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USAPhO 2009 F=MA exam #16, (angular frequency of a spring)

  1. Jan 27, 2014 #1
    http://www.aapt.org/physicsteam/2010/upload/2009_F-maSolutions.pdf

    1. The problem statement, all variables and given/known data

    Two identical objects of mass m are placed at either end of a spring of spring constant k and the whole system is placed on a horizontal frictionless surface. At what angular frequency ω does the system oscillate?

    2. Relevant equations

    f = (1/2∏) √(m/k)

    Some equation involving angular frequency and springs would help here...I cannot recall anything

    3. The attempt at a solution

    This problem seemed to be more of a reasoning problem than a mathematical one. I tried manipulating the frequency formula. However, the answer doesn't even have a 2∏ term in it! This boggled me.

    The answer is √(2k/m)
     
    Last edited: Jan 27, 2014
  2. jcsd
  3. Jan 27, 2014 #2

    Doc Al

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    Staff: Mentor

    They ask for angular frequency ω, which is not the same as f. Hint: ω is measured in radians/sec.
     
  4. Jan 27, 2014 #3
    Okay, I've done some googling and wikipedaing, and angular frequency is the magnitude of angular velocity (in general). For springs, and I would never have guessed this and I don't know how this was even made up, angular frequency ω = √(k/m). Wha...? How is that even related to the first definition of angular frequency?

    Okay, that takes me a step closer. I understand why there are no pi terms. But it is still a reasoning problem. Why, when two masses are put on the two ends, does k get doubled??? √(k/2m) would've been my most educated guess.

    Btw I've edited the original question to include the link to the problem in case you want to see all the possible answers.
     
  5. Jan 27, 2014 #4

    Doc Al

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    Frequency is cycles per second. Each 'cycle' has 2π radians, so ω = 2πf. (Not related to angular velocity.)

    Hint: The center of the spring is stationary, thus you can consider each mass as oscillating at the end of half a spring.
     
  6. Jan 27, 2014 #5
    So does it just turn into a system of two springs, each with spring constant k and mass m?

    If it does, adding up their frequencies would get me 2√(k/m)

    But I have a feeling I don't add up the individual frequencies to get the total frequency...
     
  7. Jan 27, 2014 #6

    Doc Al

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    No. If you cut a spring in half what happens to its spring constant?

    No, you don't add them! ("Total" frequency doesn't make sense.) The frequency of each mass is the frequency.
     
  8. Jan 27, 2014 #7
    Thank you for your patience with me, Doc Al!

    Yes, total frequency doesn't make sense, does it...

    And, why, believe it or not, the spring constant doubles! So √(2k/m).

    Now this question is resolved, but my knowledge of springs has been proven to be lacking. May I ask one question, just for the fun of it?

    What if this system was changed so one side had a mass m and the other side had a mass 2m? Or 3m, 4m, etc.? Would it evolve into complex, ugly oscillations, or is there a simple, elegant answer for the new angular frequency?
     
  9. Jan 28, 2014 #8

    Doc Al

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    Hint: Consider the center of mass.
     
  10. Jan 28, 2014 #9
    The new spring constant k becomes (m2 / m1 + m2)-1k + (m1 / m1 + m2)-1k I believe.
     
  11. Jan 28, 2014 #10

    Doc Al

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    Not exactly. Each segment of spring, measured from the center of mass, will have its own spring constant. While the spring constants are different, the frequency of oscillation, which depends on k/m, is the same.
     
  12. Jan 28, 2014 #11
    Say one end is 3m and the other is m. The center of mass will be 3/4 down the spring length, close to the big mass. The spring constant on the big mass's side becomes 4k because the spring there is cut into a fourth. The spring constant on the other side is 4/3 k because the spring is 3/4 the original length. Adding these up, you get (16/3)k. The angular frequency is √(16k / 12m) = √(4k / 3m)

    If I did the same process with a spring with 4m on one side and m on the other, the angular frequency would be
    √(5k / 4m).

    Am I correct? I see the pattern.
     
  13. Jan 28, 2014 #12

    Doc Al

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    So far, so good.

    Why in the world did you add them up?

    Instead, treat each mass as if it were on its own spring. The ω for each mass is given by √(ki/mi).
     
  14. Jan 28, 2014 #13
    Ohhhh I see. I thought in the first example we did (√(2k / m) we got the result by adding up the spring constants.

    So the ω for each mass in the 3m&m example is √(4k / 3m). Still. After isolating both sides. Perhaps, then, what I did (adding up the total spring constant and the total mass) was okay in this case?
     
  15. Jan 29, 2014 #14

    Doc Al

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    No, the result was gotten by analyzing each mass separately. Remember that you are asked to find the angular frequency at which the masses oscillate.

    While it is interesting that [itex](k_1+k_2)/(m_1+m_2) = k_1/m_1 = k_2/m_2[/itex], I fail to see the physical justification for adding the spring constants and masses in this manner.
     
  16. Nov 6, 2014 #15
    I think the expression for the angular frequency in all problems of this sort is ω = √( (M + m) * k / (M*m) )
     
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