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

From equation (2), we can express ( \omega ):

[ \omega = \frac{v_{max}}{A} = \frac{4}{A} ]

Substituting this into equation (1):

[ 6 = \left(\frac{4}{A}\right)^2 A ]

[ 6 = \frac{16}{A} \implies A = \frac{16}{6} = \frac{8}{3} \text{ m} ]

Now substituting ( A ) back into the equation for ( \omega ):

[ \omega = \frac{4}{\frac{8}{3}} = \frac{4 \cdot 3}{8} = \frac{12}{8} = \frac{3}{2} \text{ rad/s} ]

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

[ T = \frac{2\pi}{\omega} ]

Substituting our value for ( \omega ):

[ T = \frac{2\pi}{\frac{3}{2}} = \frac{2\pi \cdot 2}{3} = \frac{4\pi}{3} ext{ seconds} ]

Thus, the period of the motion is ( \frac{4\pi}{3} ), which corresponds to option D.