Question

A 1200 kg car carrying four 80 kg people travels over a rough "washboard" dirt road with corrugations 4.0 m apart which causes the car to bounce on its spring suspension. The car bounces with maximum amplitude when its speed is 15 km/h. The car now stops, and the four people get out. By how much does the car body rise on its suspension owing to this decrease in weight?

Answer 1

Answer:

ΔX = 0.0483 m

Explanation:

Let's analyze the problem, the car oscillates in the direction y and advances with constant speed in the direction x

The car can be described with a spring mass system that is represented by the expression

     y = A cos (wt + φ)

The speed can be found by derivatives

     $v_{y}$ = dy / dt

    $v_{y}$  = - A w sin (wt + φ

So that the amplitude is maximum without (wt + fi) = + -1

      $v_{y}$  = A w

X axis

Let's reduce to the SI system

     vₓ = 15 km / h (1000 m / 1 km) (1h / 3600s) = 4.17 m / s

As the car speed is constant

     vₓ = d / t

      t = d / v ₓ

      t = 4 / 4.17

      t = 0.96 s

This is the time between running two maximums, which is equivalent to a full period

     w = 2π f = 2π / T

     w = 2π / 0.96

     w = 6.545 rad / s

We have the angular velocity we can find the spring constant

     w² = k / m

    m = 1200 + 4 80

    m = 1520 m

     k = w² m

     k = 6.545² 1520

     k = 65112 N / m

Let's use Newton's second law

    F - W = 0

    F = W

    k x = W

    x = mg / k

Case 1  when loaded with people

   x₁ = 1520 9.8 / 65112

   x₁ = 0.22878 m

Case 2 when empty

   x₂ = 1200 9.8 / 65112

   x₂ = 0.18061 m

The height variation is

    ΔX = x₁ -x₂  

    ΔX = 0.22878 - 0.18061

    ΔX = 0.0483 m