Question

show that in a closed end organ pipe only odd harmonies can be produced.​

Answer 1

hlo mate your solution

Consider a closed pipe of length L, which encloses certain specific amount of air called as air column. This air column is set into vibrations by holding a vibrating tuning fork near its mouth. These longitudinal waves of frequency u wavelength l travel with a

velocity of v through the pipe and gets reflected at the other end, because the other end acts like a boundary and this reflected wave traveling in opposite direction superpose with incident wave forming a stationary wave such that a node is formed at the closed end and an antinodes at the open end, which is characterized by a set of natural

frequencies called as normal modes of oscillation. The first 3 nodes of oscillations are:-

The stationary waves formed is given by

y = 2a sin kx cos w t where 2a sin Kx gives its amplitude, therefore the positions of nodes

are given by sin Kx = 0

and hence Kx = np where n = 0, 1, 2, 3,

since K =2p/ l; we get , x =nl/2

where n = 0, 1, 2, 3,

In the same way, the positions of the antinodes are given by

Sin Kx = 1

hence Kx =(n+1/2) p where n = 0, 1, 2, 3, .

Since K= 2p/l ; we get, x = (n+1/2) l/2 where n = 0, 1, 2, 3,

Taking the closed end of the pipe to be x = 0, the condition for node is satisfied and the other end of the pipe to be x = L where an antinodes is formed, requires that the length L to be related to wavelength l given by L = (n+1/2) l /2 for n = 0, 1, 2, 3.

Thus, the possible wavelengths of stationary wave formed in different modes of oscillation are given by

l =2L/(n+1/2)

where n = 0, 1, 2, 3, ..and corresponding frequencies are

given by u = (n+1/2)v/2L

where n = 0, 1, 2, 3,.

Thus, for fundamental mode of oscillation (or) I harmonic n = 0, and

corresponding frequency is given by

u1 =V/4L

For second mode of oscillation n =1, corresponding frequency is given by u2 =3V/4L ; this is called as I overtone (or) III harmonic.

Similarly, for third mode of oscillation n =2, corresponding frequency is given by u3 =5V/4L ; this is called as II overtone (or) V harmonic and so on

Therefore, u1 : u2 : u3 : .= 1 : 3 : 5:.

Thus, a closed pipe produces odd harmonics i.e., the ratio of frequencies of overtone to that of fundamental frequency are odd natural numbers.

Answer 2

HLO MY FRND HERE IS UR ANSWER....

I THINK U SHOULD CORRECT UR QUESTION..

THERE IS A ERROR.... THAT'S NOT HARMONIES THAT IS HARMONICS ..

THEN UR ANSWER..

In a closed pipe system,the harmonics of a sound wave will always end at a node, just like the harmonicson a string end at a node, it is impossible for a standing wave to end at its antinodes,it makes no sense unless its an opened end pipe. Therefore the lengthsof harmonics of closed pipes are as follows:

Harmonic 1: L=lamda/4

Harmonic 2: L=3lamda/4

Harmonic 3: L=5lamda/4

For the first harmonic,the node appears at a quarter of the wavelength.

For the second harmonic,the node appears at three quarters of the wavelength and so on...

The frequency (f) from v=lamda(f) is equal to f=v/lamda.

For the first harmonic ,L=lamda/4 then lamda=4L.

f_1=lamda/4L.

For the second harmonic, L=3lamda/4 so lamda=4L/3

f_2=lamda/(4L/3)=3lamda/4L

You notice a pattern as you go on,we can write the harmonics of a closed pipe as

f_n= n lamda/4L.

where, n=2k+1 and k is any positive integer.

HOPE THIS HELPS U I THINK....!!!