### Stationery waves in closed and open organ pipe

closed pipe is a pipe which is closed at one end and open at other end. The sound wave gets reflected at each end of the pipe and   travels back through the pipe. This waves travelling in the opposite direction superposed and the standing waves are set up. The wavelength required for such match corresponds to the resonant frequency of the pipe.

### Harmonics and overtone in close organ pipe

The resonant vibration occurs in closed organ pipe of length ‘ L‘,when the frequency of source  equals to the  natural frequency .In closed organ pipe nodes forms at closed end and anti nodes forms at open end .

a. Fundamental mode (1st harmonic):

In this modes of vibration, nodes lies at closed end and anti nodes lies at open end.

L = λ14λ14  where λ is the wavelength of the wave.

λ1=4L

If ‘vv’ be the velocity of sound and f1f1to be the frequency of wave in this mode of vibration. Then,

We have v= f1 λ1

f1= v/ λ

On putting the value of the λ we have

f1= v/4L

This is called fundamental frequency or first harmonics.

b. Second mode (3rd harmonics)

In this mode of vibration, one more nodes lies between the nodes and antinodes as shown in figure

L = 3λ243λ24  and f2= 3v4l3v4l = 3f1

∴ f2 = 3f­1

This is called third harmonics or first overtone.

c. Third mode (5th harmonics)

In this mode of vibration, three nodes and three antinodes are formed as shown in figure.

L = 5λ245λ24   and   f2= 5v4l5v4l  = 5f1

∴ f3 = 5f­1

This is called fifth harmonics or second overtone.

Similarly it can be shown that f= 7f1, f5 = 9f­1 ……and so on.Only odd harmonics are present in the even harmonics are absent.

### Open organ pipe

An open organ pipe is pipe which is open at both ends.The air column in the tube is set into vibration such that at least antinodes lie at open end.

### Harmonics and overtone in open organ pipe

#### Fundamental mode of vibration (1st harmonic)

In the mode of vibration in the organ pipe, two antinodes are formed at two open ends and one node is formed in between them. If length of the pipe be ‘L’ and λ0 be the wavelength of wave emitted in this mode of vibration. Then,

L=2 λ0/4

Or, λ0=2L

if ‘vv’ be the velocity of sound and f1f1to be the frequency of wave in this mode of vibration. Then,

we have v= f1 λ0

f1= v/ λ0

on putting the value of the λ0 we have

f1= v/2L

This is the expression of the fundamental mode or first harmonic.

b. Second mode of vibration (2ndharmonics)

In the second mode of vibration in the open organ pipe, there antinodes are formed at two ends and two nodes between them.

If ‘L’ be the length of the pipe and λ0 be the wavelength of wave emitted in this mode of vibration. Then,

L= λ0

If vv be the velocity of sound and f2 to be the frequency of wave in this mode of vibration, then,

Then we have v= f2 λ0

Or, f2=v/ λ0

On putting the value of the λ0 we get, f2= v/L=2v/2L=2f1

This is the frequency of first overtone or 2nd harmonics.

c. Third mode of vibration:In the third mode of vibration in the open organ pipe, four antinodes are formed and three nodes between them. If ‘L’ be the length of the pipe and λ0be the wavelength of wave emitted in this mode of vibration then

We have L=3 λ0/2

Or, λ0=2L/3

ifvv be the velocity of sound and f3 be the frequency of wavelength on this mode of vibration, then,

we have v = f3 λ0

Or, f3= v/ λ0

On putting the value of λ0 we get

f3= 3v/2L=3f1and so on.

This is the fundamental of 2nd overtone & 3rd harmonics.Similarly, frequency of nth mode of the vibration of air column in an open pipe can be written as fn=nf1fn=nf1

Resonance tube experiment

This experiment is used to determine the velocity of sound in air at any temperature and pressure.

