A body of mass `m` is suspended from a spring fixed to the ceiling of an elevator car. The stiffness of the spring is `x`. At the moment `t=0` the car starts going up with an acceleration`w`. Neglecting the mass of the spring , find the law of motion `y(t0` of the body relative to the elevator car if `y(0)-0` and `y(0)=0`. Consider the following two cases `:`
`(a)` `w=const`,
`(b)` `w=alphat, ` where `alpha `is a constant.

Explanation

As the elevator car is a translating non`-` inertial frame, therefore the body `m` will experience an inertial force `m w` directed downward in addition to the real forces in the elevator's frame. From the Newton's second law in projection from
`F_(y)=m w_(y)` for the body in the frame of elevator car `:`
`-k((mg)/(k)+y)+mg + mw=mddot (y)` `....(A)`
(Because the initial elongation in the spring is `mg//k)`
so, `m ddot(y)=-ky+m w =-k(y-(mw)/(k))`
or, `(d^(2))/(dt^(2))(y-(mw)/(k))=-(k)/(m)(y-(mw)/(k)) ...(1)`
Eqn. `(1)` shows that the motion of the body `m` is S.H.M. and its solutino becomes
`y-(mw)/(k)=asin (sqrt((k)/(m))t+alpha) ....(2)`
Differentiating Eqn `(2)` w.r.t. time
`dot(y)=a sqrt((k)/(m)) cos (sqrt((k)/(m))t+alpha) ....(3)`
Using the initial considiton `y(0)-0` in Eqn `(2)`, we get `:`
`a sin alpha=-(mw)/(k)`
and using the other initial condition `dot(y)(0)=0` in Eqn `(3)`
we get `a sqrt((k)/(m))cos alpha =0`
Thus `alpha=-alpha//2 `and `a=(mw)/(k)`
Hence using these values in Eqn `(2)`, we get
`y=(mw)/(k)(1-cossqrt((k)/(m))t)`
`(b)` Proceed up to Eqn `(1)` . The solution of this differential Eqn be of the form `:`
`y-(mw)/(k)=a sin (ssqrt((k)/(m))t+delta)`
or,`y-(alphat)/(k//m)=a sin (sqrt((k)/(m))t+delta)`
or, `y=(alphat)/(omega_(0)^(2))= a sin (omega_(0)t+delta)` (where `omega_(0)=sqrt((k)/(m))) ...(4)`
From the initial condition that at `t=0, y(0)`, so `0=a sin delta` or `delta=0`
Thus Eqn, `(4)` takes the from `: y-(alphat)/(omega_(0)^(2))=a sin omega_(0)t ...(5)`
Differentiating Eqn. `(5)`, we get `: dot(y)-(alpha)/(omega_(0)^(2))=a omega_(0)cos omegat ....(6)`
But from the other initial condition `dot(y0(0)=0` at `t=0`
So, from Eqn. (6) `-(alpha)/(omega_(0)^(2))=a omega_(0)` or `a=-alpha//omega_(0)^(3)`
Putting the value of `a` in Eqn. `(5)`, we get the sought `y(t).e.i,`
`y-(alphat)/(omega_(0)^(t))=-(alpha)/(omega_(0)^(3))sin omega_(0)t` or `y=(alpha)/(omega_(0)^(3))(omega_(0)t- sin omega_(0)t)`