Chapter 19.5, Problem 19.151P

The suspension of an automobile can be approximated by the simplified spring-and-dashpot system shown. (a) Write the differential equation defining the vertical displacement of the mass m when the system moves at a speed v over a road with a sinusoidal cross-section of amplitude δ_m and wavelength L. (b) Derive an expression for the amplitude of the vertical displacement of the mass m.

Fig. P19.151

To determine

Write the differential equation defining the vertical displacement of the mass m when the system moves at a speed v over a road with a sinusoidal cross section of amplitude ${\delta }_m$ and wavelength L.

Answer to Problem 19.151P

The differential equation defining the vertical displacement of the mass m when the system moves at a speed v over a road with a sinusoidal cross section of amplitude ${\delta }_m$ and wavelength L is $\underset{\_}{m\frac{d^{2}x}{d{t}^2}+kx+c\frac{dx}{dt}={\delta }_m\left(k\sin {\omega }_{f}t+c\frac{2\pi v}{L}\cos {\omega }_{f}t\right)}$.

Explanation of Solution

Calculation:

Show the free body diagram of the system of automobile, spring and dashpot as in Figure (1).

The expression for the weight of the automobile (W) as follows:

$\begin{array}{l}k=\frac{W}{{\delta }_{st}}\\ W=k{\delta }_{st}\end{array}$

Here, ${\delta }_{st}$ is the static deflection and k is the spring constant.

The expression for the acceleration of the automobile (a) as follows:

$a=\frac{d^{2}x}{d{t}^2}$

Refer Figure (1),

The expression for the force by considering the vertical equilibrium condition as follows;

$\begin{array}{l}{\displaystyle \sum F}=ma\\ W-k\left({\delta }_{st}+x-\delta \right)-c\left(\frac{dx}{dt}-\frac{d\delta }{dt}\right)=ma\end{array}$

Substitute $\frac{d^{2}x}{d{t}^2}$ for a.

$W-k\left({\delta }_{st}+x-\delta \right)-c\left(\frac{dx}{dt}-\frac{d\delta }{dt}\right)=m\frac{d^{2}x}{d{t}^2}$

Substitute $k{\delta }_{st}$ for W.

$\begin{array}{l}k{\delta }_{st}-k\left({\delta }_{st}+x-\delta \right)-c\left(\frac{dx}{dt}-\frac{d\delta }{dt}\right)=m\frac{d^{2}x}{d{t}^2}\\ k{\delta }_{st}-k{\delta }_{st}-kx+k\delta -c\frac{dx}{dt}+c\frac{d\delta }{dt}=m\frac{d^{2}x}{d{t}^2}\\ -kx+k\delta -c\frac{dx}{dt}+c\frac{d\delta }{dt}=m\frac{d^{2}x}{d{t}^2}\\ m\frac{d^{2}x}{d{t}^2}+kx+c\frac{dx}{dt}=k\delta +c\frac{d\delta }{dt}\end{array}$ (1)

The expression for the time interval needed to travel $\left({\tau }_f\right)$ a distance (L) at a speed (v) as follows:

${\tau }_f=\frac{L}{v}$

The expression for the forced circular frequency $\left({\omega }_f\right)$ as follows:

${\omega }_f=\frac{2\pi }{{\tau }_f}$

Substitute $\frac{L}{v}$ for ${\tau }_f$.

$\begin{array}{c}{\omega }_f=\frac{2\pi }{\frac{L}{v}}\\ =\frac{2\pi v}{L}\end{array}$

The expression for the motion of the wheel which is sine curve $\left(\delta \right)$ as follows:

$\delta ={\delta }_m\sin {\omega }_{f}t$

Differentiate the above equation with respect to time ‘t’.

$\begin{array}{c}\frac{d\delta }{dt}=\frac{d}{dt}\left({\delta }_m\sin {\omega }_{f}t\right)\\ ={\delta }_m{\omega }_f\cos {\omega }_{f}t\end{array}$

Substitute $\frac{2\pi v}{L}$ for ${\omega }_f$.

$\frac{d\delta }{dt}={\delta }_m\frac{2\pi v}{L}\cos {\omega }_{f}t$

Substitute ${\delta }_m\frac{2\pi v}{L}\cos {\omega }_{f}t$ for $\frac{d\delta }{dt}$ and ${\delta }_m\sin {\omega }_{f}t$ for $\delta$ in the equation (1).

