Damped Oscillatory Motion with Varying Bump Timing for Control

[Adrian Simons]

Homework Statement

The following problem comes from the Testbank associated with the textbook by Paul A. Tipler & Gene Mosca, "Physics for Scientists and Engineers":

When driving over a washboard speed bumps, certain speeds make the bumps less “bumpy.” Suppose your car has mass of 1000 kg and the critical damping constant of the shock absorbers is 12000 kg/s, what is the “ideal speed” to drive across the bumps assuming that the bumps are 0.3 m apart? Hint: Find the decay time for the critically damped shock absorbers.

Answers:

a) 1.2 m/s b) 2.4 m/s c) 3.6 m/s d) 4.8 m/s e) 6 m/s

Correct Answer: b

The figure for the problem is shown below.

Relevant Equations

The textbook defines the decay time as m/b, where m is mass and b is the damping coefficient. Thus, the decay time is the time it takes for the energy of the oscillation to fall by a factor of 1/e.

First of all, the problem is not clearly defined as they don't specify if the given mass is the total mass of the car, or just the sprung mass of the car, which is really what's relevant. In any case, with the limited information given, it seems like one is forced to make the assumption that the sprung mass per wheel is 250 kg, and the damping constant for each shock absorber is 12,000 kg/s.

However, from here, I'm not sure what to do. In reality, this should be a critically damped system (so that b = 2mω₀, where ω₀ is the natural angular frequency of the system), with a forcing function whose frequency is given by 1/T, where T is the time it takes to drive over a bump, so that the answer to the problem is v = L/T, where L = 0.3m. But it's unclear what this time should be. If I choose a speed that corresponds to the natural frequency, that doesn't give any of the answers listed. Besides, it seems like that would drive the system into resonance rather than damping out the vibrations.

I also don't understand the hint given for the problem. For a critically damped system, we have b = 2mω₀, so the decay time is 2ω₀, which doesn't seem to directly relate to anything of significance regarding the problem at hand.

Can anyone elucidate?

 

[anuttarasammyak]

With no knowledge of the system we can get a quantity with dimension of L/T from the quantities mentioned in the problem as
$$\frac{12000}{1000}*0.3=3.6 \ m/s$$
You say the correct answer is 2.4 m/s so the factor 2/3 should be applied to the above. I am curious how this factor comes.

 

[haruspex]

b = 2mω0, where ω0 is the natural angular frequency of the system

Yes, but to find the ideal speed you need the damped frequency.
http://spiff.rit.edu/classes/phys312/workshops/w5b/eqn_l.gif

 

[Adrian Simons]

Yes, but to find the ideal speed you need the damped frequency.
http://spiff.rit.edu/classes/phys312/workshops/w5b/eqn_l.gif

We know the "damped frequency" by the fact that the system is critically damped, in which case, ω=0.

With no knowledge of the system we can get a quantity with dimension of L/T from the quantities mentioned in the problem as
$$\frac{12000}{1000}*0.3=3.6 \ m/s$$
You say the correct answer is 2.4 m/s so the factor 2/3 should be applied to the above. I am curious how this factor comes.

I don't know that diddling around to arrive at some value that has the correct units is a valid method to solve a physics problem. It may be a guide that your answer isn't invalid, but it doesn't reveal the underlying physics. And then even so, there's that factor of 2/3 to account for.

 

[haruspex]

We know the "damped frequency" by the fact that the system is critically damped, in which case, ω=0.

I'm questioning whether critical damping as derived from the natural frequency is the ideal here. This is a forced damped system.

 

[hutchphd]

As I recall, a critically damped system driven at "natural" resonance will be phase shifted by $\pi/2$. It is too far into the evening to work magnitude out. I surely don't see the 2/3 factor.

 

[Adrian Simons]

I'm questioning whether critical damping as derived from the natural frequency is the ideal here. This is a forced damped system.

Shock absorbers in an automobile are typically critically damped according to the natural frequency of the suspension springs in the automobile suspension. So while I agree that the bumps in the road play the part of a driving force, I don't think that changes the situation.

 

[haruspex]

Shock absorbers in an automobile are typically critically damped according to the natural frequency of the suspension springs in the automobile suspension. So while I agree that the bumps in the road play the part of a driving force, I don't think that changes the situation.

I disagree.
Look at 23.6.6 in https://phys.libretexts.org/Bookshe...rmonic_Motion/23.06:_Forced_Damped_Oscillator.
Given that the suspension is critically damped, the denominator reduces to $\omega_0^2+\omega^2$. To minimise the amplitude you should drive as fast as possible!