The glass tube is connected with lower end board by a rubber tube to reservoir which can be freely moved as our desire on up and down direction. Glass tube and a small portion of reservoir is filled with water. Now we take a tuning fork of known frequency and set into vibration by holding horizontally on the open mouth of the tube. The vibration of each of the projection pointed part of a fork force the air of glass tube to vibrate. Hence vibration is forced, the intensity of the sound heard. Now to hear loud sound we adjusted the water level by running reservoir up or down. As we obtained the loud sound at this condition, the frequency of fork is same as the fundamental frequency of the pipe and the length of air column in the tube at that condition is noted. This noted length is the first resonating length. Let L1 be the first resonating length and  c be the end correction and at the fundamental mode of vibration,

Similarly another process is repeated to obtain another loud sound which have same frequency that of tuning fork. This is called the second resonance. This is also called first overtone. In that condition the length of air column for this resonance is three times the length of the first resonance, and we have

L2+c=3λ/4…………….2

Subtracting eq. 1 from 2 we get

L2-L1=λ/2………3

Since we have v = f λ, v/f=λ then from equation 3we have

2f (L2-L1)=v.

Hence, determining the first and second resonating length and the frequency we can determine the velocity of sound in an air at given temperature.

The velocity of sound at NTP is given by

VNTP=Vt(P−0.375)fP−−−−−−−√(1−γt2)VNTP=Vt(P−0.375)fP(1−γt2)

Velocity of the transverse wave in the string

When a string of mass per unit length (m) under the action of tension; T; is set in to transverse vibration the velocity of the wave propagation through the string is given by V = Tm−−−√Tmwhere,m=massperunitlengthwhere,m=massperunitlength

In such a vibration of the string the disturbance produced at one fixed end travels along the string and gets reflected back at the other end, since the original wave and reflect wave have the same frequency and amplitude, they superimpose to produce stationary transverse wave.

Vibration in string and overtones

A string can be set in to the vibration if it ids rigidly fixed at two ends.

#### a.First mode of vibration:-

In this mode of vibration, the string vibration in one segment.This there are two nodes at fixed ends and an antinodes in between them. If ‘λ’ be the length of string and λ0 be the wave length of wave in this mode of    vibration.

If ‘V ‘ be the velocity of wave and to f0f0 be the frequency of wave in this mode of vibration.

f1=Vλ0=V2lf1=Vλ0=V2l  and  f0=12lTM−−√f0=12lTM

This is the fundamental frequency or frequency of 1st harmonics.

#### b. Second mode of vibration:

In this mode of vibration the string vibration in two segments.Thus, there are three nodes and two antinodes between of string and l1 be the wavelength of wave in this mode of vibration.

l = λ1

If Vbe the velocity of wave and f1 be the frequency of wave in this mode of vibration.

f1=Vλ1=2V2lf1=Vλ1=2V2l

f1=Vλ1=2V2lf1=Vλ1=2V2lAndf1=1lTM−−√f1=1lTM

∴f1=2f0

This is the frequency of 1st overtone and 2nd harmonics.

#### Third mode of vibration:

There are four nodes and three antinodes.

l=3λ22l=3λ22

λ2=2l3λ2=2l3

f2=32lTM−−√f2=32lTM

f2=3f0f2=3f0

This is the frequency of 2nd overtone or 3rd harmonics

In this way in the transverse string all the harmonics are present.

#### Laws of transverse vibration of string

The velocity of a transverse wave vibrating in a stretched string is given by v = T/μ−−−−√T/μ

Where T =tension of string and μ= mass per unit length. Since the frequency, f= v/2L in fundamental mode, then from above expression we have

f=12LT/μ−−−−√f=12LT/μ ………………….1

From equation 1 we have followed the relation and this relation brings laws of transverse vibration of stretched string.

a. law of length (first law): fundamental frequency is inversely proportional to the resonating length L of the string  i.e. f∝1L∝1L

b. The law of the tension (second law): fundamental frequency is proportional to the square root of the stretching force or tension. f∝T√f∝T

c. The law of mass (third law): the fundamental frequency is inversely proportional to the square root of the mass per unit length.f∝1μ√f∝1μ.

#### Experimental verification of the laws of transverse vibration

We verify these laws by using a sonometer. Sonometer is device consists of a wire under tension which is arranged in a hollow wooden board as shown in figure.

