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Consider a string of arbitrary length. One end is fixed and the other is in your hand. Bring the end to some point away from the equilibrium (creating a transverse wave). Perform the same action but 'faster' relative to the last wave. The same exact procedure is being performed but one wave has a shorter wavelength than the other.

Say $\text{A}$ is the normal trial and $\text{B}$ is the faster trial. I would like to know specifically what it is that causes this change in wavelength. My guess is: The wave speed of the medium is fixed so that in $\text{B}$ you are supplying the entire wavelength in a 'shorter' amount of time relative to the wave speed than in $\text{A}$. If this is the right way to think about it then what would be the cause of the wave speed here? (I am looking for a complete description)

edit: Actually I think my guess was wrong. I don't believe $\text{A}$ and $\text{B}$ would be travelling at the same speed with different wavelengths. I think that $\text{B}$ would be faster. (mentioned in the comments) This is because $\text{B}$ supplies a greater initial velocity to the wave. My new guess is that the dependence of the wavelength is to the ratio of [the individual wave speeds] to [the rate at which you're moving the string with respect to the wave speed]. $\text{B}$ then has a smaller ratio than in $\text{A}$ so that $\text{A}$ has a longer wavelength.

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  • $\begingroup$ Usually for a string, $v = \sqrt{\frac{T}{\mu}}$ where $T$ is the tension, $\mu$ is mass per unit length. $\endgroup$ – philip_0008 Oct 3 '16 at 18:06
  • $\begingroup$ I know this. I also know in general that $v = \sqrt{\frac{\text{elastic property}}{\text{inertial property}}}$. So, does this mean that in $\text{B}$ the tension is greater and thus the wave speed is as well? But, what causes the tension to be greater? $\endgroup$ – obliv Oct 3 '16 at 18:21
  • $\begingroup$ Just going to answer for myself: I think the tension is greater because of a greater initial energy given to the wave. $\endgroup$ – obliv Oct 3 '16 at 18:45
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as philip_0008 said, the wave speed in a string is usually given by $v=\sqrt{\frac{T}{\mu}}$ where $T$ is the tension and $\mu$ is the mass per unit length. This is the speed of the disturbance along the string. As soon as you displace one end of the sting that disturbance will begin to propagate along the string at that rate regardless of how quickly you are moving your hand.

As you say in your original question, the wave length of the wave produced depends on the time it takes your hand to complete one whole vibration. It just follows the wave equation $v=f\lambda$, or, if you like, $v=\frac{\lambda}{T}$. Since $v$ is fixed and determined by properties of the medium shorter $T$ gives shorter $\lambda$.

The derivation of the speed equation above (which you'll find in most first year texts, or here on wikipedia), assumes the x-component of the tension (along the string) is constant. The y-component, transverse to the string, of tension is not constant. It's zero at the extreme positions of the wave and at a maximum at the points where the string is at the equilibrium position.

If you want to generate a bigger wave with you hand you apply a larger tension which accelerates the string more in the transverse direction. So moving you hand faster does make the string move faster transverse to its length but it does not make the disturbance move any faster along the string.

Now, if the amplitude of the wave on the sting is large, then the approximation that the horizontal component of the tension is constant does not hold and the equation above for the speed of the wave is not completely accurate. Also, if the rope is not perfectly flexible but rather has appreciable stiffness then the frequency of the wave will have an effect on its speed.

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  • $\begingroup$ I will accept this if you explain why the disturbance will propagate at a fixed rate regardless of how fast you move your hand. Won't the tension be greater in the string the faster you displace it? $\endgroup$ – obliv Oct 3 '16 at 19:24
  • $\begingroup$ I have added some more detail on the tension in the string. I hope it helps. $\endgroup$ – M. Enns Oct 3 '16 at 20:00
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My thought is, assuming you use the same amplitude i.e. distance of 'swinging back and forth' for both trials:
As you make the frequency faster, given that the string has fixed total length, you shake it with the same amplitude, and the distance of your hand from the other end is the same, since more string length is needed to accommodate the same amplitude but shorter wavelength, the string will react, tending to compress the string, producing more tension.

string wave

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  • $\begingroup$ can you include a description of why the linear mass density $\mu$ is necessarily a factor in this wave speed as well? $\endgroup$ – obliv Oct 3 '16 at 20:09
  • $\begingroup$ The mass density acts as the inertial property of the string, preventing the string from going back to equilibrium too fast after being perturbed, or quickly following the direction of transverse forces. It has the effect of slowing down the wave when $\mu$ is greater, making it faster if $\mu$ is lesser. $\endgroup$ – philip_0008 Oct 3 '16 at 20:24

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