Answer
(a) $T = 1.33~s$
(b) $A = 0.764~cm$
(c) $a_{max} = 16.9~cm/s^2$
(d) $k = 11.1~N/m$
Work Step by Step
(a) $v_x(t) = A~\omega~sin(\omega~t+\phi)$
$v_x(t) = -(3.60~cm/s)~sin[(4.71~rad/s)~t-(\frac{\pi}{2})]$
From the equation, we can see that $\omega = 4.71~rad/s$
We can find the period.
$T = \frac{2\pi}{\omega}$
$T = \frac{2\pi}{4.71~rad/s}$
$T = 1.33~s$
(b) We can find the amplitude.
$A~\omega = 3.60~cm/s$
$A = \frac{3.60~cm/s}{\omega}$
$A = \frac{3.60~cm/s}{4.71~rad/s}$
$A = 0.764~cm$
(c) We can find the maximum acceleration.
$a_{max} = \omega^2~A$
$a_{max} = (4.71~rad/s)^2(0.764~cm)$
$a_{max} = 16.9~cm/s^2$
(d) We can find the force constant of the spring.
$\frac{k}{m} = \omega^2$
$k = m~\omega^2$
$k = (0.500~kg)(4.71~rad/s)^2$
$k = 11.1~N/m$
