Q75 / 90 · JEE Main 2024 (31 Jan Shift 1) · Physics

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 $\frac{n A}{3}$ . The value of $n$ is _____.

Correct answer: 7

Solution

Speed in SHM in terms of displacement is given by, $v = ω \sqrt{A^{2} - x^{2}}$ .

at $x = \frac{2 A}{3}$ , we get

$v = ω \sqrt{A^{2} - {(\frac{2 A}{3})}^{2}} = \frac{\sqrt{5} A ω}{3}$

Let the new amplitude be $= A'$

Then,

$v' = 3 v \Rightarrow ω \sqrt{{(A')}^{2} - {(\frac{2 A}{3})}^{2}} = \sqrt{5} A ω$

$\Rightarrow A' = \frac{7}{3} A$

Therefore, $n = 7$ .

Hence, the correct answer is 7.