JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Elasticity & SHM

A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is

\frac{2 A}{3}

. The new amplitude of motion is

\frac{nA}{3}

. The value of n is ____.

Answer: 7

Solution

For SHM,

v = ω A^{2} - x^{2}

.

At

x = \frac{2 A}{3}

v = ω A^{2} - \frac{4 A ^{2}}{9} = \frac{ω A 5}{3}

The new speed is

3 v

at the same displacement, now with amplitude

A^{'}

(3 v)^{2} = ω^{2} (A^{'2} - \frac{4 A ^{2}}{9}) ⟹ 5 A^{2} + \frac{4 A ^{2}}{9} = A^{'2} ⟹ A^{'2} = \frac{49 A ^{2}}{9}

A^{'} = \frac{7 A}{3}

n = 7

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Integer answer