# Spring elevator question

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 dont 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?

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Doc Al
Mentor
Looks good to me. (Yes, $\phi = \pi$.)

quasar987
Homework Helper
Gold Member
Could you explain to me what you did in part b. please?

Doc Al
Mentor
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
Homework Helper
Gold Member
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
Mentor
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