If a mass–spring system with an iron ball of weight \(W = 98\)N can be regarded as undamped, and
the spring is such that the ball stretches it \(1.09\)m, how many cycles per minute will the system
execute? What will its motion be if we pull the ball down from rest by \(16\)cm and let it start with
zero initial velocity
Before solving the problem, we need to find a few values from the given data,
$$F = -kx$$
Plugging in \(x = 1.09\)m in our equation, where the force against gravity is \(F=98\)N
We get, $$k = \frac{98}{1.09} = 89.9\frac{N}{m}$$
The mass \(m = 98/9.8\) would be \(10\)Kg.
Hence by \(\omega = \sqrt{\frac{k}{m}}\), we get,
$$f = \frac{\omega}{2\pi} = 0.483\frac{\mathrm{cycles}}{\mathrm{sec}}$$
Provided that we have the amplitude \(A = 0.16\) and \(\omega = 3\), we can write the
equation of motion as, $$\mathbf{x(t) = 0.16\sin(3t + \frac{\pi}{2})}$$