Simple Harmonic Motion of a Spring Elevator

[UrbanXrisis]

the question is http://home.earthlink.net/~urban-xrisis/phy002.jpg

a.
\omega = \sqrt{\frac{k}{m}}
\omega = \sqrt{\frac{500}{2}}

b.
F=kx
\frac{g}{3} * 2kg=500x
x=1.31cm

c.
since the block displaces 1.31 cm then that is it's amplitude

I don't undersatnd what they mean by inital phase angle. The initial phase would be the mass at -1.31cm moving upwards. I would just be...
x=- \omega^2 A cos(\omega t + \phi)
\phi=0

or... I think this might be a valid answer...

x_i=Acos\phi
-1.31=1.31cos\phi
\phi = \pi

am I doing this all correctly?

 

[Doc Al]

Looks good to me. (Yes, \phi = \pi.)

 

[quasar987]

Could you explain to me what you did in part b. please?

 

[Doc Al]

When the car is accelerating, the spring must exert an additional force equal to ma = mg/3, thus the spring must stretch an additional x = mg/(3k).

 

[quasar987]

But isn't this just the new equilibrium position of the block? After this new position has been attained, the sum of the force of the block is zero but it will have picked a speed in going from rest to this point while accelerating according to

a = -kx/m + g/3

Therefor it will oscilate around x = mg/(3k).

 

[Doc Al]

With no acceleration, the equilibrium point will be where the spring is stretched an amount mg/k; when the car is accelerating at g/3, the equilibrium position will be at a stretch of mg/k + mg/(3k).

The initial condition is that the car is accelerating, thus the initial position of the mass is at x_0 = -mg/k -mg/(3k). At t=0, the acceleration stops (but the speed remains at whatever it is). So, at t=0 the equilbrium position is mg/k, thus the mass is displaced mg/(3k) below that equilibrium point. It will oscillate about the equilibrium point with an amplitude of mg/(3k).

 

[quasar987]

Oops, I had misread the question. I tought the block was initially at rest, THEN the elevator started accelerating, THEN it sudently stopped. Sowwy.