Let a standing wave of length L go, get traveling wave of length 2L?

In summary: The period of oscillation of the first harmonic is then 2 sec. So, the right node is released at t = 2 sec.The next animation is for the case where the string is initially oscillating in the second harmonic between x = 0 and x = 1. The period of oscillation of the initial standing wave is now 1 sec. The node at x = 1 is released at t = 1. The final animation is for the case where the string is initially oscillating in the third harmonic between x = 0 and x = 1. The period of oscillation of the standing wave is now 2/3 s or about 0.7 sec. So, the right node is released at about
  • #1

Spinnor

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Let a standing wave of length L go, get traveling wave of length 2L?
Consider a very long string between fixed supports of mass density rho and tension T. At a distance 1 meter from one support pinch the string. The pinching does not change the tension. Adjust the mass density or tension so that when we add energy to this section of string we produce the first harmonic. The string oscillates between all kinetic and then all potential energy. Just when the string is all kinetic energy you release the string pinch point.

After a short while will you have a traveling wave pulse that is 2 meters in length? Is there a simple way to argue why this is or is not true?

Thanks.
 
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  • #2
Spinnor said:
Adjust the mass density or tension so that when we add energy to this section of string we produce the first harmonic.
Is the first harmonic the fundamental, f, or twice the fundamental, 2f = second harmonic ?
 
  • #3
Baluncore said:
Is the first harmonic the fundamental, f, or twice the fundamental, 2f = second harmonic ?
standing.gif


If the above is correct (found by Google image search "standing wave first harmonic") I was picturing letting the first harmonic go, though you might ask what if we let higher order harmonics go, I think we get the same result, a standing wave of length L goes to traveling wave of length 2L?

Thanks.

Edit, it was not clear that we were to adjust the mass density and tension to the whole string and then pinch the string 1 meter from one support.
 
  • #4
Spinnor said:
... a standing wave of length L goes to traveling wave of length 2L?
Think of a standing wave as being two travelling waves, travelling in opposite directions along the line. Those two waves are trapped in a resonator, being reflected between the two nodes.

If a pinch on the right is removed, the standing wave energy is no longer reflected on the right, so it propagates away to the right.
At the same time, the standing wave is reflected from the left, so it immediately follows the other travelling wave up the line, making a travelling wave having twice the length.
 
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  • #5
I was losing sleep trying to visualize the shape of the string after the release of one of the nodes o0).
So, I tried making some animations based on some equations in an old book from my college days. I thought I'd share them. I certainly could have made some errors. Let me know if something doesn't look right. They look reasonable to me.

The first animation is for the case where the string is initially oscillating as a standing wave in the first harmonic between x = 0 and x = 1. In the animation, the string makes one complete oscillation as a standing wave before the right node is released. Time is indicated at the top of the graph. (The control tabs are not available in these animations.) I chose the wave speed to be 1 m/s. The period of oscillation of the first harmonic is then 2 sec. So, the right node is released at t = 2 sec.

wav1.gif
In the next animation, the string is initially oscillating in the second harmonic between x = 0 and x = 1. The period of oscillation of the initial standing wave is now 1 sec. The node at x = 1 is released at t = 1.
[Edit: This animation has been updated to correct a scaling error.]

wav2.gif
Finally, animation for the third harmonic. The period of oscillation of the standing wave is now 2/3 s or about 0.7 sec. So, the right node is released at about t = 0.7.
[Edit: This animation has been updated to correct a scaling error.]
wav3.gif


In all cases, the final total width of the traveling wave is 2 meters. This agrees with @Baluncore's prediction in post #4. :woot:
 
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  • #6
You need to consider whether the node that remains at the origin, will invert the amplitude of the reflected travelling wave, the part that follows the direct wave to the right.
 
  • #7
TSny said:
I was losing sleep trying to visualize the shape of the string after the release of one of the nodes o0).
So, I tried making some animations based on some equations in an old book from my college days. I thought I'd share them. I certainly could have made some errors. Let me know if something doesn't look right. They look reasonable to me.

The first animation is for the case where the string is initially oscillating as a standing wave in the first harmonic between x = 0 and x = 1. In the animation, the string makes one complete oscillation as a standing wave before the right node is released. Time is indicated at the top of the graph. (The control tabs are not available in these animations.) I chose the wave speed to be 1 m/s. The period of oscillation of the first harmonic is then 2 sec. So, the right node is released at t = 2 sec.

View attachment 320849In the next animation, the string is initially oscillating in the second harmonic between x = 0 and x = 1. The period of oscillation of the initial standing wave is now 1 sec. The node at x = 1 is released at t = 1.

View attachment 320851Finally, animation for the third harmonic. The period of oscillation of the standing wave is now 2/3 s or about 0.7 sec. So, the right node is released at about t = 0.7.
View attachment 320854

In all cases, the final total width of the traveling wave is 2 meters. This agrees with @Baluncore's prediction in post #4. :woot:
Shouldn't you just run the simulations a bit longer. E.g., in the 1st animation when running the simulation further you should get a reflected wave moving to the left. The same with the other ones too.
 
  • #8
vanhees71 said:
Shouldn't you just run the simulations a bit longer. E.g., in the 1st animation when running the simulation further you should get a reflected wave moving to the left.
Spinnor said:
Consider a very long string between fixed supports of mass density rho and tension T.
A very long line suggests a pseudo-infinite length. There can be no significant reflection from a very long line because line losses will have dissipated or dispersed the energy.
 
