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What is the equation of wave velocity for an extended spring?

  1. Oct 4, 2013 #1
    For a free spring, the wave velocity is v = √(k / m) x L, where L is the displacement of one end right?

    And what if I have a spring which both of its end is extended and held by "oscillator" to produce standing wave. In this case how would I find the wave velocity? I see some sources state that the equation i mentioned above is applicable. But wasn't L is the displacement? Or I did get the wrong idea about L ?
     
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  3. Oct 4, 2013 #2
    I don't know if this i correct way to approach your problem, but we can calculate the potential energy of the spring from the spring constant and the displacement (L) from [itex]PE=\frac {kL^{2}}{2}[/itex]

    We then can then solve the equation [itex]PE=mv^2[/itex] for the velocity. I think the math is correct but I'm not sure it answers your question.

    EDIT: I'm thinking of displacement as the difference in the length of the relaxed spring and the length of the extended or compressed spring.
     
    Last edited: Oct 4, 2013
  4. Oct 4, 2013 #3

    AlephZero

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    L is probably the length of the spring.

    You also need to check exactly what "m" is here. Is it the mass per unit length, or the mass of the whole spring?

    If you are talking about axial waves in the spring, the wave velocity is independent of the tension in the spring, unless you are considering very large axial strains (i.e. stretching a floppy spring by a large amount). But to model that effect, you would need to consider the details of how the spring was made, e.g. a coil spring would be different from a flexible elastic band.

    If you are talking about transverse waves, the tension in the spring does change the wave velocity - but the formula in your OP doesn't make much sense for transverse waves.
     
  5. Oct 4, 2013 #4
    I defined L as the difference between the relaxed spring and the extended or compressed spring for calculating potential energy (following the OP's example). For a relaxed spring, PE=0. However, the convention may be to use "x" for displacement and L for the length of the relaxed spring. You can extract "velocity" from k, x and m, but I'm not sure how velocity, so calculated, applies in this case.
     
    Last edited: Oct 4, 2013
  6. Oct 4, 2013 #5

    AlephZero

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    OK, but you calculated the potential energy of the spring when it is in equilibrium (and stretched). For the vibration frequency you want the change PE for a small displacement from the equilibrium position.

    Reading your post again, I think you are assuming the OP means a light spring with a mass on the end. I was assuming a heavy spring with no additional mass.

    But either way, it's not very clear to me what the OP's formula is meant to be. The frequency is √(k/m) rad/sec, with the appropriate definitions of k and m for the physical situation. But to turn that into a wave velocity for a spring of length L, shouldn't there be a 2π somewhere in the formula? And if we assume L is something to do with the amplitude of the wave, it shouldn't be in the formula at all.

    Unless "wave velocity" means the velocity of one point on the spring, and not the velocity of the travelling waves in it .... :confused::confused::confused:
     
  7. Oct 5, 2013 #6
    v = √(k / m) x L, k is the spring constant , m is the mass of the spring and L is the length of the spring. This formula looks like this one:

    v = (+/-)√(k / m) √(A2-x2), where A is the amplitude and x is the displacement from equilibrium point of an oscillating object at the end of a spring.

    How to relate this to the wave velocity of an stretched + vibrating spring? And yeah this is not a heavy spring
     
  8. Oct 5, 2013 #7
    The second formula contains the variable 'x' (displacement) so it must apply to a stretched (or compressed) spring already if x is non zero. If the formula is correct and you are given k, x, the amplitude and mass, it seems you get a velocity from this formula.

    If you measure the tension T (in Newtons) on the spring directly you should be able to use this formula:

    [tex] v=\sqrt\frac{T}{m}[/tex]

    You are essentially treating the spring like any elastic "string" by knowing the tension. An oscillating mass at the end of the string will produce longitudinal waves and you would use the mean tension since it varies with the oscillations.
     
    Last edited: Oct 5, 2013
  9. Oct 5, 2013 #8
    so do u know what is the working principle behind this formula ? v = √(k / m) x L, i just can't link how the √(A2-x2) can become the stretched length of spring
     
  10. Oct 5, 2013 #9
    I am not sure if anyone has pointed this out:

    This is NOT the wave velocity. What you wrote is the velocity of the end of the spring.
    For a longitudinal wave in a spring the points of the spring may oscillate with velocities given by similar formulas, where L is replaced by displacement from equilibrium position. But this will not be the wave's velocity. They are quite different things.
     
  11. Oct 7, 2013 #10
    I would like an answer to this question myself. Consider the OP's example of an oscillating suspended weight at the end of a spring. What is the velocity of the waves produced in the spring?

    First, are "waves" produced in the spring from the oscillations? It seems the spring would just stretch and recoil by a fixed amount over the whole length of the spring. However, if the spring were compressed over a length 'x' < L, and this compressed segment were released, I could imagine a longitudinal wave propagating along a sufficiently long spring. It seems the easiest way to calculate the velocity of this wave is to use the tension ("T") formula:

    [tex] v=\sqrt {\frac{T}{\mu}}[/tex]

    Tension has the dimension of force and [itex]\mu[/itex] is the mass per unit length of the spring. Is this correct or is it at least correct regardless of how such a wave is generated?

    EDiT: The above formula can also be expressed in terms of the spring constant as T=kx.
     
    Last edited: Oct 7, 2013
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