Answer
(a) $A = 0.0406~m$
(b) $v_{max} = 1.21~m/s$
(c) $\omega = 29.8~rad/s$
Work Step by Step
(a) We can find the total energy $E$.
$E = \frac{1}{2}kx^2+\frac{1}{2}mv^2$
$E = \frac{1}{2}(155~N/m)(0.0300~m)^2+\frac{1}{2}(0.175~kg)(0.815~m/s)^2$
$E = 0.128~J$
We can find the amplitude.
$\frac{1}{2}kA^2 = E$
$A = \sqrt{\frac{2E}{k}}$
$A = \sqrt{\frac{(2)(0.128~J)}{155~N/m}}$
$A = 0.0406~m$
(b) We can find the maximum speed.
$\frac{1}{2}mv_{max}^2 = E$
$v_{max} = \sqrt{\frac{2E}{m}}$
$v_{max} = \sqrt{\frac{(2)(0.128~J)}{0.175~kg}}$
$v_{max} = 1.21~m/s$
(c) We can find the angular frequency.
$\omega = \sqrt{\frac{k}{m}}$
$\omega = \sqrt{\frac{155~N/m}{0.175~kg}}$
$\omega = 29.8~rad/s$
