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Resonance Frequency

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A certain spring elongates 9:0mm when it is suspended vertically and a block of mass M is hung on it. If driven by a force of variable frequency, at what frequency would resonance be obtained?


    a. is 0:088 rad/s
    b. is 200 rad/s
    c. is 33 rad/s
    d. is 1140 rad/s
    e. cannot be computed unless the value of M is given


    2. Relevant equations


    3. The attempt at a solution
    I really don't know how to attempt this question.
     
  2. jcsd
  3. Feb 18, 2015 #2

    Nathanael

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    If you have a driving force at various frequencies, about which frequency will give you resonance?
     
  4. Feb 18, 2015 #3
    Isn't it when the frequency is one of the natural frequencies?
     
  5. Feb 18, 2015 #4

    Nathanael

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    Right. And what is the natural frequency of this system?
     
  6. Feb 18, 2015 #5
    Can't it be anything?
     
  7. Feb 18, 2015 #6

    Nathanael

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    When you pull the mass down slightly and let it go, allowing the system to oscillate (with no driving force) at what frequency will it oscillate? This is the natural frequency of the system. Are you saying you can get it to oscillate freely at any frequency you want?
     
  8. Feb 18, 2015 #7
    No it can't oscillate freely at any frequency but I don't know what frequency this would oscillate it in, sorry. I want to say the fundamental frequency but that is just a guess, no reasoning. Can you please explain it to me?
     
  9. Feb 18, 2015 #8

    Nathanael

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    "The fundamental frequency" doesn't apply. (That has to do with standing waves in which there are multiple frequencies that cause a standing wave.)

    Consider a function which describes the displacement of the block as a function of time. Do you know what this function might look like? (Use arbitrary constants for the unknowns.)
     
  10. Feb 18, 2015 #9
    It would be a block in SHM, characterized by y=Acos(wt+Φ), where A is 9.0 mm?
     
  11. Feb 18, 2015 #10

    Nathanael

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    Right now we're finding what the frequency would be if we pull the mass down and let go (the natural frequency). The amplitude would be however far we pull it down. It could be 9mm or anything else; it's irrelevant (because frequency is independent of amplitude).

    Your equation is correct though. We can ignore the phase constant Φ. The phase constant Φ just shifts the "time zero"... clearly this will not effect the frequency.

    Now, differentiate the function twice to get the acceleration function. Consider the time when the acceleration is maximum; what acceleration does this produce? What will be the displacement at this time? What acceleration does hooke's law produce? What must ω be in order for the accelerations to agree? Then, what is the natural frequency of system?
     
  12. Feb 18, 2015 #11
    When deriving it, I get a= -ω^2Acos(ωt). Its a maximum when t=0. Won't the displacement at this time just be A? Acceleration by hooke's law is -kx/m =a. So ω must be the square root of the spring constant over the mass. I am sorry but I am not getting it:S
     
  13. Feb 18, 2015 #12

    Nathanael

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    What do you mean you're not getting it... [itex]ω=\sqrt{\frac{k}{m}}[/itex]... that's exactly right! :smile:

    So then, what is the natural frequency of the object (in terms of k). And then can you find k from the information in the problem? (Find k in terms of the mass of the object.)
     
  14. Feb 18, 2015 #13

    Nathanael

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    Some people may say it's worth memorizing ω=√(k/m) ... I say it's only worth memorizing what you unintentionally memorize.

    At any rate, if you ever forget that the angular frequency ω of a mass on a spring is ω=√(k/m), then the way I showed you would be the way to derive it. Namely, differentiate the displacement function twice with respect to time, and then compare that to hookes law and solve for ω.

    It's not usually worth memorizing any results, but it can sometimes be useful to 'memorize' how to derive the result.
     
  15. Feb 20, 2015 #14
    So I'm not too sure if I understant this still but this is where I am at. According to hooke's law. -kx=ma ===> k=ma/x (1), where x is 9 mm. Acceleration is equal to
    ω^2A (2). Plugging in equation 1 into ω=√ k/m i get ω=√[(ma/x)/m], which simplifies to ω=√(a/x).Substituting Equation 2 into the equation results in 1= ω=√(A/x). But I don't understand how to go beyond this point. Also, why are we only considering maximum acceleration?
     
  16. Feb 20, 2015 #15

    Nathanael

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    If x is 9 mm then what is "a"? It can't be the acceleration of the block because the spring constant is... constant.

    Once ω is known, how can you find the frequency? (Find the cycles per second; not the radians per second.)

    We only did that to derive ω2=k/m. We didn't have to choose the time when acceleration is maximum, any time would have worked.
    This is how it would've gone for an arbitrary time:

    Suppose the position at a particular time is ... xp=A*cos(ωT+Φ) ... where A is the position amplitude, and T is some point in time.
    The acceleration at this particular time will be ... ap=-(ω2)*A*cos(ωT+Φ) ... (differentiating twice just brings out a factor of -ω2)
    We can see that ap=-ω2xp
    Hooke's law says map=-kxp which means that ω2=k/m

    So when I said, "consider the time when the acceleration is maximum," I was just choosing a particular time T.
    I could've chose any time, so I just chose the time such that cos(ωT+Φ)=1
     
  17. Feb 20, 2015 #16
    But what is ω equal to, how do find it when we don't know k?
     
  18. Feb 20, 2015 #17

    Nathanael

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    That's the next part of the problem: we need to find k.



    Look at this part of the question:
    Can you find k from that?
     
  19. Feb 20, 2015 #18
    OH! I was circling around the same thing. Instead of doing mg=kx,i kept on doing ma=kx. Thank you for bearing through this with me! I finally get it!
     
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