Natural frequency of a crane dropping a car

[Dustinsfl]

An electromagnet weighing \(3000\) lb is at rest while holding an automobile of weight \(2000\) lb in a junkyard. The electric current is turned off, and the automobile is dropped. Assuming that the crane and the supporting cable have an equivalent spring constant of \(10000\) lb/in, find the following:
(1)the natural frequency of vibration of the electromagnet, (2)the resulting motion of the electromagnet, and (3)the maximum tension developed in the cable during motion.

For (1), would it be \(\omega_n = \sqrt{\frac{10000}{5000}}\) or \(\omega_n = \sqrt{\frac{10000}{3000}}\)? Do we look at the mass right before release or right after release?
For (2), this is a simple harmonic oscillator so \(x(t) = A\cos(\omega_nt) + B\sin(\omega_nt)\).
Not sure how to do (3)

 

[jacobi1]

For (1), you would look at the mass right after release, since this is the mass that is being accelerated by the cable.
For (2), it would be $$x(t)=-B \cos (\omega t)$$, since the mass starts from the lowest point at t=0.
For (3), remember that the maximum force (tension) occurs when the acceleration is maximum.
Can you proceed?

 

[jacobi1]

Alternatively for (3), the maximum tension will occur when the mass is at rest (this is just a restatement of the maximum acceleration criterion).

 

[Dustinsfl]

For (1), you would look at the mass right after release, since this is the mass that is being accelerated by the cable.
For (2), it would be $$x(t)=-B \cos (\omega t)$$, since the mass starts from the lowest point at t=0.
For (3), remember that the maximum force (tension) occurs when the acceleration is maximum.
Can you proceed?

Thanks but I figured out everything yesterday.