Answer
(a) $E = 0.0336~J$
(b) $A = 0.0150~m$
(c) $v_{max} = 0.669~m/s$
Work Step by Step
(a) We can find the total energy $E$.
$E = \frac{1}{2}kx^2+\frac{1}{2}mv^2$
$E = \frac{1}{2}(300~N/m)(0.0120~m)^2+\frac{1}{2}(0.150~kg)(0.400~m/s)^2$
$E = 0.0336~J$
(b) We can find the amplitude.
$\frac{1}{2}kA^2 = E$
$A = \sqrt{\frac{2E}{k}}$
$A = \sqrt{\frac{(2)(0.0336~J)}{300~N/m}}$
$A = 0.0150~m$
(c) We can find the maximum speed.
$\frac{1}{2}mv_{max}^2 = E$
$v_{max} = \sqrt{\frac{2E}{m}}$
$v_{max} = \sqrt{\frac{(2)(0.0336~J)}{0.150~kg}}$
$v_{max} = 0.669~m/s$
