Question

The suspension system of a$\mathbf{2000}\mathbf{kg}$automobile “sags”$\mathbf{10}\mathbf{cm}$when the chassis is placed on it. Also, the oscillation amplitude decreases by 50% each cycle.

1. Estimate the value of the spring constant K.
2. Calculate the damping constantfor the spring and shock absorber system of one wheel, assuming each wheel supports$\mathbf{500}\mathbf{kg}$.

Short answer

1. The spring constant for each wheel,$\text{k}=4.9\times {10}^4\text{N}/\text{m}$
2. The damping constant for the spring and shock absorber system,$\text{b}=1.1\times {10}^3\text{\hspace{0.17em}kg}/\text{s}$

Step-by-step solution

Given

1. The mass supported by one wheel is$m=500\text{kg}$

1. The mass of automobile is,$M=2000\text{kg}$
2. The compression of spring or suspension system,$x=10\text{cm}=0.1\text{m}$
3. The oscillation amplitude decreases by 50% each cycle.

Understanding the concept

Use the equation of Hook’s law to get the spring constant. To find the damping constant, we can use the equation for damping factor and the period of damped oscillation.

Hooke’s law is given as-

$\text{F}=\text{kx}$

The angular frequency for damped oscillation is given as-

$$\begin{gathered} {\omega }^{\text{'}}=\sqrt{\frac{\text{k}}{\text{m}}-\frac{{\text{b}}^{\text{2}}}{4{\text{m}}^2}} \\ \text{dampingfacto}r=e^{-\frac{\text{bt}}{2\text{m}}} \end{gathered}$$

(a) Estimate the value of the spring constant

Using the equation of Hook’s law, we can write

$\begin{array}{l}\text{k}=\frac{\text{F}}{\text{x}}\\ =\frac{500\text{\hspace{0.17em}kg}\times 9.8\text{\hspace{0.17em}m}/{\text{s}}^2}{0.1\text{\hspace{0.17em}s}}\\ =4.9\times {10}^4\text{\hspace{0.17em}N}/\text{m}\end{array}$

(b) Calculate the damping constant for the spring and shock absorber system of one wheel, assuming each wheel supports500 kg .

We calculate the time for which damping factor is 1/2.

So,

$e^{-\frac{bT}{2m}}=\frac{1}{2}$

Taking natural logarithm on both sides

$$\begin{gathered} -\frac{bT}{2m}=\ln \left(\frac{1}{2}\right) \\ bT=2m\times \ln \left(2\right) \end{gathered}$$

Here, the equation for period of damped oscillation is

$T=\frac{2\pi }{{\omega }^{\text{'}}}$

Where,$\omega \text{'}$is the angular frequency of damped oscillation and is given by

${\omega }^{\text{'}}=\sqrt{\frac{\text{k}}{\text{m}}-\frac{{\text{b}}^{\text{2}}}{4{\text{m}}^2}}$

So, the above equation becomes

$\frac{b\times 2\pi }{\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}}=2m\times \ln \left(2\right)$

By rearranging this equation for the damping constant, we get

$$\begin{gathered} b=\frac{2\times \ln \left(2\right)\sqrt{mk}}{\sqrt{\ln {\left(2\right)}^2+4{\pi }^2}} \\ \Rightarrow \text{b}=\frac{2\times \text{ln}\left(2\right)\sqrt{500\text{\hspace{0.17em}kg}\times 4.9\times {10}^4\text{\hspace{0.17em}N}/\text{m}}}{\sqrt{\text{ln}{\left(2\right)}^2+4{\pi }^2}} \end{gathered}$$

Using the given values in this equation, we get

$\text{b}=1.1\times {10}^3\text{\hspace{0.17em}kg}/\text{s}$

So, the damping constant is given as- $\text{b}=1.1\times {10}^3\text{\hspace{0.17em}kg}/\text{s}$.