A string can be described by the wave equation $$ (\partial_t^2 - \partial_x^2) \, \varphi = 0 $$ while a string attached to a spring (i.e. with a harmonic restoring force) at each location is described by the wave equation with dispersion term: $$ (\partial_t^2 - \partial_x^2 + m^2) \, \varphi = 0 $$


An important difference between the two systems is that without a dispersion term ($m=0$) all plane wave solutions have exactly the same phase velocity. In contrast, if $m\neq 0$ the phase velocity depends on the wave length.

While it is straightforward to derive this mathematically, I'm struggling to understand why this is the case. Is there some intuitive way to understand why for the string with restoring force the phase velocity ($v = \omega /k)$ depends on the wave length?


If we consider an ordinary string and focus on one specific point on it, say one maximum, it will travel during the time interval $\Delta t$ the distance $\Delta x$. This fact is independent of the wave length of the wave in question. Thus the phase velocity is always $v = \frac{\Delta x}{\Delta t}$.

enter image description here

In contrast, for a string attached to springs, this is no longer the case. During a fixed time interval $\Delta t$, the maximum of a short wave travels a different distance $\Delta x'$ than then maximum of a long wave. This means that the phase velocity of short and long waves is different: $$v' = \frac{\Delta x'}{\Delta t} \neq = \frac{\Delta x''}{\Delta t}= v''$$

enter image description here

So the answer must have something to do with the fact that for a long wave there are more springs involved per wave form. However, the whole issue is still far from obvious for me. In particular, I'm not even sure how to determine (without calculating anything) whether the phase velocity of a long wave is faster or slower than for a short wave.

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    $\begingroup$ There even a much more important distinction: for the case with restoring force there is no traveling wave if the frequency is too low. Do you have an intuitive understanding of this fact? (starting from this, we have no wave for $\omega < \omega_* = m$, zero velocity for $\omega = \omega_*$, small velocity for $\omega \gtrsim \omega_*$, velocity 1 for $\omega \gg \omega_*$) $\endgroup$ – Fabian Oct 5 '19 at 5:43
  • $\begingroup$ @Fabian No I wasn't aware of this and I currently have no intuitive understanding of it. But it sounds like a great place to start thinking about this problem $\endgroup$ – jak Oct 5 '19 at 9:08
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    $\begingroup$ @jak Well, I did start from this fact for my new answer, see below $\endgroup$ – Alfred Oct 13 '19 at 5:03
  • $\begingroup$ @Fabian See the end of my main answer. Not QUITE the end, the end of the recent part of the answer, I dd not delete the old one, which is after it, $\endgroup$ – Alfred Oct 13 '19 at 10:03
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    $\begingroup$ The dispersion equation is $\omega=sqrt(k^2+m^2)$ so the velocity is $v=sqrt(1+m^2/k^2)$ lalwys larger then one and going to infinity as $k$ goes to zero. But it has nothing to do with the number of springs per wavelength. Please read my long answer below. $\endgroup$ – Alfred Oct 13 '19 at 14:12

The extraneous springs cause faster vertical oscillations on top of what is already caused by the non-dispersive medium (the rope), thereby propagating the waves faster and making the dispersion relation non-linear.

For large $\lambda$, the spring energy dominates the oscillation because the string is stretched in a rather smooth way, so less contribution from potential energy of the stretched string, and we have $\omega \sim \sqrt{\chi/\mu} \sim m$. In the case of high $k$, the string is quite distorted and the stretched string contributes much more to the oscillation than the extraneous springs, and one therefore retrieves the case of linear dispersion $\omega \sim k$ as in the case of string without springs.

A Discrete Model & Its Continuum Limit

The string in the case $m=0$ can be modeled as a 1D chain of masses (mass $\mu$ each), with horizontal spacing $a$, neighbouring masses connected to each other by massless strings under tension $T$. In order to simulate waves on a rope in the continuum limit, one then needs to take the limit $\mu \to 0$ and $a \to 0$ such that $\mu/a$ tends to the mass density of the rope, and speed of wave propagation, $c = \sqrt{\frac{T}{\mu/a}}$, is finite. In your case $c=1$, so we can choose $T = \mu/a$. For small, smooth vertical oscillations of the chain, this system is then described by the wave equation without the mass term. (eg. cf. Greiner, Classical Mechanics; or David Morin, Classical Mechanics)

Wave-like Solutions

Propagating wave-like solutions are then guaranteed by Bloch's theorem $-$ in the limit $a \to 0$, the size of the unit cell shrinks to zero and we've pure exponential (sine/cosine) waves as solutions.

