Question

A car with a total mass of 1190 kg (including passengers) is driving down a washboard road with bumps spaced 5.0 m apart. The ride is roughest—that is, the car bounces up and down with the maximum amplitude—when the car is traveling at 5.3 m/s. The maximum net force of the car is 1.5*10^3 what is the maximum amplitude of the oscillation?

Answer 1

x = 2.84 cm

Step-by-step explanation:

First we can find the frequency of the oscillation given as

`f = (v)/(d)`

here we know that

d = 5.0 m

v = 5.3 m/s

now we have

`f = (5.3)/(5)`

`f = 1.06 Hz`

now we know that

`f = (1)/(2\pi) \sqrt{(k)/(m)}`

so we have

`1.06 = (1)/(2\pi) \sqrt{(k)/(1190)}`

`k = 5.28 * 10^4 N/m`

Now by the formula of spring force we know that

`F = kx`

so we have

`1.5 * 10^3 = (5.28 * 10^4) x`

`x = 0.0284 m`

`x = 2.84 cm`