Correct Answer - A::D
(A) The net force on the three cars is zero before the cable breaks. There are three forces on the cars: the weight, `W`, a normal force, `N`, and the upward force from the cable, `f`. Then
`F = W sin theta = 3 mg sin theta`
This force is from the elastic properties of the cable,
so `k = (F)/(x) = (3mg sin theta)/(x)`
The frequency of oscillation of the remaining two can after the bottom car is released is
`f = (2)/(2pi) sqrt((k)/(2m)) = (1)/(2pi) sqrt((3mg sin theta)/(2mx)) = (1)/(2x) sqrt((3g sin theta)/(2x))`
(B) Each car contributes equally to the stretching of the cable, so one car causes the cable to stretch `(x)/(3)`