# Damped oscillation of a car on a road: velocity calculation

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1. Jun 28, 2017

### Granger

1. The problem statement, all variables and given/known data
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.

2. Relevant 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 $$\omega=\sqrt{\frac{k}{m}}$$

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

3. 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!

2. Jun 28, 2017

### mjc123

Why x(T) = 5m? Is x the horizontal distance or the vertical displacement?

3. Jun 28, 2017

### BvU

Do you really think the whole car is going up and down with an amplitude of 5 m ?

4. Jun 28, 2017

### Granger

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...

5. Jun 28, 2017

### BvU

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.

6. Jun 28, 2017

### Granger

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.

7. Jun 29, 2017

### haruspex

You calculated λ as 0.375 s-1, but:
How many cars? How many wheels?

8. Jun 29, 2017

### BvU

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.

9. Jun 29, 2017

### Granger

´

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. Jun 29, 2017

### BvU

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 ...

 -- sorry, the 'but...' reservation was overhasty. You do have the right expression with the 2$\lambda^2$.

Last edited: Jun 29, 2017
11. Jun 29, 2017

### Granger

Right I made confusion between both lambdas. Thanks!

12. Jun 29, 2017

### BvU

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 !