And that is for the case where the forcing is fixed as a sine wave of given amplitude. In the washboard road example you won't descend as far into the dip if you go faster.
This conforms to my own experience cycling over cattle grids. By going fast, you don't fall far between bumps, so the bumps are effectively smaller.

That said, amplitude might not be the only consideration. The more you succeed in leaping from peak to peak, the less the fraction of time spent touching the ground, so the greater the normal force at each peak.

 

[anuttarasammyak]

You say the correct answer is 2.4 m/s so the factor 2/3 should be applied to the above. I am curious how this factor comes.

This is my guess. We get 1/12 second =T for resonance. For anti-resonance we should make bump to bump time be $\frac{2n+1}{2}*\frac{1}{12}$ second where n is non negative integer, so speed should be $\frac{2}{2n+1}*3.6$ m/s = 7.2, 2.4, 1.44, ... .

 

[haruspex]

This is my guess. We get 1/12 second =T for resonance. For anti-resonance we should make bump to bump time be $\frac{2n+1}{2}*\frac{1}{12}$ second where n is non negative integer, so speed should be $\frac{2}{2n+1}*3.6$ m/s = 7.2, 2.4, 1.44, ... .

Promising approach, but it leads to full washboard amplitude as the speed tends to zero. The suspension isn't doing anything there.

 

[Adrian Simons]

I disagree.
Look at 23.6.6 in https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Book:_Classical_Mechanics_(Dourmashkin)/23:_Simple_Harmonic_Motion/23.06:_Forced_Damped_Oscillator.
Given that the suspension is critically damped, the denominator reduces to $\omega_0^2+\omega^2$. To minimise the amplitude you should drive as fast as possible!

And that is for the case where the forcing is fixed as a sine wave of given amplitude. In the washboard road example you won't descend as far into the dip if you go faster.
This conforms to my own experience cycling over cattle grids. By going fast, you don't fall far between bumps, so the bumps are effectively smaller.

That said, amplitude might not be the only consideration. The more you succeed in leaping from peak to peak, the less the fraction of time spent touching the ground, so the greater the normal force at each peak.

Yes, I got the same result. I fully agree with your analysis that if you only consider the steady-state solution, going as fast as possible should minimize the amplitude of the vibrations. As I said in my original post, driving at a velocity that would shift the driving force towards the natural frequency of the system would drive it into resonance, so you want to be as far away from that condition as possible. However, v=∞ isn't one of the answers listed, either. As far as not falling far into the bumps, the actual situation is obviously more complex than what's included in the problem. But in general, physics problems at this level are just idealizations. I doubt anything along those lines was intended as part of the solution. Thanks for your contribution, though. It confirms my thoughts.

 

[anuttarasammyak]

The suspension isn't doing anything there.

May we interpret proper period $T=\frac{\omega}{2\pi}$ of the mass-suspension spring system is 1/12 second thinking that amplitude damping does not affect $\omega$ ?

 

[haruspex]

May we interpret proper period $T=\frac{\omega}{2\pi}$ of the mass-suspension spring system is 1/12 second thinking that amplitude damping does not affect $\omega$ ?

Not sure what you are asking.
My point is that if you go arbitrarily slowly over the bumps then oscillation of the suspension ceases. Any transients die away. If we pretend the tyres have zero radius, the car will rise and fall the full height of the washboard bumps. And if we don't make that pretence, the standard forced-damped model does not really apply.

 

[anuttarasammyak]

Not sure what you are asking.

A bump push up the wheel. The spring starts vibrating from the top position. After time T the spring comes back to the top position ( with or without amplitude damping during time T ). At the same time next bump hits and push up the wheel. The spring starts vibrating from the higher position and so on. This is my scenario of resonance.

A bump push up the wheel. The spring starts vibrating from the top position. After time T/2 the spring comes to its lowest position. At the same time next bump hits and push up the wheel. The spring starts vibrating from the middle height position and so on. This is my scenario of anti-resonance.

Such a control of bump timing would control enhancement/reduction of the oscillation.
We can extend these T and T/2 to longer times $nT$ and $(n+1/2)T$ expecting that amplitude damping mechanism does not alter $\omega=\frac{2\pi}{T}$ or damping mechanism is already incorporated in constant $\omega$.

I agree your point saying in my words that the larger n the calmer spring oscillates for all resonance , anti-resonance and the other timing cases.