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 ___________.

<p>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.</p><p>In SHM, the velocity $v$ of a particle at a displacement $x$ from the mean position can be given by the formula:</p><p>$v = \omega \sqrt{A^2 - x^2}$</p><p>where:</p><ul><li>$\omega$ is the angular frequency of the motion,</li><li>$A$ is the amplitude, and</li><li>$x$ is the displacement at that instance.</li></ul><p>Given:</p><ul><li>Displacement at the instance, $x = \frac{2A}{3}$,</li><li>Initial velocity is increased to three times at this displacement.</li></ul><p>Thus, let's find the initial velocity $v$ at $x = \frac{2A}{3}$:</p><p>$$v = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\sqrt{5}A\omega}{3}$$</p><p>With the velocity increased to three times, the new velocity $v'$ becomes:</p><p>$v' = 3v = 3 \times \frac{\sqrt{5}A\omega}{3} = \sqrt{5}A\omega$</p><p>For the new amplitude $A'$, the velocity $v'$ at the same displacement $x$ is:</p><p>$v' = \omega \sqrt{{A'}^2 - \left(\frac{2A}{3}\right)^2}$</p><p>Setting the expressions for $v'$ equal gives:</p><p>$\sqrt{5}A\omega = \omega \sqrt{{A'}^2 - \frac{4A^2}{9}}$</p><p>$\sqrt{5}A = \sqrt{{A'}^2 - \frac{4A^2}{9}}$</p><p>Solving for $A'$ gives:</p><p>${A'}^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9} = \frac{49A^2}{9}$</p><p>$A' = \sqrt{\frac{49A^2}{9}} = \frac{7A}{3}$</p><p>Therefore, the new amplitude of the motion is $\frac{7A}{3}$, which means the value of $n$ is 7.</p>