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Damped oscillation of a car on a road: velocity calculation

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  1. Jun 28, 2017 #1
    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. jcsd
  3. Jun 28, 2017 #2
    Why x(T) = 5m? Is x the horizontal distance or the vertical displacement?
     
  4. Jun 28, 2017 #3

    BvU

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    Do you really think the whole car is going up and down with an amplitude of 5 m ?
     
  5. Jun 28, 2017 #4
    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...
     
  6. Jun 28, 2017 #5

    BvU

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    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.
     
  7. Jun 28, 2017 #6
    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.
     
  8. Jun 29, 2017 #7

    haruspex

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    You calculated λ as 0.375 s-1, but:
    How many cars? How many wheels?
     
  9. Jun 29, 2017 #8

    BvU

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    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.
     
  10. Jun 29, 2017 #9
    ´

    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...
     
  11. Jun 29, 2017 #10

    BvU

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    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: Jun 29, 2017
  12. Jun 29, 2017 #11
    Right I made confusion between both lambdas. Thanks!
     
  13. Jun 29, 2017 #12

    BvU

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    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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