Answer:
the spring constant k = [tex]5.409*10^4 \ N/m[/tex]
the value for the damping constant [tex]\\ \\b = 1.518 *10^3 \ kg/s[/tex]
Explanation:
From Hooke's Law
[tex]F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m[/tex]
Thus; the spring constant k = [tex]5.409*10^4 \ N/m[/tex]
The amplitude is decreasing 37% during one period of the motion
[tex]e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}[/tex]
[tex]b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s[/tex]
Therefore; the value for the damping constant [tex]\\ \\b = 1.518 *10^3 \ kg/s[/tex]
