A 1000 kgcar carrying four$\mathbf{82}\mathbf{kg}$people travel over a “washboard” dirt road with corrugations$$" width="9">4.0 mapart. The car bounces with maximum amplitude when its speed is 16 km/h. When the car stops, and the people get out, by how much does the car body rise on its suspension?

1. Mass of the car$:M=1000\text{\hspace{0.17em}kg}$
2. Mass of each person in the car$:m=82\text{\hspace{0.17em}kg}$
3. The distance between corrugations$x=4.0\text{\hspace{0.17em}m}$
4. The speed at which amplitude is maximum:$v_m=16\text{\hspace{0.17em}km}/\text{hr}=4.44\text{m}/\text{s}$

The spring of the car suspension gets compressed whenever it passes over the corrugations. Thus, the spring exhibits simple harmonic motion as it passes over the washboard dirt road.

The restoring force developed during the deformation:$\mathbf{F}\mathbf{=}\mathbf{-}\mathbf{kx}$

The time period of an oscillation produced in the car:$\mathbf{T}\mathbf{=}\frac{2\pi }{\omega }$

The angular frequency of oscillation,$\mathbf{\omega }\mathbf{=}\mathbf{}\sqrt{\frac{k}{m}}$

When the car along with the people travels over the corrugations, the compression force on the suspension obeys Hooke’s law. Hence, we can write the equation for the suspension as

$F=-k{x}_i$

Here,$F=(M+4m)g$and$x_i$= compression when people are inside the car

So, we have $(M+4m)g=-k{x}_i$

Giving us

$\begin{array}{l}{x}_i=\frac{F}{k}\\ =\frac{(M+4m)g}{k}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \mathrm{..}\left(1\right)\end{array}$

Similarly, when people get off the car, the car travels, again we write the equation as

$F^{\text{'}}=-k{x}_f$

where$F’=Mg$ and$x_f=$compression when onlythecar is travelling

So, we have

$Mg=-k{x}_f$

Giving us

$\begin{array}{l}{x}_f=\frac{F}{k}\\ =\frac{Mg}{k}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \mathrm{..}\dots \dots \dots \dots \dots \mathrm{..}\left(2\right)\end{array}$

When the car travels on the corrugations, suspension gets periodically compressed. The distance between the two consecutive impulses isthesame as the distance between the corrugations. i.e.$d=4.0\text{m}$.

Hence, the period of the oscillations of the suspension

$T=\frac{d}{v}$

v is the speed of the car.

The angular frequency of the car

$\begin{array}{l}\omega =\frac{2\pi }{T}\\ =\frac{2\pi v}{d}\end{array}$

And we have

$\omega =\sqrt{\frac{k}{(M+4m)}}$

As the car is carrying people

Hence, we determine the expression for k as…

$\begin{array}{l}k=(M+4m){\omega }^2\\ =(M+4m){\left(\frac{2\pi v}{d}\right)}^2\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(3\right)\end{array}$

To determine the rise in suspension, we use equations (1), (2) and (3) as.

$\begin{array}{l}Rise=x_i–x_f\\ =\frac{(M+4m)g}{k}-\frac{Mg}{k}\\ =\frac{4mg}{k}\\ =\frac{4mg}{(M+4m)}{\left(\frac{d}{2\pi v}\right)}^2\end{array}$

For the given values, the rise of suspension is given as-

$\begin{array}{l}\text{Rise}=\frac{4\times 82\text{\hspace{0.17em}kg}\times 9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}{(1000\text{\hspace{0.17em}kg}+(4\times 82\text{\hspace{0.17em}kg}))}{\left(\frac{4.0\text{\hspace{0.17em}m\hspace{0.17em}}}{2\times 3.14\times 4.4\text{\hspace{0.17em}m}/\text{s}}\right)}^2\\ =0.05\text{m}\\ =5.0\text{cm}\end{array}$

The suspension rises by $0.05\text{\hspace{0.17em}m}$.