$\begin{array}{c}m\frac{d^{2}x}{d{t}^2}+kx+c\frac{dx}{dt}=k{\delta }_m\sin {\omega }_{f}t+c{\delta }_m\frac{2\pi v}{L}\cos {\omega }_{f}t\\ ={\delta }_m\left(k\sin {\omega }_{f}t+c\frac{2\pi v}{L}\cos {\omega }_{f}t\right)\end{array}$ (2)

Therefore, the differential equation defining the vertical displacement of the mass m when the system moves at a speed v over a road with a sinusoidal cross section of amplitude ${\delta }_m$ and wavelength L is $\underset{\_}{m\frac{d^{2}x}{d{t}^2}+kx+c\frac{dx}{dt}={\delta }_m\left(k\sin {\omega }_{f}t+c\frac{2\pi v}{L}\cos {\omega }_{f}t\right)}$.

To determine

Derive an expression for the amplitude of the vertical displacement of the mass m.

Answer to Problem 19.151P

The expression for the amplitude of the vertical displacement of the mass m is $x_m\sin \left({\omega }_{f}t-\varphi +\psi \right)$, $x_m=\frac{{\delta }_m\sqrt{k^2+{\left(c{\omega }_f\right)}^2}}{\sqrt{{\left(k-m{\omega }_f^2\right)}^2+{\left(c{\omega }_f\right)}^2}}$, $\tan \varphi =\frac{c{\omega }_f}{k-m{\omega }_f^2}$, and $\tan \psi =\frac{c{\omega }_f}{k}$.

Explanation of Solution

Calculation:

The expression for the general solution from the identity as follows:

$A\sin y+B\cos y=\sqrt{A^2+B^2}\sin \left(y+\psi \right)$

Here, $\psi$ is the Eulerian angle.

The expression for the force transmitted (F) to the automobile as follows:

$F=\left[\sqrt{k^2+{\left(c{\omega }_f\right)}^2}\right]\sin \left({\omega }_{f}t+\psi \right)$

Substitute $\left[\sqrt{k^2+{\left(c{\omega }_f\right)}^2}\right]\sin \left({\omega }_{f}t+\psi \right)$ for F in equation (2).

$m\frac{d^{2}x}{d{t}^2}+kx+c\frac{dx}{dt}={\delta }_m\left[\sqrt{k^2+{\left(c{\omega }_f\right)}^2}\right]\sin \left({\omega }_{f}t+\psi \right)$ (3)

The expression for the differential equation of the motion for the damped forced vibration as follows:

$m\ddot{x}+c\dot{x}+kx=P_m\sin {\omega }_{f}t$ (4)

Compare the equation (3) and (4).

$P_m={\delta }_m\sqrt{k^2+{\left(c{\omega }_f\right)}^2}$

The expression for the steady state of motion of the system as follows:

$x=x_m\sin \left({\omega }_{f}t-\varphi +\psi \right)$

The expression for the steady state of motion of the system as follows:

$x_m=\frac{P_m}{\sqrt{{\left(k-m{\omega }_f^2\right)}^2+{\left(c{\omega }_f\right)}^2}}$

Substitute ${\delta }_m\sqrt{k^2+{\left(c{\omega }_f\right)}^2}$ for $P_m$.

$x_m=\frac{{\delta }_m\sqrt{k^2+{\left(c{\omega }_f\right)}^2}}{\sqrt{{\left(k-m{\omega }_f^2\right)}^2+{\left(c{\omega }_f\right)}^2}}$

The expression for the phase angle $\left(\varphi \right)$ as follows:

$\tan \varphi =\frac{c{\omega }_f}{k-m{\omega }_f^2}$

The expression for the Eulerian angle $\left(\psi \right)$ as follows:

$\tan \psi =\frac{c{\omega }_f}{k}$

Therefore, the expression for the amplitude of the vertical displacement of the mass m is $x_m\sin \left({\omega }_{f}t-\varphi +\psi \right)$, $x_m=\frac{{\delta }_m\sqrt{k^2+{\left(c{\omega }_f\right)}^2}}{\sqrt{{\left(k-m{\omega }_f^2\right)}^2+{\left(c{\omega }_f\right)}^2}}$, $\tan \varphi =\frac{c{\omega }_f}{k-m{\omega }_f^2}$, and $\tan \psi =\frac{c{\omega }_f}{k}$.