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MIT open course question. Harmonic oscilators

  1. Jul 12, 2012 #1
    Hello everybody. Here is a ( I hope) simple question that cant let me sleep. The question is : At the course of physics one of the greatest profesors. Dr. Lewin drew a simple equation that 10Tm(2)= √m(2)/M(1) times 10Tm(1) . It was a harmonic oscilator composed by two object placed on a frictionless surface (device working as a hovecraft). This is whole note form this course

    ( Tm(2) - means period of mass 2, not times mass 2 )

    m(1) = 186 +/- 1g

    A=20 cm 10 T = 15.16 +/- 0.1 sec ( observed while experiments work)
    40 cm 10T= 15.13 +/- 0.1 s (same)

    THen

    m(2) = 372 +/- 1g

    10Tm(2)=√m(2)/m(1) 10Tm(2)

    Please help ":))))
     
  2. jcsd
  3. Jul 12, 2012 #2
    what is the question?
     
  4. Jul 12, 2012 #3
    If you are asking how this formula is deduced, I will try to write a way down, if someone knows a faster way to explain -right now I cannot think properly...haha

    [itex]\frac {(T_{1})^{2}}{m_{1}}=\frac {(T_{2})^{2}}{m_{2}}[/itex]

    but T=2π/ω

    [itex]\frac {(2π/ω_{1})^{2}}{m_{1}}=\frac {(2π/ω_{2})^{2}}{m_{2}}[/itex]


    [itex]\frac {1}{m_{1} ω_{1}^{2}}=\frac {1}{m_{2} ω_{2}^{2}}[/itex]


    [itex]m_{1} ω_{1}^{2}=m_{2} ω_{2}^{2}[/itex]

    Now if you know that this is k, you are fine. Otherwise you can write [itex]ω=\sqrt{\frac {k}{m}}[/itex]

    [itex]m_{1} \frac {k_{1}}{m_{1}}=m_{2} \frac {k_{2}}{m_{2}}[/itex]

    [itex]k_{1}=k_{2}[/itex]

    So if your oscilator is the same, the above equation holds right....
     
  5. Jul 12, 2012 #4
    Ok but How This can be if ( as Lewin said ) The period of an oscilation is independent of his mass. Here, because its equation it's impossible because to make


    [itex]\frac {(T_{1})^{2}}{m_{1}}=\frac {(T_{2})^{2}}{m_{2}}[/itex]
    equal we have to increase Period if... we have increased mass. Sory, maybe i see it wrong way but... unfortunately... I dont get it.
     
  6. Jul 12, 2012 #5

    haruspex

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    Can you explain a bit more? What is the set-up - how are the objects connected? What is "10T"? Ten periods??
     
  7. Jul 15, 2012 #6
  8. Jul 15, 2012 #7

    vanhees71

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    You can derive this from the equation of motion of the mass tied to the springs, which we assume to be ideal ones obeying Hook's linear law, i.e., the force is proportional to the elongation out of equilibrium. Then you get the equation of motion
    [tex]m \ddot{x}=-D x.[/tex]
    It is easy to see that the general solution of this equation is given by
    [tex]x(t)=A \cos(\omega t)+B \sin(\omega t) \quad \text{with} \quad \omega=\sqrt{\frac{D}{m}}.[/tex]
    Thus, the motion is periodical with a period of
    [tex]T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{D}}.[/tex]
    Now in Lewin's Lecture (which I think are extremely good, as far as I can judge from watching several minutes around the point you gave), he just has doubled the mass and left the springs the same, i.e., [itex]D[/itex] is kept the same as before. Then the previous formula tells you that the period of the motion is proprotional to the square root of the mass, i.e., the new period is
    [tex]T_{\text{new}}=\sqrt {\frac{m_{\text{new}}}{m_{\text{old}}}} T_{\text{old}}.[/tex]
    Then he also gave an error estimation from the errors of the quantities going into this equation and then did the experiment. That's how physics works :-)).
     
  9. Jul 15, 2012 #8
    You claim that Mr. Lewin contradicted himself. However, you didn't specify what he said clearly. What precisely did Mr. Lewin say?
    Maybe you meant to write.
    The period of an oscillator is independent of its mass.
    Then what is the "its"? A period doesn't have a mass.
    The critical part of this problem is what Lewin actually said in this sentence. The equations are straightforward. If there is a contradiction, then it caused by the words that Lewin said. Something Lewin said meant something different from what you thought he meant. Maybe you left a word out of the sentence (like scale). Or maybe in meant something different from what Lewin thought he meant (maybe he thought "scale" but didn't say it).
    Here are three possibilities:
    1) Maybe Lewin was talking about a different type of oscillator. The period of a pendulum is independent of the mass of the bob, because of the principle of equivalence.
    2) Maybe Lewin said that the period of the oscillator is independent of the units of mass. In other words, it doesn't matter if the mass is described in kilograms, grams or slugs. The period will be the same.
    3) Equivalently, the period of an oscillator is independent of the mass scale. The word scale refers to a generalization of the word units.
    4) Maybe Lewin said that the ratio between periods was independent of mass.
     
  10. Jul 15, 2012 #9
    I ve seen a spring with 15.5 kg ball that had a period the same as this ball woth Mr. Lewin on it. which means mass has no matter. My english may be not as good as rest of my languges but I merely remarquing upon the fakt that I see that T1/m1 is not the same as T2/m2... maybe this formula is not a formula of this experiment.. thats why im asking If I have had make a mistake sory . I didnt want it
     
  11. Jul 15, 2012 #10

    TSny

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    At what time of the video does Lewin make the statement that the period is independent of the mass? Is he still talking about a mass on a spring or is he talking about a pendulum?

    Mass on spring: T= 2[itex]\pi[/itex][itex]\sqrt{m/k}[/itex] (period depends on mass)

    Simple Pendulum (small oscillations): T = 2[itex]\pi[/itex][itex]\sqrt{L/g}[/itex] (period does not depend on mass)
     
  12. Jul 15, 2012 #11
    about pendulum. and I cant understand what in this second case when as vanhees said he ve doubled mass . its a device 2 objects on frictionless surface. woth springs on both sides. and he just moved it.. and wrote this formula
     
  13. Jul 15, 2012 #12
    Maybe the OP could accept that as an answer. When Lewin says that the period is independent of the mass, he is only talking about the simple pendulum.
     
  14. Jul 16, 2012 #13

    vanhees71

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    Now I'm confused. I watched the video around minute 24, and there was no pendulum but a little wagon on an airtrack fixed with two springs. Of course, within the linear regime of elongations, this leads to harmonic motion with frequency [itex]\omega=\sqrt{D/m}[/itex], where [itex]D[/itex] is the spring constant in Hook's Law and [itex]m[/itex] is the mass of the oscillating body. See also my previous posting. Of course, then the period is
    [tex]T=\frac{2 \pi}{\omega}=\frac{1}{2 \pi} \sqrt{\frac{m}{D}},[/tex]
    i.e., the period is proportional to the square root of [itex]m[/itex] as long as the linear regime of the elongation-force law is not left. All this is nicely demonstrated in the video.
     
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