How Do You Calculate the Oscillation Frequency of an Engine Block on a Cable?

[Rubidium]

1. Homework Statement
A winch cable has a cross sectional area of 1.5 cm^2 and a length of 2.5 m. Young's modulus for the cable is 150 GN/m^2. A 950-kg engine block is hung from the end of the cable.
(a) By what length does the cable stretch?
(b) If we treat the cable as a piece of string, what is the oscillation frequency of the engine block at the end of the cable?




2. Homework Equations
f=1/T=$$\omega$$/2$$\pi$$




3. The Attempt at a Solution
For (a) I used Young's modulus and found that the cable stretches 0.104 cm.
For (b) I don't know how to go about finding the frequency.
Thanks.

 

[dwintz02]

Think of the cable as a spring and the engine as a mass on the end of a spring. When the cable stretches it will stretch past its equilibrium point and then compress back up, then stretch back down...etc. Given that the cable is behaving like a spring, how can you relate Young's modulus to the spring constant (usually k)?

 

[ogb p]

I would like to commend you on your attempt at finding the solution for part (a). Your use of Young's modulus to calculate the amount of stretch in the cable is correct. Now, for part (b), we can use the equation for the oscillation frequency of a simple harmonic oscillator, which is f=1/T=1/(2π√(k/m)), where k is the spring constant and m is the mass. In this case, we can treat the cable as a spring and use its effective spring constant, which is given by k=Y*A/L, where Y is Young's modulus, A is the cross sectional area, and L is the length of the cable. Plugging in the values, we get k=150 GN/m^2 * 1.5 cm^2 / 2.5 m = 90,000 N/m. The mass of the engine block is 950 kg, so the oscillation frequency can be calculated as f=1/T=1/(2π√(90,000 N/m / 950 kg))=0.236 Hz. This means that the engine block will oscillate back and forth 0.236 times per second at the end of the cable. I hope this helps!