(a) Here mass supported by each wheel, M = 750 kg and x =15 cm= 0.15 mIf k is spring constant of the spring, then restoring force developed on being compressed through a distance $x, F = -kx$If M is mass supported by each wheel, then $\therefore kx = Mg$or $k = \frac{M g}{x} = \frac{750 \times 9.8}{0.15}$ $= 4.9 \times 10^{4} \mathrm{N} \mathrm{m}^{-1}$(b) If b is damping constant for the spring and shock absorber system, then damped amplitude of oscillation is given by $A = A_0 e^{-\frac{b T}{2 M}}$where T is period of oscillation of the spring, $A_0$ the initial amplitude of the oscillations and M, the mass supported by it. From the above relation we have $A = A_0 e^{-\frac{b T}{2 M}}$As the amplitude of oscillation decreases by 50% during one complete oscillation, $\frac{1}{2} A_0 = A_0 e^{-\frac{b t}{2 M}}$$\Rightarrow 2 = e^{\frac{b t}{2 M}}$ $\Rightarrow \log_e 2 = \frac{b t}{2 M} \log_e e = \frac{b t}{2 M}$Now, $t = 2 \pi \sqrt{\frac{M}{4 R}}$ $= 2 \times \frac{22}{7} \times \sqrt{\frac{3000}{4 \times 4.9 \times 10^{4}}}$ $= \frac{44}{70} \sqrt{\frac{15}{9.8}}$$\therefore b = \frac{2 m \log_e 2}{t}$ $= \frac{2 \times 750 \times 0.6931}{ \frac{44}{70} \sqrt{\frac{15}{9.8}} }$ $= 1336.99 \mathrm{kg} \mathrm{s}^{-1}$