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.

In SHM, the velocity $v$ of a particle at a displacement $x$ from the mean position can be given by the formula:

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

where:

• $\omega$ is the angular frequency of the motion,
• $A$ is the amplitude, and
• $x$ is the displacement at that instance.

Given:

• Displacement at the instance, $x = \frac{2A}{3}$,
• Initial velocity is increased to three times at this displacement.

Thus, let's find the initial velocity $v$ at $x = \frac{2A}{3}$:

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

With the velocity increased to three times, the new velocity $v'$ becomes:

$$v' = 3v = 3 \times \frac{\sqrt{5}A\omega}{3} = \sqrt{5}A\omega$$

For the new amplitude $A'$, the velocity $v'$ at the same displacement $x$ is:

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

Setting the expressions for $v'$ equal gives:

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

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

Solving for $A'$ gives:

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

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

Therefore, the new amplitude of the motion is $\frac{7A}{3}$, which means the value of $n$ is 7.