Physics - SHM Question with Solution | TestHub

The velocity of a particle in simple harmonic motion at a distance $x$ from the equilibrium position is given by $v=\omega \sqrt{A^2-x^2}$.

At $x=\frac{2A}{3}$, the initial velocity is:

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

The speed is tripled at this instant, so the new velocity $v^{\prime }$ is:

$v^{\prime }=3v=3\left(\frac{\sqrt{5}}{3}\omega A\right)=\sqrt{5}\omega A$

Let the new amplitude be $A^{\prime }$. The angular frequency $\omega$ and the position $x=\frac{2A}{3}$ remain unchanged. Using the velocity formula for the new amplitude:

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

Substituting the values:

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

Squaring both sides:

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

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

$A^{\prime }=\frac{7A}{3}$

Answer: $\frac{7A}{3}$