Answer
(a) $k = 59~N/m$
(b) The energy involved in this motion is 0.060 J
Work Step by Step
(a) $f = \frac{1}{2\pi}~\sqrt{\frac{k}{m}}$
$k = m~(2\pi~f)^2$
$k = (0.240~kg)~[(2\pi)~(2.5~Hz)]^2$
$k = 59~N/m$
(b) We can find how much energy is involved in this motion as:
$E = \frac{1}{2}kA^2$
$E = \frac{1}{2}(59~N/m)(0.045~m)^2$
$E = 0.060~J$
The energy involved in this motion is 0.060 J.
