
\(PE=\dfrac{1}{2}kx^{2}=\dfrac{1}{2}k\dfrac{A^{2}}{4}\). \(KE=\dfrac{1}{2}k(A^{2}-x^{2})=\dfrac{1}{2}k\dfrac{3A^{2}}{4}\).
Step 4: Ratio
\(\dfrac{KE}{PE}=\dfrac{3A^{2}/4}{A^{2}/4}=3:1\).
Final Answer:
The ratio of kinetic to potential energy is 3:1. \[ \boxed{\text{(C) }3:1} \]
Starting from the mean position: \(x=A\sin\omega t\), \(\omega=\dfrac{2\pi}{T}\).
Step 2: Find the Displacement
At \(t=\dfrac{T}{12}\), \(\omega t=\dfrac{\pi}{6}\), so \(x=A\sin30^{\circ}=\dfrac{A}{2}\).
Step 3: Energies
\(PE=\dfrac{1}{2}kx^{2}=\dfrac{1}{2}k\dfrac{A^{2}}{4}\). \(KE=\dfrac{1}{2}k(A^{2}-x^{2})=\dfrac{1}{2}k\dfrac{3A^{2}}{4}\).
Step 4: Ratio
\(\dfrac{KE}{PE}=\dfrac{3A^{2}/4}{A^{2}/4}=3:1\).
Final Answer:
The ratio of kinetic to potential energy is 3:1. \[ \boxed{\text{(C) }3:1} \]
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