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 5 - 2020 - Paper 1

From the second equation, we can express A in terms of v_max:

$A = \frac{v_{max}}{ω}$

Substituting this into the first equation gives: $a_{max} = ω^2 \left( \frac{v_{max}}{ω} \right) \implies a_{max} = ω v_{max}$

Thus, $ω = \frac{a_{max}}{v_{max}} = \frac{6 ext{ m/s}^2}{4 ext{ m/s}} = 1.5 ext{ s}^{-1}$.

The period (T) is related to angular frequency (ω) by: $T = \frac{2π}{ω}$

Substituting for ω gives: $T = \frac{2π}{1.5} = \frac{4π}{3} ext{ seconds}.$

Therefore, the period of the motion is (\frac{4π}{3}).

Based on the above calculation, the correct option from the choices given is:

D. (\frac{4π}{3})