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Several people have said that end corrections occur because of acoustic radiation or something similar, where energy is used in vibrating the air outside the pipe.

How exactly does energy loss cause the frequency to be lowered? I thought that radiation would only cause the amplitude of the returning wave to be lowered, which does not change the frequency of the standing wave.

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The concept of end correction is indeed related to, as you say, "energy used in vibrating the air outside the pipe". The end correction is of real-world importance, for example, in the design of loudspeaker ports. The correction factor depends on the geometry of the exit.

Normally the problem is described in terms of acoustic impedances. So, if one end of the pipe is driven (say, inside your loudspeaker) with input impedance $Z_0$, and if the exit of the pipe is terminated with impedance $Z_\mathrm{exit}$ (this is the side that will create an end correction), then the impedances are related by

$$Z_0 = \frac{Z_\mathrm{exit}+i\tan(kL)}{1+iZ_\mathrm{exit}\tan(kL)}\; .$$

This will be recognized as the equation for a transmission line. Here, $Z_0$ and $Z_\mathrm{exit}$ are normalized (dimensionless) mechanical impedances: $$Z_0 = \frac{\mathrm{Z}_{m,0}}{\rho \, c_s S} \quad\text{and}\quad Z_\mathrm{exit} = \frac{\mathrm{Z}_{m,\mathrm{exit}}}{\rho \, c_s S} \; , $$ where $\rho$ is the density of air, $c_s$ is the sound speed, and $S$ is the cross-sectional area of the pipe. The frequenecy-dependence of $Z_\mathrm{exit}$ depends on the type of exit (fully baffled, partially baffled, open) and the calculation is quite complicated in all cases. The simplest case is the fully baffled (flanged) case, which at low frequency becomes $$Z_\mathrm{exit} \sim i \frac{8}{3\pi} k a \; ,$$ where $a$ is the radius of the pipe. In the limit of long-wavelength ($k \ll 1$), the impedance takes the simple form
$$Z_0 \sim i \left( kL + \frac{8}{3\pi} ka \right) \; .$$ Thus, at the input end, the pipe looks like it has length $$L_\mathrm{effective} = L + \frac{8a}{3\pi}$$ The second term is the end correction. In terms of the effective length, the resonant frequencies are $$f_n = n \, \frac{c_s}{2L_\mathrm{effective}} \; .$$ You can see this by retaining the $\tan$ function and solving for the zeros.

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let's use a vibrating string as an easier-to-visualize example. our string is stretched tight and fastened securely to two heavy objects so the ends cannot move. we pluck the string; it vibrates up and down everywhere along its length except at the ends, which are fixed in position. we measure its frequency. next we attach a little mechanism to one end of the string so it can slide up and down along the block while still being under tension. the mechanism has a little friction in it, so sliding it up and down requires work. we pluck the string again and notice that the vibrating string pulls the mechanism up and down a little bit and its vibrations are damped. the fact that the string termination point moves up and down at the block means that the string termination point is no longer at the block; the string is now vibrating as if its termination point is slightly "behind" the block surface i.e., the string acts "longer" and as a result the string's vibrating frequency is shifted down a bit. in the case of an organ pipe, when energy leaves the pipe's open end, the pipe plays "flat" and an "end correction" must be applied to get it to resonate at the intended pitch.

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