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Helical Spring Question!

  1. Aug 27, 2006 #1
    hi i am struggling with this question as i cant work out how to approach it. I am unsure which steps i need to take to resolve it and any help would be much appricated!

    The question is

    A mass suspended from a helical spring is set vibrating and it is found that 125 complete oscillations takes 75s. If the ampliltude of the motion is 145mm determine the distance from the equilibrium position when the velocity reaches 50% of its maximum velocity:surprised

    really stuck cheers for any help in advance
  2. jcsd
  3. Aug 27, 2006 #2


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  4. Aug 27, 2006 #3
    i dont as the question i have been given is diffrent from all my notes usually you have mass or spring constnat (k) but i just really havent got a clue where to start
  5. Aug 27, 2006 #4


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    Well then I strongly suggest you hit the books.

    Using the principle of conservation of energy, can you write an equation which links the amplitude, the speed at a given instant, and the displacement at that particular instant?
  6. Aug 27, 2006 #5
    i cant see how that helps thou i am given the amplitude motion is 145mm but i cant see how i am meant to link it. Ive work out that the angular velocity is 10.47 so i used 10.47 0.0145 x cos 10.47 x 0.6 which is wACoswT which gives the velocity i got -0.052ms-1 i just catn get that intial step and work out what i need to find out
  7. Aug 27, 2006 #6


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    145 mm = .145 m
    Could you explain the equation you just used: Why does the cosine term have the period T in it?

    T is actually not needed at all to solve this problem.

    To make your equations clearer could you use LaTeX, or just stick with symbols/letters (no numbers, please)?
    Last edited: Aug 27, 2006
  8. Aug 27, 2006 #7
    well i thought that finding the displacement of the spring would help in find the displacement at 50% of its velocity. So i used x(t) = Xm Cos wT It is used to find the elongation (x) of the spring.. The formlua i posted a minute ago is wrong i just went threw it. I know that the time is takes to complete one cycle is 0.6 seconds which is 1.6 hz. which i used to find the angular velocity... thats as far as i have got now i appricate the help
  9. Aug 27, 2006 #8


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    Ok, the equation is still wrong. It should be:
    [tex]x(t) = A \sin (\omega t)[/tex]
    (You could replace the sine with cosine.) The important part is that there is no T (well actually yes there is, "inside" w), and t is the variable.

    You earlier took the derivative of this equation, so you ended up with the velocity:
    [tex]v(t) = \omega A \cos (\omega t)[/tex]

    First of all, can you determine the maximum speed?
    Now, can you determine when the mass has the speed half the maximum?
  10. Aug 27, 2006 #9
    ok so the equation for the velocity would be
    v = 10.47 x .145 x cos (10.47 x 0.6)?

    The amplitude i think is .145 and the angular velocity is 10.47 just cant work out what i am missing here. i think i can work out the speed at half by dividing the time by amplitude by 2?
  11. Aug 27, 2006 #10


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    Again, no.
    [tex]v = \omega A \cos (\omega t)[/tex]
    where t is a variable, and has nothing to do with T, the period.

    Thus [tex]v_{max} = \omega A[/tex]
    Can you figure out why this is true, and determine an instant the mass has maximum speed?

    Now, can you determine an instant t so that [tex]v(t) = \frac{v_{max}}{2}[/tex] ?

    No, you can't.
  12. Aug 27, 2006 #11
    ah i see now because T has no effect on the velocity at which it will fall an rise! thanks so i got Vmax = 10.47 x .145 = 1.51815ms-1 and at 50% of the maxiumum velocity t = 1.15815ms-1 / 2 = 0.759075ms-1. So to find the displacement at 50 % i need to use the velocity at 50%?
  13. Aug 27, 2006 #12


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    How did you get this? It's wrong.

    You get t from the equation:
    [tex]v(t) = \frac{v_{max}}{2}[/tex]

    [tex]\omega A \cos (\omega t) = \frac{\omega A}{2}[/tex]
  14. Aug 27, 2006 #13
    have i got the angular velocity and amplitude right? did i get the max velocity right then? sorry just cant quite how you find t or is t = wA/2 which would be 10.47*.145/2 ?
    Last edited: Aug 27, 2006
  15. Aug 27, 2006 #14


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    They should all be correct. You however don't need the value of angular velocity to solve this problem.

    You need to solve the instant t the mass has the given speed of half the maximum:

    [tex]v(t) = \frac{v_{max}}{2}[/tex]

    Now substitute for v(t) and vmax for the equation above:
    [tex]\omega A \cos (\omega t) = \frac{\omega A}{2}[/tex]

    Solving for t:

    [tex]t = \frac{\arccos(1/2)}{\omega}[/tex]

    Have you got any idea what to do next?
  16. Aug 27, 2006 #15
    i cant really read what that says does it say t = arccos(1/2)/w?

    i know we need to solve t but i just cant work out the fiqures to put in this is what is making me strugle from the begining i know w is 10.47 and thats it apart form what i explained earlier either i am being stupid or just missing something. I get the steps which you are kindly explaining but just cant work out where to get the fiqures from
  17. Aug 27, 2006 #16
    i got 4.29 for t=arccos(1/2)/w
  18. Aug 27, 2006 #17


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    It does, yes. You get it from the equation above it if you solve for t.

    Again, w is rather irrelevant, as its numerical value is not needed (it will cancel out).

    By figures, do you mean like numbers? I don't usually substitute them for the symbols until the very end: this makes the solution easier to follow.

    Don't think about the steps we've taken.
    We've solved an instant t the speed is half the maximum. We are asked for the displacement when the speed is half the maximum. Do you know what to do?
  19. Aug 27, 2006 #18
    so now we have t is t the half, or do we work out the half from t? then rearrange another equation to produce x which is the extrention of the spring?
  20. Aug 27, 2006 #19


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    We already have an equation x(t).
    The t we solved is the instant the speed is half the maximum. Thus, we substitute this t to the displacement equation x(t).

    I strongly suggest revising simple harmonic motion.

    Another way of solving the problem is through the principle of conservation of energy (which I suggested in the beginning). You might want to try if you can solve the problem that way.
  21. Aug 27, 2006 #20
    so all i need to do now is rearrange the equation again to find X which is the extention ?
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