This is a practitce problem , I tried it but did wrong, can someone help correct

Question

This is a practitce problem , I tried it but did wrong, can someone help correct it Thank you!

Below is the hint from the question:

Initially, with the 3 cars at equilibrium, how does the spring force up the slope compare with the component of the gravitational force down the slope? Do you recall how to find that component? Can you now find the spring constant and the frequency? Where is the new equilibrium point? What is the distance between the two equilibrium points?

Ø Your answer ts paraly correct. Iry again. in the qure here, tree 2 1000 kq ore cars are held at rest onmine rah ay usnqcable that is paralel to the rals, which are ncined at angle 38". Ihe cable stretches 18 cm lust before the couping between the two lower cars breaks, detaching the lowest car. Assuming that the cable obeys Hooke's law, find the (a) frequency and (b) amplitude of the resulting oscllations of the remaining two cars Car that eaks fre 134 (b)NmberTO.18

Explanation / Answer

You have already solved that

Spring constant of System will be, Using

Fnet = k*x

k = Fnet/x

k = (3*m*g*sin A)/x

k = (3*21000*9.81*sin 38 deg)/0.18

k = 2113873.67 N/m

Now frequency will be for remaining two cars

f = w/(2*pi)

f = (1/(2*pi))*sqrt (k/m)

f = (1/(2*pi))*sqrt (2113873.67/(2*21000))

f = 1.13 Hz

Part B.

Theoratically

for three car system amplitude (max stretch) was 18 cm, now this will become initial stretch for two-car system.

For two car system equilibrium stretch will be 12 cm, Now difference between initial stretch and equilibrium stretch will give amplitude, So

Amplitude = 18 - 12 = 6 cm = 0.06 m

Mathematically

From Spring law

Fnet = k*x

Amplitude = xi - xf = Fi/k - Ff/k

Fi = initial force on three car system

Ff = final force on two car system

Amplitude = (3F - 2F)/k = F/k = 21000*9.81*sin 38 deg/2113873.67

x = 0.06 m

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