  • #9
Sorry, I thought the right end of the longer string is also fixed. If there's no restriction the waves shall simply move on to the right.
 
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  • #10
I did not expect the wave shape to be "rectified", all the "bumps" on one side, minus one for my intuition.

I think there must be a slight glitch in the program, maybe some physics left out. The total energy of the wave must be conserved so in the third animation there is a sudden change in the amplitude which would violate energy conservation. When the pinch point is released that information travels at most wave velocity? The motion of the string on the left can not change until a finite time after the pinch release?

Much to think about, thank you for the animations.
 
  • #11
Spinnor said:
I did not expect the wave shape to be "rectified", all the "bumps" on one side, minus one for my intuition.

I think there must be a slight glitch in the program, maybe some physics left out. The total energy of the wave must be conserved so in the third animation there is a sudden change in the amplitude which would violate energy conservation. When the pinch point is released that information travels at most wave velocity? The motion of the string on the left can not change until a finite time after the pinch release?

Much to think about, thank you for the animations.
Thank you @Spinnor. I found the scaling errors for the second and third harmonic animations and updated the animations. I think the energy is now conserved. I'm still getting the rectification of the bumps. I was a little surprised at that also, but I don't yet see any mistake in the coding. I'm getting the bumps in the traveling waves to have the shape of sine squared functions rather than sine functions.
 
  • #12
Here's how I got the expressions that I plotted in post #5.

The solution ##y(x, t)## of the wave equation in terms of initial conditions at ##t = 0## is given by d'Alembert's formula : $$y(x, t) = \frac 1 2 \left(f(x+ct) + f(x-ct)\right) + \frac 1 {2c} \int_{x-ct}^{x+ct}g(\bar x) d\bar x$$ Here, ##c## is the wave speed, ##f(x)## is the initial displacement of the string and ##g(x)## is the initial velocity profile of the string.

In our case, the string is semi-infinite along the positive x-axis and fixed at ##x = 0##. For ##t < 0##, we assume that the string is vibrating in the mth harmonic standing wave in the region between ##x = 0## and ##x = 1## with amplitude 1. We'll also take ##c = 1##. So, $$y_{sw}(x, t) = \sin(m\pi x) \sin(m\pi t) \,\,\, \text{ for } 0 \le x \le 1 \text {, 0 otherwise.} $$
At ##t = 0## when the node at ##x = 1## is released,

##\,\,\,f(x) = 0## for ##x \ge 0## and
##\,\,\,g(x) = m\pi \sin(m\pi x)## for ##0 \le x \le 1##, ##g(x) = 0## for ##x > 1##.

It can be shown that the node at ##x = 0## is accounted for in d'Alembert's formula by extending ##f(x)## and ##g(x)## to all negative values of ##x## as odd functions about ##x = 0##. Thus, in d’Alembert’s formula we use ##f(x) = 0## for all ##x## and
$$g(x) =
\left\{
\begin{array}{lr}
m\pi \sin(m\pi x), & \text{if }|x|\le 1\\
\,\,\,\,\,\,0, & \text{if }|x| >1
\end{array}
\right\}
$$

So, with ##c=1##, the solution is $$y(x, t) = \frac 1 2 \int_{x-t}^{x+t}g(\bar x) d\bar x$$

Let ##G(x)## be an antiderivative of ##g(x)##. Then we can write the solution as $$y(x, t) = \frac 1 2 \left[G(x+t) – G(x-t)\right].$$

An antiderivative of ##g(x)## is $$G(x) = \int_{-\infty}^x g(\bar x)d \bar x.$$ Evaluating this for the extended ##g(x)## yields

$$
G(x) =
\left\{
\begin{array}{lr}
(-1)^m - \cos\left(m\pi x\right), & \text{if }|x| \le 1\\
\,\,\,\,\,\,0, & \text{if }|x| >1
\end{array}
\right\}
$$

Using the Heaviside Pi function ##\Pi(x)##, ##G(x)## can be written as
$$
G(x) =
\left\{
\begin{array}{lr}
2\sin^2\large\left(\frac{m \pi x}{2}\right) \Pi(\frac x 2), & \text{if }m \text{ is even}\\
-2\cos^2\large\left(\frac{m \pi x}{2}\right) \Pi(\frac x 2), & \text{if }m \text{ is odd}
\end{array}
\right\}
$$
Using this in ##y(x, t) = \frac 1 2 \left[G(x+t) – G(x-t)\right]## gives the solution for any positive ##x## and positive ##t##.

Aside: The above solution for even ##m## will give “negative bumps” for the graphs rather than positive bumps. This is due to the fact that our expression for the standing wave ##y_{sw} (x, t) = \sin(m\pi x) \sin(m\pi t)## has the property that points of the string just to the left of ##x = 1## are moving downward at ##t = 0## for even ##m##. So, when the right node is released at ##t = 0##, the string is pulled downward as the wave propagates toward the right beyond ##x = 1##. This produces negative bumps for even harmonics. If you want the bumps to always be positive, you can change the overall sign of ##y_{sw}(x, t)## for the even harmonics. This is what I did for the m = 2 plot in post #5.

EDIT: The expression for ##G(x)## can be tidied up as a single expression for both even and odd values of ##m## as $$G(x) = -2\sin^2\left[ \frac{m \pi}{2}(x+1)\right] \Pi\left(\frac x 2\right)$$
 
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