QJEEM24Ph23df: JEE Main 2024 Physics - Class 11 JEEM24Ph23df Solution: A particle performs simple harmonic motion with amplitude . Its …

To find the new amplitude of the motion when the speed is increased to three times at a given displacement, we use the concepts of simple harmonic motion (SHM) and its formulas.

In SHM, the velocity $v$ of a particle at a displacement $x$ from the mean position can be given by the formula:

$v = \omega \sqrt{A^2 - x^2}$

where:

• $\omega$ is the angular frequency of the motion,
• $A$ is the amplitude, and
• $x$ is the displacement at that instance.

Given:

• Displacement at the instance, $x = \frac{2A}{3}$ ,
• Initial velocity is increased to three times at this displacement.

Thus, let's find the initial velocity $v$ at $x = \frac{2A}{3}$ :

$v = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\sqrt{5}A\omega}{3}$

With the velocity increased to three times, the new velocity $v'$ becomes:

$v' = 3v = 3 \times \frac{\sqrt{5}A\omega}{3} = \sqrt{5}A\omega$

For the new amplitude $A'$ , the velocity $v'$ at the same displacement $x$ is:

$v' = \omega \sqrt{{A'}^2 - \left(\frac{2A}{3}\right)^2}$

Setting the expressions for $v'$ equal gives:

$\sqrt{5}A\omega = \omega \sqrt{{A'}^2 - \frac{4A^2}{9}}$

$\sqrt{5}A = \sqrt{{A'}^2 - \frac{4A^2}{9}}$

Solving for $A'$ gives:

${A'}^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9} = \frac{49A^2}{9}$

$A' = \sqrt{\frac{49A^2}{9}} = \frac{7A}{3}$

Therefore, the new amplitude of the motion is $\frac{7A}{3}$ , which means the value of $n$ is 7.