Question

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. initially, the mass is released from rest from a point 4 inches above the equilibrium position. find the equation of motion. (use g = 32 ft/s2 for the acceleration due to gravity.)

Answer 1

When the spring is in the equilibrium we know that force of the spring is equal to the force of gravity. This allows us to find the spring constant k.
$mg=ky_0\\ k=\frac{mg}{y_0}=2304\frac{N}{in}$
When the system is not in equilibrium the force of gravity and force of the spring are not going to be the same. There is going to be a force acting on the weight.
$F=ma=mg-kl\\ ma=mg-k(y_0+y)\\ ma=mg-ky-ky_0\\ ky_0=mg\\ ma=mg-ky-mg\\ ma=-ky\\ \frac{d^2y}{dt^2}+\frac{k}{m}y=0$
This is the same differential equation we get for the horizontal spring system. The general solution to this equation is:
$y(t)=A\sin(\omega t)$
Where:
$\omega=\sqrt{\frac{k}{m}};\\ A=y(0)=4 in$
So our final equation will be;
$y(t)=4\sin(9.8 t)$