The vibration of the wire are passed by the movable bridges to the box and then, to the air inside it.

a.To verify (first law) f∝1L∝1L: For the verification of first law,we take a tuning fork of known frequency. Hanging a load of 0.5kg on the one end  of string now to find the resonating of the wire between the bridges C and D we placed small paper placed on the wire and noted the resonating length when the paper fall down from the wire by the vibration of tuning fork . Let L1 be first resonating length of the given known fork and the same process is repeated for next tuning fork having the different frequency than that of first one.  Let L2 be the second resonating length for the second tuning fork. Then it found that at the constant tension and constant mass per unit of string the product of f1xL1=f2xL2This show that the f∝∝1/L

b. To verify (second law) f∝T√∝T:  For the verification of the second law, the length of wire AB is fixed between the bridge C and D and the load W on the one end of the wire is varied to alter tension T. to measure the frequency of the vibration in this wire, an auxiliary wire PQ is used which runs parallel to the experiment wire AB. The tension on this wire is kept constant as shown in figure.

The bridge M and N is moved until the note on the length L of auxiliary wire between MN is same as that in the experiment wire CD. Since the tension in the wire MN is constant, we have f α 1/L. by varying the W, tension T in AB is varied. A graph between 1/L and  T√T is a straight line passing through the origin, which shows that f αT√T at constant μ and resonating length.

c. To verify (third law) f∝1μ√∝1μ: for the verification of the third law we take the wire of the different diameters are used under same vibrating length and same tension. Parallel to the experimental wire of AB, an auxiliary wire PQ with known tension is fixed on the sonometer as Shown in figure.

The experimental wire is plucked at the middle between two bridges C and D and the length of the auxiliary wire PQ is set into resonance by varying the position of the bridges N and M on the auxiliary string. Let the resonating length of the auxiliary wire be L1 and the frequency of the vibration of the experimental wire is proportional to this length. Second experiment wire is taken with the same load and same vibrating length and auxiliary wire is again made in resonance with the experiment wire. The resonating length is again notes as L2. When a graph between 1/L and μ√μ is plotted, a straight line is obtained which is passes through the origin. Since the frequency of auxiliary string, fα1/L, so the frequency of the experimental wire is f as both are in resonance shown that f α1/μ√μ

### Waves in Rod

#### Kund’s tube method

Kund’s tube method is based on the principle of setting the longitudinal stationery waves in

Rod in resonance. It is a small scale apparatus designed to determine the velocity of sound in solid, liquid and gases.

Kund’s tube apparatus:

It consists a glass tube AB length of 150 cm and diameter of   about 5 cm. Two caps c1 and cwith stopcock s1and sare fixed. A rod b is clamped at the middle point of the screw. As shown in figure.A air tight piston is p is capable of maintaining the lengthof air column in the tube.

To determine the velocity of the sound in the solid:

A solid is taken in the form of the rod and   inserted in the tube AB.  Thin layer of lycopodium powder   is spread along the length f glass tube. Longitudinal waves are set up in the rod by rubbing the piece of resin cloth c . Thus set up waves are transferred into an air column of glass tube and gets reflected by piston.  The piston is adjusted in such a way that   the air column resonates with vibration of rod.  At this condition, particles of lycopodium powder   begin to vibrate with maximum and minimum amplitude at antinodes and nodes respectively.  As a result small heaps are formed. The distance between two consecutive heaps gives the distance between the two consecutive nodes.

Let λaλa and λbλbbe the wavelength of the wave in air and in the rod .Land Lb be the distance between the two consecutive nodes and length of the rod respectively.

If Va and Vb the velocity of the wave in the air and rod.

La=λa2………………λa2………………( mean distance between  consecutive nodes )

Similarly   Lb=λbλb/2……………………2

We know that frequency = velocity /   wavelength

f=VaλaVaλa =VbλbVbλb………………..3

From 1, 2 and 3

Vb =LbLaLbLaVa ……………4

If we know the terms of R.H.S, we can determine the velocity of sound in rod