Answer
(a) $0.47s$
(b) $2.5cm$
(c) $4.6\frac{m}{s^2}$
Work Step by Step
(a) We know that
$T=2\pi\sqrt{\frac{m}{K}}$
We plug in the known values to obtain:
$T=2\pi\sqrt{\frac{0.85}{150}}=0.47s$
(b) The amplitude can be determined as
$A=\frac{v_{max}}{\omega}$
$\implies A=\frac{Tv_{max}}{2\pi}$
We plug in the known values to obtain:
$A=\frac{(0.4730)(0.35)}{2\pi}=2.5cm$
(c) We can find the maximum acceleration as
$a_{max}=A(\frac{2\pi}{T})^2$
We plug in the known values to obtain:
$a_{max}=0.02635(\frac{2\pi}{0.4730})^2=4.6\frac{m}{s^2}$
