Question

A car traveling on a rough and bumpy road, which is modeled with the basic displacement y (t) = (0.01) sin (5.818t) m. The car suspension system, which is modeled with 1-degree of freedom, has a rigidity of k = 4 x 10^5 N/m, and a damping coefficient of c = 40 x 10^3 kg/s, and the mass of the car m = 1007 kg. Calculate the absolute displacement amplitude of the mass of the car.

Answer 1

Here the displacement equation is given as

y(t)=0.01 sin(5.818 t)

here degree of freedom of car is 1, and stiffness is $4\times 10^5 N/m$

and damping coefficient is , c= $4\times 10^3 kg/s$ and mass of car is 1007 kg

$\omega _n= \sqrt \frac{k}{m}=\sqrt{\frac{4\times 10^5}{1007}}=19.93$

and damping coefficient is

$\zeta=\frac{c}{2\sqrt{mk}}=\frac{4\times 10^3}{2\sqrt{1007\times 4\times 10^5}}=0.9965$

Now the frequncy ratio we are getting value of frequency from equation as , $\omega=5.818$

$r=\frac{ \omega}{\omega_n}=\frac{5.818}{19.93}=0.292$

And in case of single degree of freedom for damped motion we have amplitude formula as

$X=Y\sqrt{ \frac{ 1+(2\zeta r)^2}{(1+r^2)^2 +(2 \zeta r)^2}}$

$X=0.01\sqrt{ \frac{ 1+(2\times0.997\times 0.292)^2}{(1+0.292^2)^2 +(2 \times0.997\times 0.292)^2}}=0.011 m$

So, magnitude of absolute displacement is 0.011 m