Question

Four people, each with a mass of 72.4 kg, are in a car with a mass of 1130 kg. An earthquake strikes. The vertical oscillations of the ground surface make the car bounce up and down on its suspension springs, but the driver manages to pull off the road and stop. When the frequency of the shaking is 1.80 Hz,the car exhibits maximum amplitude of vibration. The earthquake ends, and the four people leave the car as fast as they can. By what distance does the car’s undamaged suspension lift the car’s body as the people get out?

Short answer

The distance does the car’s undamaged suspension lift the car’s body as the people get out of it is $x=1.56\times {10}^{-2} \text{m}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The mass of each people is 72.4 kg.
• The mass of the car is 1130 kg
• The frequency is 1.80 Hz.

Significance of the period of the oscillation of the particle

The period of the oscillation of the particle is:

$\mathit{T}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{\omega }}\mathbf{=}\mathbf{2}\mathit{\pi }\sqrt{\frac{\mathbf{m}}{\mathbf{k}}}$

Determination of the distance

Part (a):

By using the concept and formula from step (1): we can write

$T=2\pi \sqrt{\frac{L}{g}}$

The period of the oscillation of the particle is

$$\begin{gathered} T=\frac{2\pi }{\omega } \\ T=2\pi \sqrt{\frac{m}{k}} \end{gathered}$$

Squaring on both sides we get

$$\begin{gathered} T^2=4{\pi }^2\frac{m}{k} \\ k=4{\pi }^2\frac{m}{T^2} \end{gathered}$$

But,

$T=\frac{1}{f}$

Thus,

$k=4{\pi }^2{f}^{2}m$

Substitute all the value in the above equation,

$$\begin{gathered} k=4\times 3.{14}^2\times {\left(1.80 \text{Hz}\right)}^2\times \left(4\times 72.4 \text{kg}+1130 \text{kg}\right) \\ k=1.82\times {10}^5 {\text{kgs}}^{-2} \end{gathered}$$

Due to force balancing:

$x=\frac{F}{k}$

Substitute all the value in the above equation,

$$\begin{gathered} x=\frac{4\left(72.4 \text{kg}\right)\times 9.81 m{s}^{-2}}{1.82\times {10}^5 {\text{Nm}}^{-1}} \\ x=1.56\times {10}^{-2} \text{m} \end{gathered}$$

Hence the distance does the car’s undamaged suspension lift the car’s body as the people get out of it is,$x=1.56\times {10}^{-2} \text{m}$ .