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4P

Expert-verifiedFound in: Page 436

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The springs of a certain car are adjusted so that the oscillations have a frequency of 3.00 Hz****. **

**What is the spring constant of each spring if the mass of the car is 1450 kg****and the mass is evenly distributed over the springs?****What will be the oscillation frequency if five passengers, averaging 73.0 kg****each, ride in the car with an even distribution of mass?**

- Spring constant is $1.29\times {10}^{5}\mathrm{N}/\mathrm{m}$
- Frequency of oscillation is data-custom-editor="chemistry" $2.68\mathrm{Hz}$.

- Frequency of oscillations, $\mathrm{f}=3.00\mathrm{Hz}$
- Mass of car, ${\mathrm{m}}_{\mathrm{car}}=1450\mathrm{kg}$
- Mass of each passenger, ${\mathrm{m}}_{\mathrm{passenger}}=73\mathrm{kg}$.

**The car is mounted on four springs. First, we calculate the mass supported by each spring as we have to calculate the spring constant. We can use the formula for angular frequency in terms of the spring constant and mass supported by each spring to find angular frequency and frequency.****To calculate frequency, we first calculate the total mass by considering each passenger’s mass and the car’s mass. We can find the mass supported by each spring. We can use the spring constant found in the first part to find the frequency.**

** **

** **

Formula:

Frequency relation to the spring constant of a body

$\mathrm{\omega}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$ (i)

The angular frequency of the motion is given as follows:

$\begin{array}{rcl}\mathrm{\omega}& =& 2\mathrm{\pi f}\\ & =& 2\mathrm{\pi}\times 3\\ & =& 18.85\mathrm{rad}/\mathrm{sec}\end{array}$

Mass supported by each spring is one-fourth of the mass of the car. Hence, the mass load on each spring is given as

$\begin{array}{rcl}\mathrm{m}& =& \frac{1450}{4}\\ & =& 362.5\mathrm{kg}\end{array}$

Now using equation (i), the spring constant can be derived as

$\begin{array}{rcl}18.85& =& \sqrt{\frac{\mathrm{k}}{362.5}}\\ \mathrm{k}& =& 128804.41\mathrm{N}/\mathrm{m}\\ & =& 1.29\times {10}^{5}\mathrm{N}/\mathrm{m}\end{array}$

Hence, the value of the spring constant is $1.29\times {10}^{5}\mathrm{N}/\mathrm{m}$.

Now five passengers having an average mass of 73 kg ride in the car, so the total mass is as follows:

$\begin{array}{rcl}\mathrm{M}& =& {\mathrm{m}}_{\mathrm{car}}+5\left({\mathrm{m}}_{\mathrm{passenger}}\right)\\ & =& 1450+5\left(73\right)\\ & =& 1815\mathrm{kg}\end{array}$

So, mass supported by each spring is

$\begin{array}{rcl}\mathrm{m}& =& \frac{1815}{4}\\ & =& 453.75\mathrm{kg}\end{array}$

Now, using equation (i), we get the angular frequency as follows:

$\mathrm{\omega}=\sqrt{\frac{129000}{453.75}}\phantom{\rule{0ex}{0ex}}=16.86\mathrm{rad}/\mathrm{sec}\phantom{\rule{0ex}{0ex}}2\mathrm{\pi f}=16.86\mathrm{rad}/\mathrm{sec}\left(\because \mathrm{angular}\mathrm{frequency}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{frequency},\mathrm{\omega}=2\mathrm{\pi f}\right)\phantom{\rule{0ex}{0ex}}\mathrm{f}=8.43\mathrm{Hz}$

Hence, the frequency of oscillation is $8.43\mathrm{Hz}$.

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