A vehicle with mass 1,050 kg and suspension stiffness 435 kN/m is travelling with a velocity v...


A vehicle with mass 1,050 kg and suspension stiffness{eq}435\text{ kN/m} {/eq} is travelling with a velocity {eq}v {/eq} on a sinusoidal road surface with amplitude 1 cm and a wavelength of 5.3 m.

(a) Determine the critical speed of the vehicle.

(b) Three possible values of suspension damping are {eq}12.1\text{ kN}\cdot \text{s/m, 42}\text{.7 kN}\cdot \text{s/m, and 64}\text{.1 kN}\cdot \text{s/m} {/eq}.

Assuming that the desired speed of the vehicle ranges from 11 m/s (25 mph) to 22 m/s (50 mph), which is the best choice? Justify your answer.


(a) 17.2 m/s

(b) {eq}64.1\text{ kN}\cdot \text{s/m} {/eq}

Critical speed:

Critical speed is the speed at which a rotor shaft having a maximum vibration. If we running a shaft speed beyond the critical speed then the shaft can be destructive mechanical vibration.

Answer and Explanation:

We have given that,

Mass {eq}\left( m \right) = 1060{\rm{kg}} {/eq}

Stiffness {eq}\left( k \right)435{\rm{kN/m}} {/eq}

Amplitude {eq}\left( y \right){\rm{km}} {/eq}

Wavelength {eq}\left( L \right) = 5.3{\rm{m}} {/eq}


Let us consider that the vehicle moving on a sinusoidal road surface

As the vehicle moves along the surface, the mass will move up and down

Free body diagram is drawn below

{eq}\begin{align*} m\ddot x &= - k\left( {x - y} \right) \\ m\ddot x &= - kx + ky \\ m\ddot x + kx &= ky \\ \end{align*} {/eq}

Let {eq}y = Y\sin \left( {\omega t} \right) {/eq}

Here, {eq}\omega = \dfrac{{2\pi V}}{L} {/eq}


{eq}m\ddot x + kx = kY\sin \left( {\dfrac{{2\pi V}}{L}} \right) {/eq}

By solving the above equation, we get

{eq}k = \dfrac{Y}{{1 - \left( {\dfrac{\omega }{{\omega _n }}} \right)^2 }} {/eq}

Here, {eq}\omega _n = \sqrt {\dfrac{k}{m}} {/eq}


Compute the critical speed of the vehicle

{eq}\begin{align*} \dfrac{\omega }{{\omega _n }} &= 1 \\ \omega &= \omega _n \\ \dfrac{{2\pi V}}{L} &= \sqrt {\dfrac{k}{m}} \\ V &= \dfrac{{5.3}}{{2\pi }}\sqrt {\dfrac{{435 \times 10^3 }}{{1050}}} \\ V &= 17.2{\rm{m/sec}} \\ \end{align*} {/eq}

Hence, the required critical speed of the vehicle is {eq}17.2\;{\rm{m/sec}} {/eq}


For critical condition, the damping ratio {eq}\left( r \right) {/eq} is given by

{eq}r = \dfrac{c}{{C_e }} {/eq}

Here {eq}C_e = 2\sqrt {km} {/eq}

{eq}\begin{align*} c &= \left( 1 \right)C_e \\ c &= 2\sqrt {km} \\ c &= 2\sqrt {435 \times 10^3 \times 1050} \\ c &= 42743.42 \\ c &= 42.7\;\dfrac{{{\rm{kN}}{\rm{.s}}}}{{\rm{m}}} \\ \end{align*} {/eq}

From step-2 critical speed is {eq}17.2{\rm{m/sec}} {/eq}

And the possible speed is from {eq}11{\rm{m/sec}} {/eq} to {eq}22 m/sec {/eq} Which is greater than critical speed so, we have to use {eq}64.1\dfrac{{{\rm{kN}}{\rm{.s}}}}{{\rm{m}}} {/eq} damping to avoid the speed ranging between the critical speed and {eq}22{\rm{m/sec}}{\rm{.}} {/eq}

Hence, the required answer is {eq}64.1\dfrac{{{\rm{kN}}{\rm{.s}}}}{{\rm{m}}} {/eq}

Learn more about this topic:

Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9

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