Question:
A pendulum is executing simple harmonic motion and its maximum kinetic energy is \(K_1\). If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is \(K_2\). Then :
A pendulum is executing simple harmonic motion and its maximum kinetic energy is \(K_1\). If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is \(K_2\). Then :
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Maximum KE equals the potential energy at the extreme position, \(mgL(1-\cos\theta_0)\), for the same angular amplitude.
Updated On: Nov 5, 2026
- \(K_2 = \frac{K_1}{2}\)
- \(K_2 = 2K_1\)
- \(K_2 = K_1\)
- \(K_2 = \frac{K_1}{4}\)
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The Correct Option is B
Solution and Explanation
Taking the amplitude as the (same) angular amplitude \(\theta_0\): at the extreme position the bob is raised by \(h = L(1-\cos\theta_0)\), so the maximum kinetic energy (at the mean position) is \(K = mgL(1-\cos\theta_0)\).
Thus \(K \propto L\) for fixed \(m, \theta_0\).
When \(L\) is doubled, \(K_2 = 2K_1\).
Answer: (2) \(K_2 = 2K_1\)
Thus \(K \propto L\) for fixed \(m, \theta_0\).
When \(L\) is doubled, \(K_2 = 2K_1\).
Answer: (2) \(K_2 = 2K_1\)
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