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The maximum potential energy (PE) of a particle in SHM is 2 × 10-4 J. What will be the PE of the particle when its displacement from the mean position is half the amplitude of SHM ? |
Answer» (PE)max = 1/2 kA2 , PE = 1/2 kx2 ∴ PE = (PE)max (x/A)2 = 2 × 10-4 J × (x/A)2 = 5 × 10-5 J is the required answer. |
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