Question

Three 10,000 kg ore cars are held at rest on a 30° incline on a mine railway using a cable that is parallel to the incline. The cable is observed to stretch 15cm just before the coupling between the lower two cars breaks, detaching the lowest car. Assuming the cable obeys Hooke's Law, find a) the frequency and b) the amplitude of the resulting oscillations of the remaining two cars.

Thank you!

Answer 1

find k for the cable from \(F = kx = 3mg \sin \theta, \; x = 0.15\), [m is the mass of one car] which **assumes** that equilibrium was reached before de-coupling occurred. NB if you don't assume that equilibrium was reached before de-coupling, then the cars are still moving and i am not sure how you could do this, because you do need the value of k.....

once the car is decoupled, there is \(2mg \sin \theta\) pulling it down the incline and \(3mg\), the force in the spring pulling it up. so it will move up the incline.

from here in, would look at the energy situation.

When it moves a distance \(x\) up the incline from the equilibrium extension \(x_o\), it will have spring energy \(\frac{1}{2}k(x_o-x)^2\), increased potential energy \(2mg \sin \theta \, x\) and kinetic energy \(\frac{1}{2} (2m) \dot x^2\)

really important bit which you need to check for yourself.....and we can say that \(E(t) = \frac{1}{2}k(x_o-x)^2 + 2mg \sin \theta \, x +m \dot x^2 = const = \frac{1}{2}k x_o^2\) assuming energy of the system is conserved, ie no friction etc

the rest is just maths, bit i will outline where i think it leads

so \(\dot E = k(x_o-x)(-\dot x) + 2mg \sin \theta \, \dot x +2 m \dot x \ddot x = 0\)

cancel \(\dot x\)'s

\(\ -kx_o+kx + 2mg \sin \theta +2 m \ddot x = 0\)

hence \(2m \ddot x + k x= k x_o - 2mg \sin \theta \)

but \(kx_0 = 3mg \sin \theta\)

so that becomes \(2m \ddot x + k x= mg \sin \theta \)

\(\ddot x + \frac{k}{2m} x= \frac{1}{2}g \sin \theta \)

the general solution is therefore \(x(t) = A \cos \omega t + \frac{mg \sin \theta }{k}\) with \( \omega = \sqrt{\frac{k}{2m}}\)

\(x_o = A + \frac{mg \sin \theta }{k} = 3 \frac{mg \sin \theta }{k}\) so \(x(t) = \frac{2mg \sin \theta }{k} \cos \omega t + \frac{mg \sin \theta }{k}\)

that should tell you everything you need to know about the oscillations if the assumptions are correct and i haven't once again botched the latex with typos etc....which seems to happen a lot these days

Answer 2

\[k=\frac{3mg \sin(\theta )}{x}~~~~m=10,000~~\theta=30^0,~x=0.15\]

??

Answer 3

@IrishBoy123 Thank you so much for your help!