A body of mass `m` was suspended by a non-stretched spring, and then set fre without push. The stiffness of the spring is `x`. Neglecting the mass of the spring , find `:`
`(a)` the law of motion `y(t)`, whee `y` is displacement of the body from the equilibrium position,
`(b)` the maximum and minimum tensions of the spring in the process of motion.

Explanation

Let `y(t)=` displacement of the body from the end of the unstretched position of the spring (not the equilibrium position). Then
`m ddot(y)=-ky+mg`
This equation has the soloution of the form
`y=A+B cos (omegat+alpha)`
if `- m omega^(2)B cos ( omegat +alpha)=-k[A+B cos(omegat+ alpha)]+ mg`
Then ` omega^(2)=(k)/(m)` and `A=(mg)/(k)`
we have `y=0` and `y=0` at `t=0`. So
`- omega B sin alpha=0`
` A+Bcos alpha=0`
Since `B gt 0` and `A gt 0` we have `alpha=pi`
`B=A=(mg)/(k)`
and `y=(mg)/(k)(1-cos omegat)`
`(b)` Tension in the spring is
`T=ky=mg(1-cos omegat)`
so `T_(max)=2 mg, T_(m i n)=0`