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 \(\displaystyle x(t)=-B \cos (\omega t)\), 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 \(\displaystyle x(t)=-B \cos (\omega t)\), 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.

 

[CellCoree]

For (1), the natural frequency of the crane and the car system can be calculated using the formula \(\omega_n = \sqrt{\frac{k}{m}}\), where \(k\) is the spring constant and \(m\) is the mass of the system. In this case, the mass of the system is the combined weight of the car and the electromagnet, which is \(5000\) lb. Therefore, the natural frequency is \(\omega_n = \sqrt{\frac{10000}{5000}} = \sqrt{2}\) rad/s.

For (2), the resulting motion of the electromagnet can be described by the equation \(x(t) = A\cos(\omega_nt) + B\sin(\omega_nt)\), where \(A\) and \(B\) are constants determined by the initial conditions. Since the electromagnet is initially at rest, we can assume that \(A = 0\) and \(B = -2000\) lb, since the electromagnet will move downwards with a displacement of 2000 lb. The equation becomes \(x(t) = -2000\sin(\sqrt{2}t)\).

For (3), the maximum tension developed in the cable can be calculated using the formula \(T_{max} = kx_{max}\), where \(k\) is the spring constant and \(x_{max}\) is the maximum displacement of the electromagnet. In this case, \(k = 10000\) lb/in and \(x_{max} = 2000\) lb. Therefore, the maximum tension in the cable is \(T_{max} = 10000 \times 2000 = 20000000\) lb.