# Why does the phase velocity of a string attached to springs depend on the wave length?

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

$$v=L/T=\omega/k$$.

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

$$2cos(mt)$$

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

$$2cos(kx)$$

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

$$2cos(kx)cos(mt)$$

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

$$\omega=\sqrt{m^2+ak^2}$$

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

$$\omega^2=m^2+k^2$$

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

$$\omega^2=k^2$$

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 $$\omega^2=k^2+m^2$$