Broken Symmetry (or not?)


In the case $m=0$, the equilibrium state of the system about which small oscillations are considered could be translated in the vertical direction as a whole with no energy cost and that would result in a new equilibrium state. The choice of a definite ground state then implies that we start with a symmetry broken ground state, thereby inevitably running into masless Goldstone bosons for $k\to 0$. With this one can imagine a dispersion of the form, $\omega_{k} = A k+B k^2+\dots \ $ about $k=0$, which will satisfy the wave equation only for linear dispersion, leading to dispersion-free waves.


In the other case with finite mass $m$, one needs to attach springs of stiffness $\chi = \mu m^2$ to each mass $\mu$, from say below, in order to produce the mass term in the wave equation for vertical oscillations. The ground state is now unique and we do not have Goldstone modes anymore. In fact, even in the limit $T\to 0$ when the wave effects almost cease to exist, the system still needs a finite energy cost $\omega^*\gtrsim m$ to be excited (as pointed out already by @Fabian). This means $\omega_{k}/k$ diverges as $k \to 0$ and thus the phase velocity clearly must be wavelength dependent.

Normal modes in Momentum Space

The continuum Lagrangian can be diagonalized in momentum space yielding normal modes, i.e. $$ L = \sum_{k} \frac{1}{2} \bigg[\dot{\phi}_{-k}\dot{\phi}_{k}-(k^2+m^2)\phi_{-k}\phi_{k} \bigg],$$ the normal mode frequencies simply being $\omega_{k}^2 = k^2 + m^2$.

In QFT, the $m=0$ and $m\neq0$ case would correspond to the massless and massive Klein-Gordon fields respectively; the latter would require at least the rest energy $m$ to create a particle in the momentum state $k=0$, consistent with the dispersion relation.

Now, suppose conversely the dispersion relation was linear $-$ then the wave would be photon like and it would take vanishingly small energy energy to create photons at arbitrarily small $k$, which the extraneous springs will simply not allow in the model with springs and masses.

A Classical Relook via the Virial Theorem

Finally, note that we can apply classical virial theorem on the discrete system of springs and masses in mode with wave vector $k$ as the potential energy function is a homogeneous function of degree $n=2$, and consider the limit $a\to0$, yielding \begin{align} \langle T \rangle_k = \frac{n}{2} \langle U \rangle_k = \langle U \rangle_k \\ \implies \omega_{k}^2 = k^2 + m^2. \end{align}

Further Edit

Following your edit: The fact that the phase velocity is faster for a long wavelength wave has nothing to do with the fact that there are more springs per wavelength.

In fact, if you just compute the energy of the string+spring system per wavelength, you will get that the kinetic energy contribution per wavelength is $\frac{1}{4} \mu A^2 \omega_k^2$, the potential energy of the stretched string per wavelength is $\frac{1}{4} \mu A^2 k^2$ and that of the springs is $\frac{1}{4} \mu A^2 m^2$, where $A$ is the amplitude of the wave in mode $k$. So, per wavelength everything is really independent of the wavelength.

However, from the virial theorem we know that the average potential energy must be equal to the average kinetic energy, which produces the non-linear dispersion quoted above.

I still think the most interesting conceptual part is the emergence of Goldstone modes in a string (w/o springs) for which $\omega_k \to 0$ as $k\to 0$ $-$ which is itself a consequence of broken symmetry. In a crystal which has broken translational symmetry, these Goldstone modes correspond to acoustic modes that cost vanishingly little energy for $k\to 0$ (sound waves). In a superfluid, these correspond to Bogolons. In ferromagnets, these Goldstone modes are the magnons.

