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

The maximum velocity ($v_{max}$) is given by:

$v_{max} = ext{ω} A$

Substituting $A$ from our earlier result:

$v_{max} = ext{ω} \cdot \frac{6}{ ext{ω}^2} = \frac{6}{\text{ω}}$

Now, equating to the maximum velocity:

$4 = \frac{6}{\text{ω}} \Rightarrow \text{ω} = \frac{6}{4} = \frac{3}{2}$

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

$T = \frac{2\pi}{\text{ω}}$

Substituting $ext{ω}$:

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

Thus, the period of the motion is $\frac{4\pi}{3}$.