A particle performs simple harmonic motion with amplitude $$A$$. Its speed is increased to three times at an instant when its displacement is $$\frac{2A}{3}$$. The new amplitude of motion is $$\frac{nA}{3}$$. The value of $$n$$ is _____.

A particle in SHM with amplitude $$A$$ has its speed tripled at the instant when its displacement is $$\frac{2A}{3}$$. We need to find the new amplitude $$A' = \frac{nA}{3}$$.

We start by recalling the velocity-displacement relation in SHM.

For a particle in SHM with angular frequency $$\omega$$ and amplitude $$A$$:

$$ v^2 = \omega^2(A^2 - x^2) $$

At the instant $$x = \frac{2A}{3}$$, the original speed is found as follows:

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

Next, when the speed is tripled to $$3v$$ while the displacement remains $$x = \frac{2A}{3}$$ and the angular frequency $$\omega$$ is unchanged, the new amplitude $$A'$$ satisfies:

$$ (3v)^2 = \omega^2(A'^2 - x^2) $$

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

Substituting the value of $$v^2$$ gives:

$$ 9 \times \frac{5\omega^2 A^2}{9} = \omega^2\left(A'^2 - \frac{4A^2}{9}\right) $$

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

Dividing both sides by $$\omega^2$$ leads to:

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

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

Finally, solving for $$A'$$ and determining $$n$$ gives:

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

Since $$A' = \frac{nA}{3}$$, it follows that $$n = 7$$.

The answer is $$\boxed{7}$$.