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

Answer: 0.0392 m

Explanation:

Given

Mass of the car, M = 1300 kg

Mass of each person, m = 84 kg

Distance of corrugation, d = 3.2 m

Speed of car, v = 13 km/h = 3.61 m/s

To solve this, we would be doing some derivations.

If, T = d/v, the angular frequency of Simple Harmonic Motion is,

w = √ (k / M + 4m), but we also know that,

w = 2π/T. Now we substitute for w

2π/T = √ (k / M + 4m), here again, we substitute for T

2πv/d = √ (k / M + 4m), making subject of formula, we have

k = (M + 4m) [2πv/d]²

the vertical displacement of the car with respect to the ground is given by, F = kx. We also know that the mass is M + 4m, so that

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

x (i) = (M + 4m) g / k, we can also write

Mg = kx (f)

x (f) = Mg / k

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

x (i) - x (f) = 4mg / (M + 4m) [2πv/d]²

x (i) - x (f) = 4mg/M + 4m * (d/2πv) ², now we substitute all the values into the equation to have

x (i) - x (f) = (4 * 84 * 9.8) / (1300 + 4 * 84) * (3.2/2 * 3.142 * 3.61)

x (i) - x (f) = (3292.8/1636) * (0.14) ²

x (i) - x (f) = 2 * 0.0196

x (i) - x (f) = 0.0392 m