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Amplitude of an undamped driven harmonic oscillator

  1. Sep 17, 2016 #1
    1. The problem statement, all variables and given/known data

    An automobile with a mass of 1000 kg, including passengers, settles 1.0 cm closer to the road for every 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 5.0 cm, and 20 cm, respectively. The distance between the front and back wheels is 2.4 m. Find the amplitude of oscillation of the automobile, assuming it moves vertically as an undamped driven harmonic oscillator. Neglect the mass of the wheels and springs and assume that the wheels are always in contact with the road.

    2. Relevant equations

    ## \frac {d^2x} {dt^2} + 2\beta \omega_0 \frac {dx} {dt} + \omega_0^2x = F(t) ## (Driven Harmonic Oscillation) EQ1

    ## \frac {d^2x} {dt^2} + 2\beta \omega_0 \frac {dx} {dt} + \omega_0^2x = \frac 1 m F_0 sin(\omega t)## (Sinusoidal Driving Force) EQ2

    ## \beta = 0 ## (when undamped) EQ3

    ## F_{net} = ma = \sum F_x + \sum F_y ## (Newton's Second Law) EQ4

    3. The attempt at a solution

    My main issue in this problem is first deriving the equation of motion, and expressing its general solution as well. The solution gives the answer for equation of motion as $$ m \frac {d^2 y} {dt^2} = -k(y-asin(\omega t))$$ and gives the general solution as $$ y(t) = Bcos(\omega_0 t +\beta) + \frac {a \omega_0^2} {\omega_0^2 - \omega^2} sin(\omega t)$$

    My initial thought process is, okay, so this is a Driven Harmonic Oscillator with a Sinusoidal Driving Force, yet it is undamped. So I use equation 2, with the condition of equation 3. $$ \frac {d^2x} {dt^2} + \omega_0^2x = \frac 1 m F_0 sin(\omega t) $$

    I have absolutely no idea what to do when I get here, besides it being a 2 dimensional differential equation that I don't know how to solve. So I tried using Newton's Second Law, and the forces are 0 within the x direction.

    $$ F_{net} = m \frac {d^2x} {dt^2} = -mg - ky $$

    I really feel as if this isn't right at all either. I just really need a direction to go in, and get that initial equation of motion.
     
  2. jcsd
  3. Sep 17, 2016 #2

    Simon Bridge

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    Since this is a physics homework thing, you have probably derived the required relations in class or you have them in your notes so you do not need to derive everything all over again. The structure of the problem leads me to suspect this is what you are expected to do - it's a plug and chug.

    Note:
    Your 2nd law relation is does not work for the overall equation of motion because you have failed to include the driving force. You will need it, though, to figure out some of the values you need to plug into the correct solutions. If you had it right you'd get the same as what you had from eq2, as ##\omega_0## is related to ##k## ...and the force of gravity just shifts the equilibrium point, playing no role in the oscillations (revise for mass on spring with no driving or damping).

    The DE you have is second-order non-homogeneous.
    Solve that the same way as for any such equation.
     
  4. Sep 18, 2016 #3
    I've done some thinking on the problem, and these solutions the book provides just do not seem correct.

    A is equal to ## \frac {F_0} {m} ## in the following equations.

    For one thing, what we've derived in lecture amounts to the full damped driving force equation $$ e^{-\beta t}(A_1e^{\sqrt{\beta^2-\omega_0^2}t}+A_2e^{-\sqrt{\beta^2-\omega_0^2}t}) $$, and without dampening simplifies to $$ A_1e^{\omega_0^2t}+A_2e^{-\omega_0^2t} $$.

    The particular solution simplifies from $$ \frac {A} {\sqrt{\omega_0^2-\omega^2-4\beta^2\omega^2}}cos(\omega t - \delta)$$ to $$ \frac {A} {\omega_0^2 -\omega^2} cos(\omega t - \delta) $$

    Thus making the general solution: $$ y_g = \frac {A} {\omega_0^2 -\omega^2} cos(\omega t - \delta) + A_1e^{\omega_0^2t}+A_2e^{-\omega_0^2t} $$

    The complementary solution doesn't seem to match, and it seems as if the numerator of the particular solution is either wrong, or little a has some canceling factor with ##\omega_0^2## in it, but it can't because the expression they gave for the equation of motion would then be incorrect.

    Some how it's supposed to be ## F = my'' = -k(y+asin(\omega t)) ##, where I would assume that little a is supposed to be ## \frac {F_0} {m} ##, so if you rewrite m to try to get k in the equation you get ## m = \frac {k} {\omega_0^2} ##, and the equation of motion would become ## F = my'' = -ky + \frac {\omega_0^2} {k} F_0 sin(\omega t) ## ... Pulling out a -k like they appear to do creates ## F = my'' = -k(y - \frac {\omega_0^2} {k^2} F_0 sin(\omega t)##.

    In order to make this jive with the solution they have you would have to make ## a = - \frac {\omega_0^2} {k^2} F_0 ## which would simplify it to ## F = my'' = -k(y-asin(\omega t)) ## ... the problem here is if you substitute this in for the a in the particular solution that I gave in my original post, it doesn't simplify to the solution it SHOULD be that I proposed in the post.

    Please let me know if you see my logic or not. I'm struggling here
     
  5. Sep 18, 2016 #4

    Simon Bridge

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    OK - you seem confused. Let's make sure us two are on the same page:
    I don't know what "full damped" means in this context. The usual derivation will include a damping coefficient like this:
    ##\ddot x +2b\omega_0\dot x + \omega_0^2x = \frac{1}{m}F_0\sin\omega t## ... where "b" is the damping coefficient.
    If there is no damping, what is the damping coefficient equal to?

