Question

A compact car has a mass of 1200 kg. Assume that the car has one spring on each wheel, that the springs are identical, and that the mass is equally distributed over the four springs.

Part A
What is the spring constant of each spring if the empty car bounces up and down 1.6 times each second?
Express your answer using two significant figures.in N/m.

Part B
What will be the car's oscillation frequency while carrying four 70 kg passengers?
Express in two sig figs in Hz.

Answer 1

Concepts and reason

The concepts required to solve this problem is frequency of a spring mass system.

Initially, use the expression of frequency of a spring-mass system to find the spring constant of the car’s springs.

Then, use the expression of frequency to determine the frequency of oscillation of the car’s springs when 4 passengers weighing 70 kgs each sits inside the car.

Fundamentals

The frequency of a spring-mass system can be written as follows:

$f = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}}$

Here, k is the spring constant and m is the mass.

(A)

The frequency of a spring-mass system can be written as follows:

$f = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}}$

Rearrange the above expression for k.

$\begin{array}{c}\\k = {\left( {2\pi f} \right)^2}m\\\\ = 4{\pi ^2}{f^2}m\\\end{array}$

Substitute 1.6 Hz for f and 1200 kg for m in the above expression.

$\begin{array}{c}\\k = \left( {4{\pi ^2}} \right){\left( {1.6{\rm{ Hz}}} \right)^2}\left( {1200\,{\rm{kg}}} \right)\\\\ = 121277.6989{\rm{ N/m}}\\\end{array}$

The 4 springs of the car are equally compressed so that the spring constant of each spring can be written as follows:

$k' = \frac{k}{4}$

Here, k’ is the spring constant of each spring.

Substitute 121277.6989 N/m for k in the above expression.

$\begin{array}{c}\\k' = \frac{{121277.6989\;{\rm{N/m}}}}{4}\\\\ = 30319.42473{\rm{ N/m}}\\\\ = 30000{\rm{ N/m}}\\\end{array}$

(B)

The net mass of the car is as follows:

$M = 4m' + m$

Here, m is the mass of car, m’ is the mass of passenger, and M is the net mass of the car.

Substitute 70 kg for m’, and 1200 kg for m in the above expression.

$\begin{array}{c}\\M = 4\left( {70{\rm{ kg}}} \right) + 1200\;{\rm{kg}}\\\\ = 1480\;{\rm{kg}}\\\end{array}$

The frequency of a spring-mass system can be written as follows:

$f = \frac{1}{{2\pi }}\sqrt {\frac{k}{M}}$

Substitute 121277.6989 N/m for k and 1480 kg for M in the above expression.

$\begin{array}{c}\\f = \frac{1}{{2\pi }}\sqrt {\frac{{121277.6989{\rm{ N/m}}}}{{1480{\rm{ kg}}}}} \\\\ = 1.44{\rm{ Hz}}\\\\ = 1.4{\rm{ Hz}}\\\end{array}$

Ans: Part A

The spring constant of each spring is 30000 N/m

Part B

The frequency is 1.4 Hz.