A 1300 kg car carrying four 84 kg people travels over a "washboard" dirt road with corrugations 3.2 m apart. The car bounces with maximum amplitude when its speed is 13 km/h. When the car stops, and the people get out, by how much does the car body rise on its suspension?

0.04m

Explanation:

Mass of car (M) = 1300kg

Mass of people (m) = 84kg

Distance (d) = 3.2m

Speed (v) = 13km/h

V = 13km/h = x m/s

(13 * 1000) / 3600 = 3.6m/s

The distance travelled between the two maximums is the distance travelled during the period.

T = v / d equation (i)

Angular frequency of simple harmonic motion

ω = √(k / M + 4m)

but ω = 2π /T

2π / T = √(k / M + 4m)

put T = v / d

2πv / d = √(k / M + 4m

solving for k,

K = (M + 4m) * (2πv / d)²

The vertical displacement of the car with respect to ground and force constant K =

F = Kx

F = force of gravity

F(i) = (M + 4m)g

(M + 4m)g = KX(i)

X(i) = (M + 4m)g / I

Force without people F(f) = m*g

mg = KX(f)

X(f) = Mg / k

X(i) - X(f) = [(M + 4m)g / k - Mg / k]

X(i) - X(f) = 4mg / k

But k = (M + 4m) * (2πv / d)²

X(i) - X(f) = [4mg / (M + 4m) ] * (d /2π v)²

X(i) - X(f) = [4 * 84 * 9.8 / (1300 + 4*84)] * [(3.2 / 2Π*3.6 ]²

X(i) - X(f) = (3292.8 / 1636) * 0.020

X(i) - X(f) = 2.01 * 0.020

x(i) - x(f) = 0.04m

The car body rise on its suspension by 0.04m