    Also see:
    https://en.wikipedia.org/wiki/Harmonic_oscillator#Sinusoidal_driving_force
     
    Last edited: Sep 18, 2016
  6. Sep 18, 2016 #5
    The problem states "assuming it moves vertically as an undamped driven harmonic oscillator", so I just figured the damping coefficient would be 0. I also watched an MIT driven oscillator video, and it referred to an undamped driven harmonic oscillator as of the form ## \ddot{x} + \omega_0^2x = \frac {1} {m} F_0 cos(\omega t) ##, and is how my book refers to it as well.
     
  7. Sep 18, 2016 #6
    Okay, let me rephrase my question...

    In the end for MY derived particular solution, I get $$ y_p = \frac {A} {\omega_0^2-\omega} cos(\omega t) $$ The ## \frac {A} {\omega_0^2 - \omega^2} ## should be the amplitude in that case, where ## A = .05 [m] ## according to the question, and it should also equal ## A = \frac {F_0} {m} ## according to my book.

    However, the units on this amplitude are incorrect. It comes out to units of ## \frac {m} {1/s^2} ##. Since amplitude is just in meters, I am missing some factor of units ## 1 / s^2 ## being multiplied by the A, which will cancel out, and leave meters.

    The solution proposed by the book implies my particular solution should be ## y_p = \frac {A \omega_0^2} {\omega_0^2-\omega} cos(\omega t)## This seems correct because it has the correct units.

    Question: What could I be doing wrong, that I am not getting the ## \omega_0^2 ## in the numerator of MY derived particular solution?
     
  8. Sep 18, 2016 #7

    Simon Bridge

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    I cannot tell what you have done there ... have you used "A" for two different things?
    ##A/(\omega_0^2-\omega^2)## has dimensions of length if ##A=F_0/m##.

    When I do it, I propose ##y(t)=A\sin\omega t## as a solution to ##\ddot y + \omega_0^2y = \frac{F_0}{m}\sin\omega t##
    ... and I get ##A=F_0/m(\omega_0^2-\omega^2)##. (Notice this is a different use for A.)

    You will have to look carefully through the book to see where they get that extra ##\omega_0^2## in the numerator.
    Maybe they are using "A" to mean something different?
     
  9. Sep 18, 2016 #8
    This is what is confusing me the most. My A is defined as ## A = \frac {F_0} {m} ##, it's not the amplitude. So I did get the same thing as you for my actual amplitude I just wrote it as ## \frac {A} {\omega_0^2-\omega^2} ##, but this has wrong units as far as I've checked. My issue is the solution says it should be ## \frac {A\omega_0^2} {\omega_0^2-\omega^2}##, which comes to the units I want.
     
  10. Sep 18, 2016 #9

    Simon Bridge

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    OK ... the claim is that ##F_0/m(\omega_0^2-\omega^2)## has the wrong units. Lets do the dimensional analysys:

    ##[F_0/m(\omega_0^2-\omega^2)] = [F_0]([m][\omega_0^2-\omega^2])^{-1}##
    Breaking that down:
    ##\qquad [F_0] = MLT^{-2}##
    ##\qquad [m] = M##
    ##\qquad [\omega_0^2-\omega^2] = ( [ \theta ] [t]^{-1} )^2 = T^{-2}##

    Put it all together and you have: ##[F_0/m(\omega_0^2-\omega^2)] = MLT^{-2}(MT^{-2})^{-1} = L##
    ... which is the required dimensions.

    In terms of SI units: ##[F_0/m(\omega_0^2-\omega^2)] = \text{kg.m.s}^{-2}(\text{kg.s}^{-2})^{-1} = m## ...
     
  11. Sep 18, 2016 #10
    Okay, so then I've been interpreting one of the values incorrectly.

    Using the conditions in the problem I've found ## k = 98,000 [\frac {N} {m}] ##, ## \omega_0 = \sqrt{\frac{k}{m}} = 9.9 [\frac {rad} s]##, and ## \omega = \frac {2\pi v} {\lambda} = 175 [\frac {rad} s] ##. The problem says the amplitude of the roads sine curve is ## 5.0 [cm] ##.

    In my solution ## \frac {A} {\omega_0^2-\omega^2} = \frac {F_0} {m(\omega_0^2 - \omega^2)} ##, I was under the impression that A was the ## 5.0 [cm] ## within the problem statement. Is it actually ## F_0 ## ? I thought ##F_0## was supposed to be a force, not an amplitude? I feel like this is definitely where I'm making an error.
     
  12. Sep 18, 2016 #11
    I think I've figured it out actually, so I figured I needed to find out what ## F_0 ## was then, because originally I thought it was just an amplitude, or something vague. I figured it would be the spring force of the road, which is ## kA_0 ## (A_0 will be the roads amplitude of 5 [cm]), then divided by ## m ##, we get ## \frac k m A_0 = \omega_0^2 A_0 ## Which explains how the solutions manual had the ##\omega_0^2## within their particular solution for the differential equation. This then gives me the correct answer.

    Thank you for your time, help, and patience, I was really struggling with this concept!
     
  13. Sep 18, 2016 #12
  14. Sep 18, 2016 #13

    Simon Bridge

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    No worries.
     
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