Question

A car with a total mass of 1800 kg (including passengers) is driving down a washboard road with bumps spaced 4.9 m apart. The ride is roughest—that is, the car bounces up and down with the maximum amplitude—when the car is traveling at 5.7 m/s. What is the spring constant of the car's springs? Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

k = 9.6 x 10^5 N/m or 9.6 kN/m

Explanation:

First, we need to use the expression to calculate the spring constant which is:

w² = k/m

Solving for k:

k = w²*m

To get the angular velocity:

w = 2πf

The problem is giving the linear velocity of the car which is 5.7 m/s. With this we can calculate the frequency of the car:

f = V/x

f = 5.7 / 4.9 = 1.16 Hz

Now the angular velocity:

w = 2π*1.16

w = 7.29 rad/s

Finally, solving for k:

k = (7.29)² * 1800

k = 95,659.38 N/m

In two significant figures it'll ve 9.6 kN/m

Answer 2

Answer:

k=96.16 kN/m

Explanation:

Maximum amplitude is achieved, when the system operated in the resonance- frequency of the bumps is equal to the natural frequency of the spring-mass system.

Frequency of the bumps, as an input force:

f=V/d, where d- distance between the bumps and V- velocity of the resonance.

From the natural frequency of the spring- mass system, we can get:

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

For the given problem, then the value of k, can be found as:

$k=2\pi mf^{2}=2\pi m (\frac{V}{d}) ^{2} =96158 N/m$