Answer to Question #99515 in Classical Mechanics for Steven

Question

A body of mass m is on a weightless spring (with the spring constant k)
suspended on the ceiling of an elevator car. At the time t = 0 starts
to move the car up with an acceleration a. Find that
Law of motion of the body y (t) concerning the elevator car, if ¨y (0) = 0 and
y˙ (0) = 0. Consider the two cases:
(a) a = const,
(b) a = bt, where b = const.

Answer 1

Case1:

Acceleration is constant:

a(t) = const.=w=wt⁰

v(t)= ∫ a(t)dt=∫ wt⁰dt=wt+c

at,t=0 ,v(t)=0

Thus

0=c

or c=0

Thus

v(t)=wt

Now, Again

y(t)=∫v(t)dt

y(t)=∫wtdt

y(t)=wt²/2+c

y(t)=wt²/2+c

at,t=0

y(t)=0

Thus

0=c

or c=0

Thus,

y(t)=wt²/2

Case II:

Acceleration is not constant:

a(t) = bt

v(t)= ∫ a(t)dt=∫ btdt=bt²/2+c

at,t=0,v(t)=0

Thus

0=c

or c=0

v(t)=bt²/2

Now

y(t)=∫v(t)dt

y(t)=∫bt²/2dt

y(t)=bt³/6+c

at,t=0

y(t)=0

Thus

0=c

or c=0

Thus,

y(t)=bt³/6