Car driving on washboard road

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In summary, the amplitude of oscillation for a car with a mass of 1000 kg traveling at a constant horizontal speed of 20 km/h over a washboard road with sinusoidal bumps, with an amplitude and wavelength of 5.0 cm and 20 cm respectively, can be found by first calculating the spring constant, which is equal to 9.8*10^4 N/m. Then, the frequency of bumps can be determined by multiplying the car's speed by the frequency of bumps (1 bump/0.2 m), resulting in a frequency of 27.7 bumps/sec. Using Newton's second law and setting up a differential equation, the amplitude of oscillation can be found by solving for the steady-state
  • #1

Homework Statement

A car with mass 1000 kg, including passengers, settles 1.0 cm closer to the road for ever additional 100 kg of passengers. It is driven with a constant horizontal component of speed 20 km/h over a washboard road with sinusoidal bumps. The amplitude and wavelength of the sine curve are 5.0cm and 20cm, respectively. Find the amplitude of oscillation of the automobile, assuming it moves vertically as an undamped driven harmonic oscillator. Neglect the mass of wheels and springs and assume that the wheels are always in contact with the ground.

Homework Equations

Newton's second law

The Attempt at a Solution

Alright first thing I did was figure out the spring constant.
I know that kx = mg when the car has the 1000 kg, and that k(x-0.01) = (m+100)g. So use equation one and substitute into equation 2, such that k(x-0.01) = kx + 100g. Solve for k and you get that k = 9.8*10^4 Nm.

Next thing is to solve for the frequency of bumps.
I am looking for some function [tex] y(t) = cos(\omega t) [/tex], where there would be some sort of phase that I don't really care about since the amplitude is all that matters in the end. So [tex] \omega = 2 \pi f [/tex] where f is the frequency that the car travels over the bumps. The car travels at 5.5m/s (I converted), and so at 1bump/0.2m multiply it by the velocity and the frequency is 27.7 bumps/sec; hence, [tex] \omega = 27.7*2*\pi = 1.74*10^2 rad/s[/tex] and the forcing function is [tex]y(t) = 0.05cos(174t)[/tex].

At this point, I made the differential equation based on Newton's 2nd law.
[tex] \sum_i F_i = m\frac{d^2 y}{dt} [/tex]

Figure out which forces are in the picture (gravity, spring force, and the forcing frequency).

[tex]m \ddot{y} = -mg + ky + 0.05cos(174t) [/tex]

rewrite in terms of homogeneous and particular ODEs (and divide by m)

[tex]\ddot{y} - \omega_0^2y = -g + \frac{0.05cos(174t)}{m}[/tex] where [tex]\omega_0 = \sqrt{k/m}[/tex]

then we are concerned with just the steady-state (i.e. particular) solution, and guess that it will be [tex]y(t) = A + Bsin(174t) + Ccos(174t)[/tex], which gives

[tex] -174^2 Bsin(174t) - 174^2 C cos(174t) + \omega_0^2 A - \omega_0^2 Bsin(174t) - \omega_0^2 Ccos(174t) = 5*10^-5cos(174t) - g[/tex]

relate the linear and trig terms to show that
[tex] A = - \omega_0^2 / g[/tex]
[tex] B = 0 [/tex]
[tex] -174^2C - \omega_0^2 C = 5*10^-5[/tex]
such that
[tex]C = \frac{5*10^-5}{(-174^2 + \frac{9.8*10^4}{1000}})[/tex]

C will give the amplitude of oscillations, but it seems really small (it works out to -1.64*10^-9. I'm pretty sure I did something wrong, but I don't know what.
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  • #2
The units don't come out correctly for your final answer... you should get meters, but you get [tex]\frac{meters*sec^{2}}{kg}[/tex]
Your final answer needs to be multiplied by the spring constant k = 98000 [tex]\frac{N}{m}[/tex] and then you get -1.61x10^-4 meters.

Your problem is with the forcing function: you cannot plug in 5 cm into the forcing function because the forcing function looks like y(t)=Fcos(wt), where F is FORCE.

Therefore, you need to figure out what force, hence name, forcing function, is being applied to the washboard. The force in the forcing function is the spring constant from the suspension system times the amplitude of the washboard sine wave (5 cm, they give you that). Now you got a number in Newtons, hence you get force.

If you re-calculate this, you find out that the final answer is -1.61x10^-4 meters (the units come out very nicely), which is about 0.161 mm
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