##### Closed Organ pipe (closed at one end)

The same is the case of a glass tube partially filled with water.

$ \displaystyle \lambda = \frac{4l}{2P-1} $

$ \displaystyle n = \frac{v}{\lambda} =(2P-1)\frac{v}{4l} $

Where , P = 1 , 2 , 3…

for P =1 ; n_{1} = v/4l = Fundamental frequency or frequency of Ist harmonic

for P = 2 ; n_{3} = 3(v/4l) = 3 n_{1} = frequency of 3rd harmonic

i.e. n_{1}:n_{3} : = 1:3 : . . .

Thus in this case we get only odd harmonics. Here it must be noted that in case of organ pipes longitudinal stationary waves are obtained.

The end of the air column in touch with water or base of pipe suffers no displacement or density or pressure change, because the incident and reflected compressional waves are exactly out of phase.

For this reason the pressure changes here are the largest; since the compressional part is reflected the pressure increase is doubled and when the rarefaction part is reflected, the pressure decrease is doubled.

Similarly at the open end, there is maximum density change (or displacement) and minimum pressure change (The two waves travelling in opposite directions are in phase here, so there is no pressure fluctuations).

For this reason only the closed end is always node and the open end is antinode.

__Open Organ pipe (open at both ends)__

For Pth harmonic –

Total number of antinodes between two open ends = (P + 1)

Distance between two antinodes = λ/2

$ \displaystyle \lambda = \frac{2l}{P} $

$ \displaystyle n = \frac{v}{\lambda}= P\frac{v}{2l} $

P= 1, 2 , 3 ,…

For P =1; n_{1} = v/2l = Fundamental frequency or frequency of 1st harmonic

For p = 2; n_{2} = 2(v/2l) = 2n_{1} = frequency of 2nd harmonic

Hence n_{1}: n_{2} : n_{3} : = 1:2:3: …

i.e. in case of an organ pipe open at both ends we get both (odd and even) harmonics.

We can remember it like this – both even and odd harmonics are obtained when conditions at the two boundaries are same

(e.g. in case of a stretched string both ends were clamped and in case of organ pipe both were open) and only odd harmonics are obtained when conditions at two boundaries are different (e.g. in case of an organ pipe one end is closed and the other is open).

__End Correction or Rayleigh Correction__

Due to finite momentum of air molecules in organ pipes reflection takes place not exactly at open end but some what in.

So in an organ pipe antinodes are not formed exactly at free end but at a distance e = 0.6 r, where r being the radius of pipe. Therefore for-

(a) Closed organ pipe

l_{eff} = l + 0.6r

$ \displaystyle n = P\frac{v}{4( l + 0.6r)} $

(b) for open pipe: l_{eff} = l + 2×0.6r = l + 1.2r

$ \displaystyle n = P\frac{v}{2( l + 1.2r)} $

That is why for a given V and l , narrower the pipe, higher will be the frequency or pitch and shriller will be the sound.

### Velocity of Sound (Resonance column apparatus)

By this experiment we can calculate speed of sound in air and end correction.

If a vibrating tuning fork (of known frequency) is held over the open end of the resonance tube, the resonance is obtained at some position.

If l_{1} , l_{2} and l_{3} are the first, second and third resonating lengths then,

$\large l_1 + e = \frac{\lambda}{4} $ …(i)

$\large l_2 + e = \frac{3 \lambda}{4} $ …(ii)

$\large l_3 + e = \frac{5 \lambda}{4} $ …(iii)

From (i) & (ii)

$\large l_2 -l_1 = \frac{\lambda}{2}$

$\large \lambda = 2 (l_2 -l_1) $

$\large v = \nu \lambda = 2 \nu (l_2 -l_1) $

Illustration : Where is the maximum pressure variation, when an open organ pipe of length L vibrates at in its fundamental mode ?

Solution : Maximum pressure variation is at node (However displacement is zero). Therefore maximum pressure variation is at the middle of the pipe.

Exercise : A pipe 20 cm long is closed at one end. Which harmonic mode of this pipe is resonantly excited by a 430 Hz source ? Will this same source be in resonance with the pipe whose both ends are opened ? Take speed of sound = 340 m/s.