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Speed of stationary wave in a string

  1. Dec 23, 2013 #1
    Dear all,

    In my text book it is written that when a string clamped at both ends oscillates in it's fundamental mode then the frequency of the stationary wave set up in the string is given by f=v/2l .where 'f' means frequency,'v' means speed of wave and 'l' is the length of string.following are my doubts
    1.since wave is stationary not travelling so its speed should be zero always?if it is not then what is the meaning of speed in a stationary wave and what is its formula?
    2.they used the formula v=square root of T/m.where 'T' is the tension in string and 'm' is the mass per unit length of the string.but as far as i know this formula is derived for a travelling wave not for stationary wave then why did they use it.please explain

  2. jcsd
  3. Dec 23, 2013 #2
    Think of a stationary wave as the sum of two travelling waves. If the two waves move in opposite directions and have the same frequency, the result is a stationary wave. The travelling waves have a well defined speed (or phase velocity).

    You're allowed to do this because of the principle of superposition.
  4. Dec 23, 2013 #3
    I am still not clear that why the speed of stationary wave is not zero.what i think that since energy is not travelling in stationary waves so its speed must be zero.
  5. Dec 23, 2013 #4
    Nobody's saying it isn't zero. But it can be broken down into two waves with equal speeds travelling in opposite directions, and those speeds can be used in the mathematical formula to get the correct result.
  6. Dec 23, 2013 #5


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    The pattern from two waves is what's stationary and not the waves themselves. It's just a name!!!!
  7. Dec 23, 2013 #6


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    There are two ways to look at this. One is to "do the math" and not bother too much what it means. The other is to "do the physics" and see what it means in reality.

    The math is straightforward. Start from the trig identities
    ##\sin(a-b) = \sin a\cos b - \cos a\sin b##
    ##\cos(a+b) = \sin a\cos b + \cos a\sin b##
    Adding them gives
    ##\sin(a-b) + \sin(a+b) = 2\sin a\cos b##
    Now put ##a = x## and ##b = vt##, and we have
    ##\sin(x-vt) + \sin(x+vt) = 2\sin x\cos vt##

    ##\sin(x-vt)## and ##\sin(x+vt)## are traveling waves going in opposite directions with speed ##v##.
    ##2\sin x\cos vt## is a stationary wave, like the vibration in a string of length ##\pi##.

    I don't think the physics gives much insight into what happens to the steady vibrations of a string, because "in real life" the traveling waves are reflected from the ends of the string. You might just as well set up the equations of motion and solve them. To satisfy the boundary conditions, this is an eigenvalue problem and the eigenvalues give you the vibration frequencies of the different modes, with 0, 1, 2, ... nodal positions along the length of the string.

    On the other hand there are situations where the traveling waves DO give a lot of insight. For example, think about vibrations of a circular ring of material. There are no "end points," so traveling waves can go around the ring "for ever" in either direction. This gets to be even more fun when the ring is rotating. Whether you think something is a stationary or a travelling wave depends whether you are looking at the rotating disk from a "fixed" point not on the disk, or whether you are on the disk rotating with it.....

    ... and depending on where you are looking from, the two "forwards" and "backwards" traveling waves may appear to have different speeds, and/or different frequencies ...

    ... and you can get patterns of motion that look similar to the vibration of a non-rotating ring, but the pattern "rotates" around the ring at a different angular speed from the ring's rotation.

    Now apply some forces, from ANOTHER object, that is rotating at a yet another different speed .....

    ... and keeping track of all this in terms of "traveling waves" becomes extremely useful. But you probably won't cover it in the course you are taking right now.

    (I didn't just make that up. It's what happens to a vibrating compressor or turbine disk, in a steam turbine or a jet engine).

    Maybe a better name, at least for the string, is "standing wave" not "stationary wave".
    Last edited: Dec 23, 2013
  8. Dec 23, 2013 #7


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    "Standing wave" is in fact the customary term in the US, at least in all the textbooks that I've used.
  9. Dec 24, 2013 #8


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    Standing wave was used in the UK too, then someone thought they could add 'clarity' to the situation- haha - and started using the term 'stationary'.
    Ask any Radio Engineer what SWR means.
  10. Dec 24, 2013 #9
    Imagine the first oscillation. The wave immediately starts traveling towards the end of the string, and no mistake, the speed is exactly what the formula gives you. When it comes to the end, it bounces back with the same speed. So basically standing wave which doesn't move is like illusion created by two waves moving in opposite direction
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