Speed in SHM in terms of displacement is given by, v=ωA2−x2.
at x=32A, we get
v=ωA2−(32A)2=35Aω
Let the new amplitude be=A′
Then,
v′=3v⇒ω(A′)2−(32A)2=5Aω
⇒A′=37A
Therefore, n=7.
JEE Main 2024 — Physics Waves & Oscillations
A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is 32A. The new amplitude of motion is 3nA. The value of n is _____.
Held on 31 Jan 2024 · Verified 6 Jul 2026.
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