4
$\begingroup$

I learned that, at frequencies corresponding to harmonics, standing waves are formed. But what actually happens at other frequencies? Won't the reflected wave superimpose with the original wave? Do any other phenomena happen?

$\endgroup$
  • $\begingroup$ A typical organ pipe is a flue pipe en which the air is blown over a sharp lip. Or there are reed pipes where a tongue moves. Or is your question about a simple tube? $\endgroup$ – Pieter Aug 31 '18 at 18:22
2
$\begingroup$

After the transients die out, the air in the pipe will always vibrate with the same frequency as the driving frequency, but the amplitude will be very small, essentially because your driving at one moment will cancel out the effect of the driving at the next. In practice, that means you won't hear anything at all.

It's analogous to somebody trying to pump a swing, but moving their legs faster or slower than the swing "wants" to go. All that happens is that they wiggle around a bit at the bottom.

$\endgroup$
  • $\begingroup$ How does the drivings cancel each other? Please, explain its essence. $\endgroup$ – manoj Sep 1 '18 at 3:05
0
$\begingroup$

Let the length of the pipe be $L$.

Two closed ends: The molecules at the ends cannot move freely due to the boundary imposed on them. Waves with wavelength $\lambda = 2L/n$ ($n$ an even integer) will have nodes at the ends and so will not be interrupted in their propagation by the boundary. Other wavelengths will loose energy at the boundary through molecular collisions and will dissipate over time. So non-harmonic wavelengths will simply die out (fairly quickly).

One closed end: The difference in the environment at the open end of the pipe (restricted by hard walls in two directions inside, unrestricted in all directions outside) acts as a sort of boundary. This time, the wavelength $\lambda = 2L/n$ uses odd-integer $n$. This is because one end is open, so there is an antinode at the open end (instead of a node). The behavior of non-harmonic waves would be similar to that of a pipe with two closed ends.

Dissipation: Sound waves in air are longitudinal waves (also called compression waves) made up of the repeated compression and expansion of the air molecules. That means that, although the net displacement of the air molecules is zero (they return to their equilibrium positions after the wave passes), the molecules must move in order to transmit the wave. A standing wave is a superposition of waves traveling in opposite directions and has nodes every half wavelength where the movement of the molecules in one direction is canceled by movement in the opposite direction. In other words, the molecules at the nodes don't move.

If a wave has a wavelength such that exactly $n$ wavelengths fit in your pipe of length $L$ (as is the case for harmonic wavelengths), you will have a node at the ends of the pipe. This means that the air at the ends of the pipe does not move, and all the movement happens between the ends. In this case little of the wave's energy is lost to the environment.

If a wave has a wavelength such that more or less than $n$ wavelengths fit in your pipe (as is the case for non-harmonic wavelengths), you will not have a node at the end. This means that the air molecules at the end are moving. There is a boundary here, and the air molecules collide with this boundary and transmit to it their energy. In this way the wave loses it's energy to the environment and dies out.

$\endgroup$
  • $\begingroup$ how exactly does the non-harmonics wavelength disssipate and why this dissipation doesnot happen in harmonics. Please elaborate. $\endgroup$ – manoj Sep 1 '18 at 18:34
  • $\begingroup$ @user8519056 I just edited my answer to reflect your request. Sorry for the slow reply, I usually don't get online on the weekends. $\endgroup$ – The Ledge Sep 4 '18 at 15:34
0
$\begingroup$

Here is an idealized mathematical model for how a waveguide may select certain wavenumbers $k=\frac{2\pi}{\lambda}$. Consider for simplicity a 1D periodic waveguide of length $L$ with a (complex) monochromatic wave $$ y(x,t)~=~Ae^{i(kx-\omega t)}\sum_{n\in\mathbb{Z}} e^{inkL}~=~2\pi A e^{i(kx-\omega t)}III(kL),$$ where $$III(\theta)~=~\delta(\theta+2\pi \mathbb{Z}) $$ is the Dirac comb/Shah function. We see that a wave is allowed iff the length $L$ is a multiple of the wavelength $\lambda$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.