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I am reading wave pulse on a string and their reflection/transmission, and my book says that frequency of the pulse depends only on the source of the wave pulse. But I am unable to understand why is it so. I know that $f=1/T$ but I am unable to relate that thing here. If someone can please explain why frequency in reflection and transmission does not change in a string to me, it will be really helpful.

Note: I found similar questions on this site but either they were really complicated or dealing with sound waves. But I think this question is different because I am talking about transverse wave here.

As pointed out in the comments this statement above is not always true. Some of the assumptions taken are:

  1. The tension in both the strings is same.
  2. The amplitude of the pulse is not very large, etc.
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    $\begingroup$ As far as this goes, the wave being transverse versus being longitudinal doesn't matter. If there was a satisfactory answer about why sound waves have a frequency determined by the source (which is only actually true if the wavelength is much much smaller than the size of the object supporting the wave as a medium), then that works here too. $\endgroup$ – march Jan 31 '16 at 4:50
  • $\begingroup$ You do not give a link for the context of the statement. I can not see how it could generally hold. Maybe this will help hyperphysics.phy-astr.gsu.edu/hbase/waves/wavsol.html $\endgroup$ – anna v Jan 31 '16 at 5:33
  • $\begingroup$ @annav I think that you got a bit confused in what I was asking in the question... Please point out which part of question is poorly stated...I just wanted to ask why the frequency of the wave pulse does not change when it goes from one medium to the other that is from one string to another one where both the strings are different and connected to ensure that the wave is transmitted from one string to the other that is both the strings have same tension . $\endgroup$ – Freelancer Jan 31 '16 at 6:29
  • $\begingroup$ Yes, the two strings were not evident in your question, after the edit its ok. I interpreted the "source" as "amplitude" for some reason, whereas on second reading you mean the "frequency of the source". I think the relevant formula from the link I gave is on one page before last: v=f*lamda. $\endgroup$ – anna v Jan 31 '16 at 6:38
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    $\begingroup$ the answers you have gotten have the maths $\endgroup$ – anna v Jan 31 '16 at 10:29
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There is an equation that translate properly the situation of a string under a pulse of frequency $\omega$ at the point $x_0$ of the string. $$ \rho\frac{d^2y}{dt^2}=T\frac{d^2y}{dx^2}+\kappa \delta(x-x_0)\sin(\omega t) $$ this equation is nothing more than an application of the Newton's second law. The first term is the $ma$ part of the $f=ma$. Here $\rho$ play the hole of the mass $m$, actually is the density of the string. The logic is that each point $x$ of the string can deform into $y_x(t)$, or $y(x,t)$. $\frac{d^2y}{dt^2}$ is the acceleration of the point $x$ at time $t$. The second term is the tension.

Tension is a force that any point of the string can feel if the nearby points of the string are not stretched. $T$ measures the strongness. If you deform the sting, then the tension would try to send the string back to stretchiness but, by conservation of energy, this produces an oscillation.

The last term is the source. The $\delta(x-x_0)$ says that the source is located at point $x_0$ of the string. The $\sin(\omega t)$ says that the source oscillates the point $x_0$ with frequency $\omega$.

Putting all this together we have that as the source start to oscillate the point $x_0$, the tension start to act on other points $x\neq x_0$ close to $x_0$. This eventually happens through on all the string. The result is a wave with velocity $v=\sqrt{T/\rho}$. This is so because at points $x\neq x_0$, we have: $$ \rho\frac{d^2y}{dt^2}=T\frac{d^2y}{dx^2} \rightarrow \frac{d^2y}{dt^2}=\frac{T}{\rho}\frac{d^2y}{dx^2} \rightarrow \frac{\partial^2 y}{\partial x_+ \partial x_-}=0 $$ where $x_{\pm}=x\pm vt$. The solution is a superposition of left and right waves $$ y(x,t)= f(x_+)+g(x_-) $$ The full solution need to be compatible with the oscillation of the $x_0$ with frequency $\omega$. Then we conclude that all points need to vibrate exactly at the same frequency with different phases, otherwise, $y(x,t) \neq f(x_+)+g(x_-)$. Try to visualise by taking the referential frame that make the the configuration of the function $y(x,t)$ at reast.

enter image description here

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I would say it has to do with energy/information conservation laws. You can imagine a wave (be it electromagnetic or vibrations on a string) is propagating through a medium. The energy of such a wave is given by:

  1. Classical harmonic oscillator (pendulum or vibrating string) $$ U = \frac{1}{2}kx^2 $$ where $x$ is displacement from the steady-state rest position and $k$ is some constant (spring constant) which tends to pull the system back towards the steady-state equilibrium. The frequency of oscillation here is $$f= \frac{1}{T}= \frac{1}{2\pi}\sqrt{\frac{k}{m}} $$ where $T$ is the period and $m$ the mass of the oscillator (mass of unit length of a string if you like). So you can see that the energy stored in the system is given by $$U=\frac{m}{2}(2\pi f)^2 x^2$$ So it is clear that if we assume absence of loss or gain, the frequency has to stay constant as the wave travels on the string otherwise we are gaining or loosing energy.

  2. Quantum mechanical harmonic oscillator (electromagnetic waves - photons) $$ E = hf$$ where $h$ is the plank constant. Again energy depends on the frequency and must be constant in the absence of local sources or sinks.

Now in case of transmission/reflection at an interface, it might help to think of the rope as a series of equally spaced balls, connected with mass-less springs/ropes i.e. something like $$-\cdot-\cdot-\cdot-$$ For an interface in case of a rope we can think of a thin piece of rope attached to a thick piece of rope. $$-\cdot-\cdot-\cdot-\odot-\odot-\odot-$$

Energy travels through this rope assembly via "nearest-neighbour" interactions. So say a pulse is traveling from left to right, for the 3rd ball to start to move, 1st ball has to transfer energy to the 2nd and so forth. We can think of frequency as a rate of oscillation (jiggling). The large ball sitting at the interface can't start jiggling until the little ball to it's left jiggles, and these jiggles arrive at a constant rate of $f$ per second and are passed on.

Now while the wave travels at a different velocity in the 2nd medium (by that I mean the pattern) the jiggles are transferred between individual balls at a constant rate, otherwise the flow of information will break down. This manifests in a change of wavelength in the 2nd medium. $$f= \frac{v_p}{\lambda}$$ where $v_p$ is the phase velocity (speed of wave pattern propagation) and $\lambda$ the wavelength. To keep $f$ constant, if $v_p$ increases, so should $\lambda$ and vise versa.

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  • $\begingroup$ I think that you are wrong. Conservation of energy of the harmonic oscilator only tells you that the frequency of one point of the string only changes if there is an interaction at that point. What you need to prove for answer the question is that all points of the string vibrate at the same frequency under an source that oscilates one point of the string. $\endgroup$ – Nogueira Jan 31 '16 at 22:39
  • $\begingroup$ Showing that the wave is a linear superposition of two counter propagating waves does require the frequency to remain the same at every point on the string (same harmonic), however it does little to address OP's question about what happens at a boundary. I was building up some foundation to get to the final part of the answer, regarding transmission/reflection at a boundary. $\endgroup$ – cvb0rg Feb 1 '16 at 15:02
  • $\begingroup$ Actually, by knowing this, the boundary set up the coefficients of the linear superposition. Only this. $\endgroup$ – Nogueira Feb 1 '16 at 15:05

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