Question

In Fig.15-51, three 10000 kgore cars are held at rest on a mine railway using a cable that is parallel to the rails, which are inclined at angle$\mathit{\theta }\mathbf{=}\mathbf{30}\mathbf{°}$. The cable stretches 15 cmjust before the coupling 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 oscillations of the remaining two cars.

Short answer

a) The frequency of oscillation is 1.1 Hz .

b) The amplitude of oscillation is 5.0 cm .

Step-by-step solution

The given data

• The mass of each ore car,m=10,000 kg.
• The angle between the cable and the ground,$\theta =30°$.
• The length of the stretched cable,x=15 cm or 0.15 m.

Understanding the concept of SHM

The cable gets stretched initially. When the last car is separated from the cable, the system of cable and the remaining two cars oscillates in simple harmonic motion.

Formulae:

The pulling force applied to a body,$F=-kx$ (i)

The frequency of a body in oscillation, $f=\frac{1}{2\tau \tau }\sqrt{\frac{k}{m}}$ (ii)

a) Calculation of frequency of resulting oscillations

When all the three cars are connected to the cable, the forces acting on the cars are,

1. The weight (3 mg)
2. The tension in the cable (T)
3. The normal force from the rail (N)

We write the force equation as,

T -3mg sin30=0

Since the cars are not accelerated. Thus, the net force is,

$$\begin{gathered} T=3mg\sin \left(30°\right) \\ =3\times 10000kg\times 9.8m/s^2\times \sin \left(30\right) \\ =1.47\times {10}^5\mathrm{N} \end{gathered}$$

Under the action of this force, the cable stretches by x=0.15 m .

So, using equation (i) for force only applied by the tension of the string equals to the stretching force, we can write,

$T=kx..........................\left(a\right)$

This gives the spring constant value as:

$$\begin{gathered} k=\frac{T}{x} \\ =\frac{1.47\times {10}^5\mathrm{N}}{0.15\mathrm{m}} \\ =9.8\times {10}^5\mathrm{N}/\mathrm{m} \end{gathered}$$

Now, after the last car is cut off, the cable and the remaining two cars oscillate with frequency f, using equation (ii) given as:

$$\begin{gathered} f=\frac{1}{2\tau \tau }\sqrt{\frac{9.8\times {10}^{5}N/m}{20000kg}} \\ =1.11\mathrm{Hz} \end{gathered}$$

Hence, the value of frequency of oscillations is 1.11 Hz .

b) Calculation of the amplitude of resulting oscillations

When all the three cars were connected to the cable, it gets stretched by 15 cm. So, using equation (a) we can write:

$$\begin{gathered} x_1=\frac{T_1}{k} \\ =\frac{3mg\sin 30°}{k} \end{gathered}$$

And when the two cars are connected to the cable using same equation (a), we can write,

$$\begin{gathered} x_2=\frac{T_2}{k} \\ =\frac{2mg\sin 30°}{k} \end{gathered}$$

The amplitude of the resulting oscillations = difference in stretching =$x_2-x_1$.

So, the amplitude of the resulting oscillations of the spring is given as:

$$\begin{gathered} A=\frac{\left(3m-2m\right)g\sin \left(30°\right)}{k} \\ =\frac{4.9\times {10}^{4}N}{k} \\ =0.05\mathrm{m} \\ =5.0\mathrm{cm} \end{gathered}$$

Hence, the value of amplitude is 5.0 cm .