The specific form of the dispersion law requires a knowledge of the potential energy functional of the system. In the case with strings the potential energy contribution is $\Big(\frac{\partial y}{\partial x}\Big)^2$, which in momentum space looks like $k^2 \vert y_k\vert^{2}$ and yields $\omega_k^2 \propto k^2$. If the potential energy had another form, eg $\propto C \Big(\frac{\partial^2 y}{\partial x^2}\Big)^2$, it would look like $C k^4 \vert y_k\vert^2$ in momentum space and yield a dispersion $\omega_k^2 \sim k^4$ (this is the non-relativistic dispersion formula and will eventually spread wavepackets). However, in all these cases the common theme is the emergence of a Goldstone mode.

The case of the string with extraneous springs breaks precisely this theme by forcing a unique groundstate (corresponding to the unique state when all the springs are unstretched) and creating an excitation gap $m$. For long-wavelength modes ($k \gg m$) this means $\omega_k \sim m$ (i.e. one still needs to excite the spring), and for short wavelength ($k \gg m$), $\omega_k \sim k$ to excite mode $k$ (like in case of a free string). The interpolation formula $\omega_k = \sqrt{k^2+m^2}$ does exactly this.

p.s. Thanks to Prof. Stone for the Pantograph Drag reference.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Oct 15 '19 at 1:37

OK, let's start over. I did not delete my old answer, but just ignore it.

This time I am trying to give an intuitive justification, but I have to take the problem "backwards", so to say, as I'll explain below.

You'll need the notion of stationary wave.

A stationary wave is a superposition of two travelling waves of same wavelength, frequency and amplitude, but propagating in opposite directions

$cos(kx-\omega t)+cos(kx+\omega t)$

Trigonometry tells us that this is equal to

$2cos(kx)cos(\omega t)$

So the maxima and the minima (in space) of such a wave are always at the same positions, determined by $cos(kx)$. One does not see anything actually moving in space along the direction $x$. The oscillation is in time only, and the oscillating motion is perpendicular to the direction $x$.

Still, one can determine the common phase speed (up to the sign, of course) of each of the original travelling waves by the ratio of the wavelength $L$ to the period $T$. In terms of the wavevector $k=2\pi/L$ and the pulsation (this is the french word for frequency multiplied by 2$\pi$, I did not find the english equivalent) $\omega=2\pi/T$ on gets


Now consider the following situation. Imagine a collection of springs, at distance $d$ from one to the next one along the $x$ axis, all identical, and attached to massive objects also all identical. For instance we can imagine that each of these objects is a separate little bit of string, just the length between two springs. Each spring will thus oscillate with a specific period $T$. Let us call $m$ the quantity $2\pi/T$. This $m$ is just a name (that you have chosen), it is not the value of the mass of the objects at the end of the springs. So each object oscillates as $cos(mt+\phi)$ where $\phi$ is some phase.

Now in a first case let us start all the objects at the same time (say, origin of time, $t=0$) and the same amplitude $2$. (Why $2$? Any value would do. So why not $2$?)

So each and every object will have exactly the same motion $2cos(mt)$

So if instead of having one little bit of string at the end of each spring, nothing will change if I tie all these bits, that is, replace them by a single string extending over all the springs. That string will have open ends, so no stretch. But note that, if I take each end of the string in jaws and pull them apart, so as to stretch the string, nothing changes either, provided I am careful to give to the jaws the same motion, perpendicular to the string, of amplitude $2cos(mt)$. Then the motion of a (now stretched string) can be described by


but here, $2$ is not just a number, it has acquired the status of a function of $x$, which just happens to take the value $2$ for every $x$

Now suppose that instead of a constant amplitude $2$, I start at time $t=0$ by a sinusoidal shape


If the string was cut in pieces, as I described earlier, one attached to each spring, each single one would have a motion proportional to $cos(mt)$ to give


But this does not work if the string is whole.

Let us assume that the wavelength $L=2\pi/k$ is much much larger than the distance $d$ between two strings. Otherwise it is even more complicated.

