A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance $$\frac{2A}{3}$$ from equilibrium position. The new amplitude of the motion is:

We begin with the standard results for a particle of mass $$m$$ executing simple harmonic motion (SHM) with angular frequency $$\omega$$ and amplitude $$A$$.

1. Total mechanical energy of SHM is purely potential at the extreme position, so

$$E=\tfrac12\,kA^{2},$$

where the force constant is related to the angular frequency by the familiar relation $$k=m\omega^{2}$$.

2. At any instant when the particle is at a distance $$x$$ from the mean (equilibrium) position, its speed $$v$$ is obtained from energy conservation, giving the well-known formula

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

Now, according to the statement of the problem, the particle is momentarily at

$$x=\frac{2A}{3}$$

from the mean position. Using the velocity formula, we first compute the speed it has before the sudden change.

Substituting $$x=\dfrac{2A}{3}$$ into $$v=\omega\sqrt{A^{2}-x^{2}}$$, we obtain

$$\begin{aligned} v_{1}&=\omega\sqrt{A^{2}-\left(\frac{2A}{3}\right)^{2}}\\[4pt] &=\omega\sqrt{A^{2}-\frac{4A^{2}}{9}}\\[4pt] &=\omega\sqrt{\frac{9A^{2}-4A^{2}}{9}}\\[4pt] &=\omega\sqrt{\frac{5A^{2}}{9}}\\[4pt] &=\frac{\omega A\sqrt5}{3}. \end{aligned}$$

At this very instant an external impulse suddenly triples the speed of the particle, without giving it time to change its position. Therefore the new speed is

$$v_{2}=3v_{1}=3\left(\frac{\omega A\sqrt5}{3}\right)=\omega A\sqrt5.$$

The position at that instant is still

$$x=\frac{2A}{3},$$

so the potential energy remains what it was, but the kinetic energy is altered because of the increased speed. Let us calculate the new total mechanical energy $$E_{2}$$ right after the impulse.

Kinetic energy after the impulse:

$$\begin{aligned} K_{2}&=\tfrac12\,m v_{2}^{2}\\[4pt] &=\tfrac12\,m\bigl(\omega A\sqrt5\bigr)^{2}\\[4pt] &=\tfrac12\,m\omega^{2}A^{2}\cdot5\\[4pt] &=\tfrac12\,kA^{2}\cdot5\quad(\text{since }k=m\omega^{2}). \end{aligned}$$

Potential energy at the same position:

$$\begin{aligned} U_{2}&=\tfrac12\,k x^{2}\\[4pt] &=\tfrac12\,k\left(\frac{2A}{3}\right)^{2}\\[4pt] &=\tfrac12\,k\frac{4A^{2}}{9}. \end{aligned}$$

Adding both contributions gives the new total energy:

$$\begin{aligned} E_{2}&=K_{2}+U_{2}\\[4pt] &=\tfrac12\,kA^{2}\cdot5+\tfrac12\,k\frac{4A^{2}}{9}\\[4pt] &=\tfrac12\,k\left(5A^{2}+\frac{4A^{2}}{9}\right)\\[4pt] &=\tfrac12\,k\left(\frac{45A^{2}+4A^{2}}{9}\right)\\[4pt] &=\tfrac12\,k\frac{49A^{2}}{9}. \end{aligned}$$

Because mechanical energy in SHM is again of the form $$\tfrac12\,k(\text{new amplitude})^{2}$$, let us denote the new amplitude by $$A'$$ and equate energies:

$$\tfrac12\,kA'^{2}=E_{2}=\tfrac12\,k\frac{49A^{2}}{9}.$$

Canceling the common factor $$\tfrac12\,k$$ from both sides yields

$$A'^{2}=\frac{49A^{2}}{9}.$$

Taking the positive square root (amplitude is a positive quantity), we obtain

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

Thus the amplitude of the motion after the speed is tripled becomes $$\boxed{\dfrac{7A}{3}}$$.

Hence, the correct answer is Option B.