A particle performs simple harmonic motion with amplitude A. Its speed is trippled at the instant that it is at distance 2A/3 from equilibrium position. The new amplitude of the motion is.
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Question

A particle performs simple harmonic motion with amplitude A. Its speed is trippled at the instant that it is at distance 2A/3 from equilibrium position. The new amplitude of the motion is.

G1 Physics · Accepted Answer

At x=2A/3, initial speed satisfies v²=ω²[A²−(2A/3)²]=(5/9)ω²A². Speed is suddenly tripled, so the new amplitude A′ satisfies (3v)²=ω²[A′²−(2A/3)²]. Hence 5A²=A′²−4A²/9, giving A′²=49A²/9 and A′=7A/3. Correct option: 3.