Since the position of the "heads" of two consecutive strings in the perpendicular direction depends (slowly because $d$ is much smaller than the distance $L$ on which the amplitude varies) on the position in $x$ of each string, by Pythagora's theorem you see that the distance of two heads is a teeny little bit larger than $d$. This means the string is a bit more stretched in this situation than in the constant amplitude case. And its direction is not quite parallel to the $x$ direction. The result of this is that, in addition to the restoring force of the spring (which causes oscillation $cos(mt)$), there is an extra restoring force, due to the stretch of the string, that clearly increases when $L$ decreases, that is when $k$ increases, since it is absent for a straight string ($k=0$ or $L$ infinite).

So maybe your intuition does not feel it this way, but, possibly because I am a professional physicist, my intuition tells me that the shape will remain the same function of $x$, but frequency increases. I thus expect

$2cos(kx)cos(\omega t)$

with $\omega > m$.

In fact because the points at the "heads" of the springs are special, compared to the remainder of the string, it will not be exactly the case. But if $L$ is sufficiently larger that $d$, one can indeed neglect the "separated" character of the springs and treat the entire string as a smooth object, neglecting the special character of the contact points of the springs. That will of course fail if $L$ is not large enough compared to $d$.

So, as I announced from the start, I went "backwards" with respect to your question : I did not explain intuitively the effect of adding the springs to the string alone, but rather how to treat a whole, stretched string attached to springs instead of separated identical objects, (for instance little separated pieces of string), each of which attached to its own spring.

Now since $2cos(kx)cos(\omega t)$=$cos(kx-\omega t)+cos(kx+\omega t)$

one can intuitively see that the string also supports travelling waves in each direction, for instance forwards in the $x$ direction

$cos(kx-\omega t)$

with now "visible" velocity $\omega/k > m/k$

Now one can go further, but not just by intuition. All that intuition tells me is that $\omega$ is an increasing function of $k$ that has value $m$ for $k=0$.

If I actually do the calculations, I find that the exact value of $\omega$ is given by


where $a$ is a quantity that depends on the linear mass of the string and on its stretch. Note that none of these quantities have been defined, nor have the units of length (meters? mms ? kms? inches? feet? miles?) nor time (seconds? minutes? days? years?) so with this freedom we can arrange to take $a=1$ and we recover the values you proposed yourself (for instance, time unit= one second, length unit, whatever length a wave travels in one second on the string without springs; or else length unit one furlong, time unit, whatever time it takes for a wave to travel one furlong on the string without springs; in both cases the quantity $a$ will be equal to $1$) and we recover your expression


Now one cannot take $k$ arbitrary large. If $k$ is nor very small compared to $1/d$, one cannot neglect the fact that the springs are separated, and the whole reasoning collapses, the whole thing becomes much more complicated. On the other hand, one has never said how big $m$ is compared to $1/d$. If $m$ is very very smaller than $1/d$ I can choose $k$ small enough compared to $1/d$ but still much larger than $m$. In which case $\omega$ will be very close to $k$, and the velocity very close to 1 (in the normalisation of space and time units I have chosen to fit your initial problem). Or one can remove the springs entirely, $m=0$

In this case the velocity will become independent of $k$, $v=\pm 1$, (because one must allow for waves propagating in both directions).

However, the fact that on a stretched string without attached springs, the velocity is independent of the wavevector is the result of an exact calculation. I do not see it as "intuitive".

The fact that identical springs acting on identical objects all oscillate at the same frequency is intuitive, and from there I can intuitively guess that the frequency increases when these objects are connected as a stretched string, but only if it is not straight ($k=0$), in which cas the frequency is the same as that of separate objects. But I cannot guess the exact form. I have to calculate it.

But wavevector-independent velocity on a string without springs attached ? This is not so intuitive. All waves do not have wavevector-independent velocity. Waves on a stretched string, yes, it is indeed true, sound waves in air, yes. Electromagnetic waves (like light, for instance) in vacuum, yes. But water waves (on the free surface of a body of water) have a wavevector-dependent velocity. So do electromagnetic waves propagating through matter. This is the reason why a prism separates white light into a "rainbow" spectrum, and diamonds so beautiful to look at. So intuition cannot tell you that the velocity is independent on the wavevector on a string, only exact calculations can prove it.

