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Quick SHM Question

  1. Apr 29, 2016 #1
    << Mentor Note -- thread moved from the technical forums, so no HH Template is shown >>

    Hello,

    So I've been doing old practice questions on SHM to revise, and just been frustrating myself on this one for a bit. The question is:

    "An object vibrating with simple harmonic motion has a maximum speed of 1.6 m s−1 and maximum magnitude of the acceleration of 8π m s−2. Calculate the period of the motion. "

    I can't remember why, but I've divided the speed by 2*pi and then divided that by the acceleration [(1.6*2*pi) / 8*pi], and it's given me the correct answer of 0.40s but I can't quite understand why that's correct! Lol

    Cheers,
    Niall
     
    Last edited by a moderator: Apr 29, 2016
  2. jcsd
  3. Apr 29, 2016 #2

    gneill

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    Staff: Mentor

    What relevant equations do you know that pertain to SHM? In particular, what equation is often used to write the position as a function of time?
     
  4. Apr 29, 2016 #3
    Do you mean: x(t) = Acos(ωt + φ)?
     
  5. Apr 29, 2016 #4

    gneill

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    Sure. That'll work. Now, given that function for displacement, how might you find velocity?
     
  6. Apr 29, 2016 #5
    By differentiating, giving: v(t) = Aωsin(ωt + φ), and acceleration would therefore be: a(t) = -Aω2cos(ωt + φ)
     
  7. Apr 29, 2016 #6

    gneill

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    Good, good. So you know how to move from displacement to velocity to acceleration through differentiating the function.

    Since you're given a value for maximum speed, say V = 1.6 m/s, then your velocity function might be v(t) = V sin(ωt + φ), right? Can you work from there?
     
  8. Apr 29, 2016 #7
    I got to that point previously, but just wasn't sure what to do next. I feel like it should be obvious, but it just hasn't clicked
     
  9. Apr 29, 2016 #8

    gneill

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    Differentiate the velocity expression...
     
  10. Apr 29, 2016 #9
    Ok, dv(t) / dt = -Vωcos(ωt + φ)
     
  11. Apr 29, 2016 #10

    gneill

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    What's the maximum value of the result?
     
  12. Apr 29, 2016 #11
    I would think that is when the phase is equal to zero, giving cos(ωt + φ) = 1
     
  13. Apr 29, 2016 #12

    gneill

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    Sure. And what is the maximum value? What multiplies the cosine function?
     
  14. Apr 29, 2016 #13
    dv(t)/dt = -Vω
     
  15. Apr 29, 2016 #14

    gneill

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    I should have caught this earlier: the derivative of sin is cos, not -cos. Not that it'll affect the eventual result, but I thought I should point it out.
    The magnitude of the value that multiplies the trig function is the important thing: Since sin and cos both range between +1 and -1, its the magnitude of the value multiplying the sin or cos that sets the maximum value of the function. In this case it's the maximum value of the acceleration, right? So don't write dv(t)/dt = -Vω since you've dropped the trig function and it's no longer a function of time. It's the maximum value. Instead, write A = Vω, where A is the maximum acceleration...
     
  16. Apr 29, 2016 #15
    Oh of course! Bloody hell. Yeah, thank you for going through that and having patience. I should have got this a hell of a lot easier, but it feels like one of those nights lol, thanks very much! Appreciate it.
     
  17. Apr 29, 2016 #16

    gneill

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    No worries. Glad to have helped.
     
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