Answer
(a) $28.4m$ (b) $42s$
Work Step by Step
Note that the maximum velocity of a simple harmonic oscillator is equal to $v_{max}=A\omega$ and the maximum acceleration is $a_{max}=A\omega ^2$. Maximum acceleration can be rewritten as $$a_{max}=(A\omega)\omega=v_{max}\omega$$ This means that $$0.65m/s^2=(4.3m/s)\omega$$ This yields an omega value of $
\omega=0.15rad/s$. Since $v_{max}=A\omega$, $A=\frac{v_{max}}{\omega}$. Substituting known values of $v_{max}=4.3m/s$ and $\omega=0.15rad/s$ yields an amplitude of $$A=\frac{4.3m/s}{0.15rad/s}=28.4m$$ To find the period, use a relation between angular frequency $\omega$ and period, which is $$T=\frac{2\pi}{\omega}$$ Substituting the known value of $\omega=0.15rad/s$ yields a period of $$T=\frac{2\pi}{.15rad/s}=42s$$
