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Simple Harmonic Motion question.

  1. Dec 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Two simple pendulums of length 0.40 m and 0.60 m are set off oscillating in step. Calculate (a) after what further time the two pendulums will once again be in step, (b) the number of oscillations made by each pendulum during this time. (Assume g = 10 ms-2


    2. Relevant equations

    T = 2[itex]\pi\sqrt{l/g}[/itex]

    3. The attempt at a solution

    I get the two different values for T, but I'm stuck after that. I've got the solution with me but I need someone to explain getting the answer to me step by step I suppose. If anyone needs the solution I'd be happy to type it all out. Thanks!
     
  2. jcsd
  3. Dec 27, 2011 #2

    Pengwuino

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    Yes, what is your actual attempt at the solution? We can show you the next step with it.
     
  4. Dec 27, 2011 #3
    No see MY attempt at the solution was finding T first, and after that I'm stuck. What I said above is that I've got the solution written with me in the textbook and that if you need the whole solution to tell me what's happening then I'll gladly type it all out or scan it in a while and post it here and maybe then you can help me. However it suits you :)
     
  5. Dec 29, 2011 #4

    BruceW

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    what does it mean by 'set off oscillating in step'? Are these coupled or separate pendulums? If they are separate, then it seems like a sort of trick question...
     
  6. Dec 29, 2011 #5

    Curious3141

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    Sorry, can't do the step by step, it's against board rules. But I'll give you a big hint:

    The number of cycles an oscillator has gone through after time [itex]t[/itex] can be represented by [itex]\frac{t}{T}[/itex], where [itex]T[/itex] is the period. If this number is an integer, it means it's undergone that many complete oscillations within that time. If the number is not an integer, it means that it is in the "middle" of an oscillation at the time.

    To be in sync, two oscillators which have been synchronised at the start need to have a cycle number difference which is a whole number. The smallest integer is zero (this occurs at the start). The next integer is one.

    Can you work out the solution now?
     
  7. Dec 29, 2011 #6
    [Sorry just read the rules again, didn't read 'em that well before I suppose. But I understand why you can't help me step by step and that's fine since it's the rules.]

    Secondly, YES! See I've got the solution and everything but I didn't understand why we needed to take the second oscillation as (n-1), I understand the rest of it. Was just stuck at this part. Thank you SO much!
     
  8. Dec 30, 2011 #7

    BruceW

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    what answer did you arrive at?
     
  9. Dec 31, 2011 #8
    For (a), I got 6.86 s. And for (b), the shorter pendulum makes 5.5 oscillations and the longer pendulum makes 4.5 oscillations :)
     
  10. Dec 31, 2011 #9

    BruceW

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    And this is the same as the answer given in the textbook? Strictly speaking, the two are not exactly in step again. But they are close to being in step.
     
  11. Dec 31, 2011 #10
    Yep, 'cause I gotta round 'em off to two significant figures.
     
  12. Jan 3, 2012 #11

    BruceW

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    But what did you round off to two significant figures? the number of oscillations? So the number of oscillations shouldn't be 5.5 and 4.5 ? But then they would not be in sync.

    I guess I'm trying to argue with the textbook's answer. And you've probably moved on from this question a long time ago. So don't worry about it.
     
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