Damped oscillation of a car on a road: velocity calculation

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Homework Help Overview

The discussion revolves around the damped oscillation of a car on a sinusoidal road profile, focusing on calculating the velocity of the car when vertical oscillations reach their maximum amplitude. The problem involves parameters such as the mass of the car, spring constant, damping constant, and the wavelength of the road profile.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the car's oscillation and the sinusoidal road profile, questioning the interpretation of displacement and the significance of wavelength. There is discussion on the nature of the oscillator, whether it is driven or undriven, and how to determine the driving frequency.

Discussion Status

Participants are actively engaging with the problem, raising questions about the definitions and assumptions made in the original post. Some have suggested that the original poster reconsider the relationship between velocity, wavelength, and frequency, while others have pointed out the need to clarify the distinction between different parameters involved in the calculations.

Contextual Notes

There is confusion regarding the interpretation of horizontal versus vertical displacement, as well as the number of wheels and their contribution to the overall mass. Participants emphasize the importance of checking dimensions in calculations to avoid errors.

Granger
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Homework Statement


The car circulates on a section of road whose profile can be approximated by a sinusoidal curve with the wavelength of 5.0 m. The mass of the car is 600.0 kg, and each wheel is equipped with a constant spring
k = 5000 Nm-1 and a damper with constant b = 450 Nm-1s.
Calculate the velocity of the car when the vertical oscillations have biggest amplitude.

Homework Equations



Equation of a damped oscillator
$$ x(t) = A e^{-\lambda t}\cos(\sqrt{\omega_0^2-\lambda^2}t +\phi)$$

where $$ \lambda=\frac{b}{2m} $$ and $$ \omega=\sqrt{\frac{k}{m}}$$

$$T=\frac{2\pi}{\omega_0}$$

The Attempt at a Solution



First thing I thought was that, since this is a damped oscillator, then the amplitude must be maximum at t=0.

Then since they give us the wavelength we know that x(0) = 0 and x(T) = 5.0 m

Substituting in the damped oscillator equation we get to:

$$ 0 = A\cos(\phi)$$
$$ 5= A e^{-\lambda T}\cos(\sqrt{\omega_0^2-\lambda^2}T - \phi)$$

Calculating all the known constants and solving this system of 2 equations we get to

$$\lambda=0.375$$
$$\omega_0 = 2.887$$
$$T=2.176$$
$$\phi=\pi/2$$
$$A=208.357$$

Now using the equation for velocity, by differentiating the equation of the damped oscillator:

$$ v(t) = -\lambdaA e^{-\lambda t}\cos(\sqrt{\omega_0^2-\lambda^2}t - \phi) - A \sqrt{\omega_0^2-\lambda^2} e^{-\lambda t}\cos(\sqrt{\omega_0^2-\lambda^2}t - \phi)$$

Substituting the know values we get to $$v=-596.43$$

The answer should be $$v=4.3$$

Can someone help me understand what am I doing wrong?

Thanks!
 
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Why x(T) = 5m? Is x the horizontal distance or the vertical displacement?
 
Do you really think the whole car is going up and down with an amplitude of 5 m ?
 
mjc123 said:
Why x(T) = 5m? Is x the horizontal distance or the vertical displacement?

Oh you're both right confused both things (horizontal and vertical displacement). Then the way I'm approaching the problem doesn't work because I don't know the value of x(T)... However the fact that the wavelength is given must serve to apply a condition but I'm not seeing what...
 
You have a driven damped harmonic oscillator. If you drive at 10 m/s your driving force has a frequency of 2 Hz, for example.
 
BvU said:
You have a driven damped harmonic oscillator. If you drive at 10 m/s your driving force has a frequency of 2 Hz, for example.

Oh so we use the relation between velocity, wavelength and frequency is $$v=\lambda f$$.
But how to choose f?
If I just use the angular frequency (dividing it by 2pi) I obtain 2.3 and not 4.3.
 
Granger said:
Oh so we use the relation between velocity, wavelength and frequency is $$v=\lambda f$$.
But how to choose f?
If I just use the angular frequency (dividing it by 2pi) I obtain 2.3 and not 4.3.
You calculated λ as 0.375 s-1, but:
Granger said:
mass of the car is 600.0 kg
Granger said:
each wheel is equipped with...
How many cars? How many wheels?
 
Granger said:
But how to choose f?
If I just use the angular frequency (dividing it by 2pi) I obtain 2.3 and not 4.3.
Haru helps you out with a big correction.
You don't choose f, you determine it -- that is the core of the exercise. The driving force has a frequency v/(5 m) and your job is to find the frequency that gives the greatest amplitude.
Be sure to post calculations (in terms of formulas with symbols, stepwise if useful), not just a numerical outcome.
 
haruspex said:
You calculated λ as 0.375 s-1, but:How many cars? How many wheels?
´

BvU said:
Haru helps you out with a big correction.
You don't choose f, you determine it -- that is the core of the exercise. The driving force has a frequency v/(5 m) and your job is to find the frequency that gives the greatest amplitude.
Be sure to post calculations (in terms of formulas with symbols, stepwise if useful), not just a numerical outcome.

How I get it now! I didn't understand this was a driven oscillator was a forced oscillator. Then the maximum amplitude or ressonance of amplitude.
Also I need to treat just one wheel of the car (which supports a quarter of its weight -> a quarter of its mass),

Then
$$\omega_0=\sqrt{\frac{k}{m}}=\sqrt{\frac{5000}{600/4}}=5.774$$
$$\lambda=\frac{b}{2m}=\frac{450}{2\times 600/4}=1.5$$
$$\omega_f=\sqrt{\omega_0^2 -2 \lambda ^2}=\sqrt{(5.774)^2 -2 (1.5) ^2}=5.37$$

Now the velocity:
$$v=\lambda f = 1.5 \times \frac{\omega_f}{2\pi}=1.3$$

Still not the correct result...
 
  • #10
There is a distinction between the ##\lambda = 1.5##/s that follows from ##b\over 2m## and the ##\lambda = 5## m to convert speed to driving force frequency.
I agree with the ##\omega_f## value you found, but ...

[edit] -- sorry, the 'but...' reservation was overhasty. You do have the right expression with the 2##\lambda^2##.
 
Last edited:
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  • #11
Right I made confusion between both lambdas. Thanks!
 
  • #12
I want to make use of the opportunity to stress that you ALWAYS want to check dimensions in your expressions. It can save you from errors big time !
 
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