91影视

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.

Use the equation of Hook鈥檚 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鈥檚 law is given as-

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

The angular frequency for damped oscillation is given as-

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

Using the equation of Hook鈥檚 law, we can write

$\begin{array}{l}\text{k}=\frac{\text{F}}{\text{x}}\\ =\frac{500\text{鈥塳驳}脳9.8\text{鈥尘}/{\text{s}}^2}{0.1\text{鈥塻}}\\ =4.9脳{10}^4\text{鈥塏}/\text{m}\end{array}$

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

So,

$e^{鈭�\frac{bT}{2m}}=\frac{1}{2}$

Taking natural logarithm on both sides

$$\begin{gathered} 鈭�\frac{bT}{2m}=\ln \left(\frac{1}{2}\right) \\ bT=2m脳\ln \left(2\right) \end{gathered}$$

Here, the equation for period of damped oscillation is

$T=\frac{2蟺}{{蝇}^{\text{'}}}$

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

${蝇}^{\text{'}}=\sqrt{\frac{\text{k}}{\text{m}}鈭�\frac{{\text{b}}^{\text{2}}}{4{\text{m}}^2}}$

So, the above equation becomes

$\frac{b脳2蟺}{\sqrt{\frac{k}{m}鈭�\frac{b^2}{4m^2}}}=2m脳\ln \left(2\right)$

By rearranging this equation for the damping constant, we get

$$\begin{gathered} b=\frac{2脳\ln \left(2\right)\sqrt{mk}}{\sqrt{\ln {\left(2\right)}^2+4{蟺}^2}} \\ 鈬�\text{b}=\frac{2脳\text{ln}\left(2\right)\sqrt{500\text{鈥塳驳}脳4.9脳{10}^4\text{鈥塏}/\text{m}}}{\sqrt{\text{ln}{\left(2\right)}^2+4{蟺}^2}} \end{gathered}$$

Using the given values in this equation, we get

$\text{b}=1.1脳{10}^3\text{鈥塳驳}/\text{s}$

So, the damping constant is given as- $\text{b}=1.1脳{10}^3\text{鈥塳驳}/\text{s}$.