Physics - SHM Question with Solution | TestHub

The velocity of a particle in simple harmonic motion is given by $v=\omega \sqrt{A^2-x^2}$

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

$v_1=\omega \sqrt{A^2-{\left(\frac{2A}{3}\right)}^2}=\omega \sqrt{A^2-\frac{4A^2}{9}}=\frac{\sqrt{5}}{3}\omega A$

The speed is tripled at this instant, so the new velocity is:

$v_2=3v_1=3\left(\frac{\sqrt{5}}{3}\omega A\right)=\sqrt{5}\omega A$

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

$v_2^2={\omega }^2(A^{\prime 2}-x^2)$

Substituting the values of $v_2$ and $x$:

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

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

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

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

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