Question

The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude decreases by 37% each cycle. Estimate the values of

a. The spring constant k.
b. The damping constant b for the spring and shock absorber system of one wheel, assuming each wheel supports 425 kg.

Answer 1

To estimate the values of k and b for the spring and shock absorber system of one wheel, we can use the given information. For k, we can calculate it using the formula for the sag of the suspension system. For b, we can use the formula for decay of oscillation amplitude.

Step-by-step explanation:

To estimate the values of k and b for the spring and shock absorber system of one wheel, we can use the information given.

First, let's calculate the spring constant k:

Using the formula for the sag of the suspension system, we can relate the weight of the car to the spring constant:

sag = (F_weight) / k

Where F_weight is the weight of the car. Rearranging the formula, we can solve for k:

k = (F_weight) / sag

Given that the weight of the car is 1700 kg * 9.8 m/s^2 and the sag is 7.7 cm, we can calculate k.

Next, let's calculate the damping constant b:

Given that each wheel supports 425 kg and the oscillation amplitude decreases by 37% each cycle, we can use the formula for decay of oscillation amplitude to determine the damping constant:

amplitude_n+1 = 0.63 * amplitude_n

Using this formula and the mass of each wheel, we can calculate b.

Answer 2

the spring constant k =
`5.409*10^4 \ N/m`

the value for the damping constant
`\\ \\b = 1.518 *10^3 \ kg/s`

Step-by-step explanation:

From Hooke's Law

`F = kx\\\\k =(F)/(x)\\\\where \ F = mg\\\\k = (mg)/(x)\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = (425 \ kg * 9.8 \ m/s^2)/(0.077 \ m)\\\\k = 5.409*10^4 \ N/m`

Thus; the spring constant k =
`5.409*10^4 \ N/m`

The amplitude is decreasing 37% during one period of the motion

`e^{(-bT)/(2m)}= (37)/(100)\\\\e^{(-bT)/(2m)}= 0.37\\\\(-bT)/(2m) = In(0.37)\\\\(-bT)/(2m) = -0.9943\\\\b = (2m(0.9943))/(T)\\\\b = (2m(0.9943))/((2 \pi)/(\omega))\\\\b = (m(0.9943) \ ( \omega) ))/( \pi)`

`b = \frac{m(0.9943)(\sqrt{(k)/(m))}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{(5.409*10^4)/(425)) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s`

Therefore; the value for the damping constant
`\\ \\b = 1.518 *10^3 \ kg/s`