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Simple Harmonic Motion and the Reference Circle

  1. Jan 24, 2009 #1
    1. The problem statement, all variables and given/known data

    A loudspeaker diaphragm is producing a sound for 2.5 s by moving back and forth in simple harmonic motion. The angular frequency of the motion is 7.54 x 10^4 rad/s. How many times does the diaphragm move back and forth?



    2. Relevant equations

    angular freq. (W for omega) = change in theta/ change in T or W = 2 (pi) (f)

    W = 2pi/ T



    3. The attempt at a solution

    Ok, first of all, I need some help clarifying exactly what I'm looking for. Is it the number of times it goes around the circle? So, every 2 pi would be one time?? If so, I think I need to use W = 2 (pi) f, but I have frequency and W, so how do I manipulate this so I can "count" each time it goes around in a circle? (Sorry for the naive-ness of this. This topic confuses the heck out of me)

    Thanks in advance :)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 24, 2009 #2

    LowlyPion

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    Homework Helper

    If you thought of moving "back and forth" as 1 complete revolution, and I asked you how many revolutions it would make in 2.5 seconds, I doubt you would have a problem.
     
  4. Jan 24, 2009 #3
    So, would it make 1 revolution (2 pi) in 2.5 seconds? I'm really confused I guess- what I just said doesn't make sense, becuase if it did, then the answer would just be 1, which didn't take the ang. frequency into account.

    What about this:

    W = 2 pi/T so 7.54 x 10^4 rad/sec (2.5 sec) = 2 pi

    so, seconds cancel and Im left with radians....

    Perhaps that doesn't work either, becase I'm confused as to what it really means.

    I know there are 2 pi radians in one circle. And- one circle = 1 back and forth motion of the diaphragm....
     
  5. Jan 24, 2009 #4
    Think of a spring oscillating back and forth. One cycle of the spring can be from its fully compressed state to its fully stretched state and back to its fully compressed state. This is similar to a ball traveling in a circular path. Place the circle's center at the center of an x-y coordinate system. Allow the ball to move around the circular at a constant velocity with a light shining from above and casting a shadow of the ball on the x axis. The shadow will move back and forth in simple harmonic motion as the ball moves around the circle. So, the ball will rotate 2pi radians and take T seconds to do this for one cycle. So, the angular frequency is

    [tex]\omega=\frac{2\pi}{T}\frac{radians}{sec}[/tex]

    where T is the period or time for one cycle. Frequency and period are related by

    [tex]f=\frac{1}{T}\frac{cycles}{sec}[/tex]

    so

    [tex]\omega=2\pi \mbox{f}[/tex]

    Radians and cycles are unitless.
     
  6. Jan 24, 2009 #5
    Ok, so when I plug numbers into W = 2pi (f) - or do it the way I stated in my previous post (same thing) I end up with this numerical value: 1.89 x 10^5 rad = 2 pi rad Is this correct? Can you please help me understand what this means? Do I need to divide through on the left by 2 pi? If so, what does that mean?
     
  7. Jan 24, 2009 #6
    You are given the angular frequency and the time. The product of these two, which is what you did, gives the total angular rotation. You are on the right track. You have radians on the right side of the equation and 2 pi radian/cycle on the left side. Divide both sides by 2 pi radians per cycle and you answer will be the total number of cycles. The total number of cycles is the total number of times the speaker goes back and forth.
     
  8. Jan 24, 2009 #7
    Ok one last question- how exactly do the units work themselves out?

    This is what I have:

    1.89 x 10^5 rad = 2 pi rad/cycle

    To isolate cycle don't I have to multiply by 2 pi on each side? Even then I would be left with 1/cycle? You said to divide by 2 pi rad/ cycle, but then what's left on the right? Nothing- so it would be zero??
     
  9. Jan 24, 2009 #8

    LowlyPion

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    Homework Helper

    Make it easy on yourself.

    It's going at 75,400 radians per second.

    How many radians then in 2.5 sec?

    Divide that by 2π and that's the number of revolutions isn't it?
     
  10. Jan 24, 2009 #9
    Ok- I guess that makes sense. However, in other contexts I would probably still be confused with the cycles, revolutions, etc... I just can't wrap my head around it- silly I know :shy: I have some more problems to practice with.

    Thank you both, chrisk and LP- :smile:
     
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