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Homework Help: Help with right or wrong question in simple harmonic motion

  1. Apr 13, 2009 #1
    I have this right or wrong question in physics that i know the answer to but not the way to solve it can you help me with it?:

    2 identical springs were put Vertically. if one was pulled 5cm (spring1) and the other was pulled 10cm(spring2) and were left to move a simple harmonic motion would the periodic time for (spring2)be double the periodic time for (spring1).

    The answer is wrong i think but i have to mathematically prove it so my teacher don't think i am doing it randomly
    i think its wrong because according to the law for the periodic time
    T=[tex]2\pi\ [/tex] x [tex]\frac{\sqrt{m}}{\sqrt{k}}[/tex]
    and k =f/x and since x is in the root it even if we double it we wont get double T or something like that?
  2. jcsd
  3. Apr 13, 2009 #2
    look ,the equation for the periodic time You wrote is correct.
    It's stated that the springs are identical! means that the K is the same, K is constant for a spring, no matter how much you stretch it(edit-until you overstretch it,so hooke's law doesn't work).
    although k=f*x and X is doubled, It only means that the f will be twice as less, the f is not constant!!! this is why the statement is wrong, They'll have the same periodic time
    hope You understood,I'm kinda new to S.H.O.
    good luck.
  4. Apr 13, 2009 #3
    so your telling me if we have 2 springs with the same constant and we pull one 1million meter and the other 1 centimeters they will have the same periodic time???
    edit: oh and isn't hook's law f= -k * x
    Last edited: Apr 13, 2009
  5. Apr 13, 2009 #4
    Well,Lets stay in proportion.
    You'll kill the spring if you pull it 2 much, but lets say you pull 1 spring 1 centimeter, and the other one 20 centimeters(and they made by the same manufacturer,by the same engineer and the same metal and the same bluh bluh bluh) yes, their periodic time will be the same,that's what I think, when you pull the spring more, it has more Elastic force( cause F=k*x and x is bigger now) so when it reaches the point where x is 1 centimeter(where the other spring started) it has a speed V and the force that acts in this position is like the forces acted on the other spring at first, but it has a speed, so it's not so "NOT OBVIOUS" when You think about it... but maybe I'm wrong(although I'm 99% tight this time).
    Hope You understood.
  6. Apr 13, 2009 #5
    but in simple harmonic motion when the force increase so will the displacement right?
    The restoring force must be proportional to the displacement so if the force increase so will the displacement and the amplitude right? so if we have bigger amplitude wont we need more time for one oscillation?
  7. Apr 13, 2009 #6
    "so if we have bigger amplitude wont we need more time for one oscillation?"
    I don't think so,if it would be like you stated, where is the A(amplitude) in the formula?
    the amplitude is the most far point from the equilibrium if I'm not mistaken,It doesn't affect the T.
    let's forget about springs,take a pendulum,It's periodic time is not dependent on the Amplitude..
  8. Apr 13, 2009 #7
    ok then assuming you are right how can i write all of that mathematically i mean the question also says if you were to put right or wrong you must give a mathematical reason
  9. Apr 13, 2009 #8
    Assuming I'm right:
    T1-T of spring one
    T2-T of spring two
    K=Both springs' constant.

    T1=2pi *sqrt(m/k)
    T2= 2pi*sqrt(m/k)

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