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:
Explanation:
The frequency is calculated as:
[tex]f=\frac{8.0m/s}{4.2m}\\ f=1.905Hz[/tex]
As we know that[tex]k=257.6kN/m[/tex]
[tex]f=\frac{1}{2\pi } \sqrt{\frac{k}{m} }\\ k=m(2\pi f)^{2}\\ k=1800kg[2\pi (1.905Hz)]^{2} \\k=257.6kN/m[/tex]