Incomplete question.The complete is here
A car with a total mass of 1800 kg (including passengers) is driving down a washboard road with bumps spaced 4.2 m apart. The ride is roughest—that is, the car bounces up and down with the maximum amplitude—when the car is traveling at 8.0 m/s.What is the spring constant of the car's springs?
Answer:
Step-by-step explanation:
The frequency is calculated as:
![f=(8.0m/s)/(4.2m)\\ f=1.905Hz](https://img.qammunity.org/2021/formulas/physics/high-school/ta4rqltc260cb8io4mari9241q1vgid56x.png)
As we know that
![k=257.6kN/m](https://img.qammunity.org/2021/formulas/physics/high-school/d255jp2tptpx3dicdqpskxpc5niq73b6w7.png)
![f=(1)/(2\pi ) \sqrt{(k)/(m) }\\ k=m(2\pi f)^(2)\\ k=1800kg[2\pi (1.905Hz)]^(2) \\k=257.6kN/m](https://img.qammunity.org/2021/formulas/physics/high-school/nxbxprdph9b00cq9j3airpeadfpqmj4a2c.png)