This is essentially why I had a reluctant attitude in my first answer. You wanted an intuitive explanation of a problem that was more complicated than the basic one (for $m=0$ but $k$ nonzero) that already did not have an intuitive solution. When you spoke of actually tying springs to a string, I thought of cutting the string into individual pieces, one for each spring, and the intuition came to me.

Now something amusing.

Suppose I get hold of one end of the string (say, the end at negative $x$) and I shake it with a pulsation ($2\pi$ times frequency) $\omega$ larger than $m$. What will happen is that a wave will propagate away from that end, towards positive $x$,

$Acos(kx-\omega t)$

with some amplitude $A$ and wavevector $k=\sqrt{\omega^2-m^2}$

What if I shake that end with a $\omega$ less than $m$ ? No wave can propagate with $\omega$ less than $m$, so what happens ?

In that case the effect of the shaking will not propagate away, it will remain in the neighborhood of the shaked end, and will decrease exponentially from there

$A exp(-\kappa x)cos(\omega t)$

where $\kappa^2=m^2-\omega^2$

This will be true only provided the penetration length $1/\kappa$ is large compared to the distance $d$ between the strings.

If $m$ is large and $\omega$ not sufficiently close to $m$, so that $1/\kappa$ is not large enough compared to $d$ the problem will be more complicated because then one cannot ignore the fact that we are dealing with separate springs.

Please ignore what is below this line

I don't think you'll be happy with my answer.

I find the physical problem you consider a bit strange. You start with a string of given linear mass density and given stretch. The wave equation is indeed the one you wrote first This leads to waves propagating in each direction with a velocity independent of the wavelength. You chose to normalize the absolute value of this velocity to unity


so $v=\omega/k=\pm1$

So far, so good.

Then you proceed to attaching each little bit of your string to its average position by a spring of given strength. This add a harmonic restoring force that you represent by a factor $m^2$. Writing the equation of motion is simple, this is your second equation. Solving it is also simple, you get dispersive waves satisfying the dispersion relation $\omega^2=k^2+m^2$ So, in particular the absolute value of $\omega$ is never less than $m$.

Mathematically, there is no difficulty. But I have never seen such a physical situation. Can you give any example of an actual system looking like that ?

To have, as you are asking for, an intuitive understanding of a system, one needs some intuitive connection with such a system. And I regret to say that I do not, because I cannot imagine such a system "for real". So my answer is that I do not have any intuitive understanding to offer you, but for the dispersion relation $\omega^2=k^2+m^2$

  • $\begingroup$ Thanks for your answer. I think, however, that the system is already as physical as it gets. We simply take a rope and then attach springs to it. This is an experiment every highschool student can perform. $\endgroup$ – jak Oct 11 '19 at 8:36
  • $\begingroup$ @jak Please note that I have a completely new answer ! $\endgroup$ – Alfred Oct 13 '19 at 5:04

This is a good model for the overhead power line for electric trains. For such trains it is essential that there being no travelling-wave solution for phase velocities less that that of the train, or else the train will leave a "wake" that can make the overhead cable disconnect from the power pick-up pantagraph. When the French TGV went for the electric locomative speed record a few years ago they engineers had to increase the cable tension to ensure this. Anyway see problem 1 in homework set https://courses.physics.illinois.edu/phys508/fa2019/508hw9.pdf for some explanation.

  • $\begingroup$ Showing a different physical situation leading to the same equation is certainly very nice, per se. But it does not address the "intuitive" request of the OP, does it ? $\endgroup$ – Alfred Oct 13 '19 at 4:52

In the first case with mass=0,the spring will oscillate with phase velocity which is only dependant on the amount of displacement.In the second case it also depends on the mass that is suspended.The mass of the system along with the spring constant decides the wavelength and the phase velocity is also dependant on mass because of the changing potential energy of the mass.Hence both phase velocity and wavelength are interdependant

  • $\begingroup$ The quantity the OP calls $m$ is a NOT a mass. It is a frequncy multiplied by $2 pi$. Your answer does not mean anything. $\endgroup$ – Alfred Oct 12 '19 at 23:21

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