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Simple Harmonic Motion

  1. Jan 22, 2013 #1
    1. The problem statement, all variables and given/known data

    A mass is attached to a spring with a force constant of 32N/m.
    The spring and the mass are set into simple harmonic motion on a frictionless, horizontal surface.
    The period of vibration of this mass is 0.4 seconds.

    a)
    Calculate the object's mass
    b)
    Calculate the frequency of the oscillations in the simple harmonic motion


    2. Relevant equations

    T=period (seconds)
    f=frequency (Hz)
    m=mass (kg)
    k= force constant (N/m)

    T=2pi√(m/k)
    f=(1/2pi)√(k/m)


    3. The attempt at a solution


    a)
    k(0.4/2pi)2=m
    k(0.394784176)=m
    m = 12.63309363 kg

    two sig digits: m = 13


    b)
    f=1.570796327√(k/m)
    f=2.5 Hz


    Here's the thing:
    According to the answers in the book, part B is correct. However, they say that the answer to part A is [ m = 0.13kg ].

    This isn't the first time that the book's answer has differed from my own by 101. Its actually the third time. However, previously I just wrote it off as a misprint.
    Now, I'm not so sure. So I'm here to see if I'm doing anything wrong.
    The strange thing is that the answer to part B meshes with the book's answer. I find that odd because the answer to part B is dependant on the answer yielded from part A (mass value).

    Insights?
     
  2. jcsd
  3. Jan 22, 2013 #2
    Instead of calculating ##\frac{0.4}{2\pi}## as you should be, you are calculating ##\frac{0.4}{2}\pi##. This is why it's very important to use brackets correctly. It's probably safe to say this is (at least) the third time you've made this mistake, if it's the third time you've had this problem. Your answer to part b is coming out correctly because you are probably making the same mistake again and it's fortuitously cancelling out the first mistake.
     
    Last edited: Jan 22, 2013
  4. Jan 22, 2013 #3
    You made a very basic mistake. You wrote (0.4/2pi). which you then interpreted as (0.4/2)*(pi), while it should be (0.4/(2*pi)). The interpretation is actually correct, the original writing (0.4/2pi) is not, because you really divided (0.4) by (2pi).

    Then you made the same mistake when computing the frequency, which compensated for the earlier mistake, thus yielding the correct result.

    Use brackets to avoid such mistakes.
     
  5. Jan 22, 2013 #4
    Ahhhhhhhh. Thanks guys.

    I meant to have the pi in the denominator... but when I went to punch it into my calculator, I simply typed "0.4/2pi." I assumed that the multiplication algorithm in the calculator would multiply it by the denominator.

    Simple enough mistake, I guess.

    Thanks again!
     
  6. Jan 22, 2013 #5
    The calculator follows the common mathematical conventions. The convention is that the operations of division and multiplication have the same order of precedence, and are evaluated from left to right. 0.4/2pi means (0.4/2)pi = 0.2pi, you MUST use brackets if you want 0.4/(2pi).
     
  7. Jan 22, 2013 #6
    Depends on how fancy the calculator is. It's also possible the calculator just follows the (appropriately named) 'calculator order of operations', which means it just evaluates everything in the order you enter it. Either case yields the same (incorrect) answer for this computation, but it's something else to keep in mind. It's generally unwise to rely on BEDMAS (or PEMDAS or whatever other acronym one may have learned in elementary school) to ensure something is evaluated in the desired order. One should always use copious bracketing to remove any ambiguity.
     
  8. Jan 22, 2013 #7
    This is really beside my point. My point is that the expression was transformed incorrectly even before it was fed to the calculator. And the calculator did happen to use the common convention, so the result was incorrect.

    Bracketing must be used not just to satisfy the calculator, it must be used to keep the symbolic manipulations correct (and understandable by others).
     
  9. Jan 22, 2013 #8
    I'm making a broader point in the interest of helping the OP in the future.

    As I said, you have no way of knowing which convention the calculator used since both give the same answer for the given input.

    That is exactly what I said, so I have no idea why you apparently find my comment objectionable.
     
  10. Jan 22, 2013 #9
    It just seemed to me that you emphasized the calculator part of the story, while the error was in the symbolic manipulation.
     
  11. Jan 22, 2013 #10
    I think it was mainly a misunderstanding about how the calculator would interpret it since it was clear from the OP's reply that s/he understood the correct form. In any case, I don't think we actually disagree about anything so there's not much point in this.
     
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