Answer
The amplitude of the oscillations is $~~5.0~cm$
Work Step by Step
We can find $k$:
$kx = mg sin \theta$
$k = \frac{mg sin \theta}{x}$
$k = \frac{(30,000~kg)(9.8~m/s^2) (sin 30^{\circ})}{0.15~m}$
$k = 980,000~N/m$
We can find the equilibrium stretch distance after one car breaks off:
$kx = mg sin \theta$
$x = \frac{mg sin \theta}{k}$
$x = \frac{(20,000~kg)(9.8~m/s^2) (sin ~30^{\circ})}{980,000~N/m}$
$x = 0.10~m$
$x = 10~cm$
The system will oscillate about the equilibrium point.
Since the system started oscillating when the stretch distance was $15~cm$, the amplitude of the oscillations is $~~5.0~cm$
