A particle undergoing simple harmonic motion has a maximum acceleration of 6 m/s² and a maximum velocity of 4 m/s - HSC - SSCE Mathematics Extension 2 - Question 8 - 2020 - Paper 1

Now substituting $A$ into the acceleration equation:

$6 = \frac{(4)^2}{\frac{4}{A}}$

Thus,

$6A = 16\implies A = \frac{16}{6} = \frac{8}{3}$

Next we find $\omega$ in terms of $A$:

Using: $4 = \omega A$

We can solve for $\omega$: $\omega = \frac{4}{\frac{8}{3}} = \frac{4 \cdot 3}{8} = \frac{12}{8} = \frac{3}{2}$

The period ($T$) is related to the angular frequency by the formula:

$T = \frac{2\pi}{\omega}$

Substituting in the value of $\omega$:

$T = \frac{2\pi}{\frac{3}{2}} = \frac{2\pi \cdot 2}{3} = \frac{4\pi}{3}$

Thus, the final answer is:

D. (\frac{4\pi}{3})