Problem 12.38

12.38. a)
$\omega = \sqrt{\frac{k}{m}}$
$\omega = 15.8 rad/s$

b)
Summing the forces while the elevator is going up yields
F_spring - mg = ma
so the force due to the spring is
$F_{spring} = \frac{4}{3} mg ~=~ 26.1 N$
so the equillibrium length for the spring when the elevator is moving up is
$x_{moving} = \frac{F_{spring}}{k}$
x_moving = 5.23 cm

c)
Now, when the elevator stops the mass is still 5.23 cm displaced, but the force is different. The equillibrium position is now
$x_{still} = \frac{mg}{k}$
x_still = 3.92 cm
so, the mass is displaced
5.23 cm - 3.92 cm = 1.31 cm from equillibrium.
So, the amplitude is given by
A = 1.31 cm
but the mass starts oscillating 1.31 cm below equillibrium, so
$\phi = \pi$
where the phase constant is $\phi$
and the motion of the block is given by
$x(t) = 1.31 \cos (15.8t + \pi)$.