Waves in pipes and strings - Class 12 Physics

Open Organ Pipe and   Correction in Pipes

Stationary Waves in an Open Organ Pipe

An open pipe is one which is opened at both ends. When air was blown into the pipe through one end, a wave travels through the tube to the next end from where it is reflected. Due to a superposition of the incident and reflected waves, a stationary wave is set up in the air in the pipe.

Fundamental Mode (First Harmonic)

Let length of pipe be L and the velocity of sound in air is v. in this mode of vibration, there are antinodes at open end and one node at the middle of the pipe as shown in the figure. Let λ be the wavelength of the waves. Then,

$$ \begin{align*} L &= \frac {\lambda }{2} \\ \text {or,} \: \lambda &= 2L \\ \text {Thus the frequency of fundamental mode or first harmonic is} \\ \text {or,} \: f_1 &= \frac {v}{\lambda } = \frac {v}{2L} \dots (i) \\ \end{align*} $$

Overtones in Open Organ Pipe

If a stronger blast of air is blown into the pipe, notes of higher frequencies are obtained which are called overtones. Two overtones are discussed below.

Second Mode (First Overtone)

In this mode of vibration, two antinodes are at the open ends but inside the pipe, there are two nodes and one antinode as shown in figure. If λ be the wave length of the wave, then

$$ \begin{align*} L &= \lambda \\ \text {Thus, the frequency of first over tone or second harmonic is} \\ f &= \frac {v}{\lambda } = \frac {v}{L} = 2\frac {v}{2L} = 2f_1 \\ \therefore f_2 &= f_1 \dots (ii) \\ \end{align*} $$ 

Third Mode {Second Overtone)

In this mode of vibration, two antinodes are produced at both open ends and inside the pipe there are three nodes and two antinodes. If λ be the wave length of the wave then,

$$ \begin{align*} L &= \frac {3\lambda }{2} \\ \text {or,} \: \lambda &= \frac {2L}{3} \\ \text {Thus, the frequency of second overtone or third harmonic is } \\ f_3 &= \frac {v}{\lambda } = \frac {3v}{2L} = 3 \frac {v}{2L} \\ \therefore f_3 &= 3f_1 \dots (iii) \\ \end{align*} $$

In this way, other higher modes of vibration can be obtained. From equation (i), (ii) and (iii), it is found that frequencies of higher modes of vibration are integral multiple of fundamental frequency f₁. So, all harmonics are possible in an open pipe.

Conclusions

1. The frequencies of various modes of vibration are an integral multiple of the fundamental frequency.
2. The frequency of the first overtone is two times the fundamental frequency; the frequency of the second overtone is three times the fundamental frequency. So the frequency of n^th overtone is (n + 1) times the fundamental frequency.
3. All harmonics are present.
4. Since sound produced by open end organ pipe contains all harmonics, so it is richer in quality than produced by closed end organ pipe.
5. The fundamental frequency of an open pipe is twice that of a closed of the same length.
6. The note from an open pipe is richer than that of a closed pipe owing to the presence of extra overtone.

End Correction in Pipes

In organ pipes, we take the antinode exactly at the open end and calculation of L is made accordingly. However, the antinode actually lies a little outside the open end because the air just outside the open end is set into vibration. So the displacement antinode of a stationary wave occurs a little distance beyond the end. The distance between the antinode and the open end of the pipe is called the end correction. The end correction, denoted by ‘e’, in a closed pipe as in figure is given by

$$ \begin{align*} L + e &= \frac {\lambda }{4} \\ \text {or,} \: \lambda &= 4(L + e) \\ \text {where L is length of the pipe.} \\ \text {In case of an open organ pipe sounding in fundamental mode as in figure,} \\ \text {The end correction is given by} \\ L + e + e &= \frac {\lambda }{2} \\ \text {or,} \: L + 2e = \frac {\lambda }{2} \\ \text {or,} \: \lambda &= 2(L + 2e) \\ \end{align*} $$

The end correction is needed at two ends in an open pipe. The end correction determined mathematically is e = 0.58 r or 0.6 r ; where r is radius. The temperature and end correction affect on the frequency of sound in a pipe. For example, take a closed pipe sounding in a fundamental mode. The frequency of the mode is

$$ \begin{align*} f &= \frac {v}{\lambda } = \frac {v}{4(L + e)} \\ \text {The velocity of sound v, at temperature} \: \theta \: \text {in terms of velocity} \: v_o \: \text {at} \: 0^o\: C\: \\\text {given by} \\ \frac {v}{v_o} &= \sqrt {\frac {T_o}{T}} \\ &= \sqrt {\frac {273 + \theta }{273}} \\ &= \sqrt {1 + \frac {\theta }{273}} \\ \text {Then,} f &= \frac {4}{4(L + e)} \\ &= \frac {v_o}{4(L + e)} \sqrt {1 + \frac {\theta }{273}} \\ \end{align*} $$

So, the frequency of fundamental mode increases as the temperature rises. It follows that for a given temperature and the length of pipe, the frequency decreases as e increases.

Resonance and Velocity of Transverse Wave along a Stretched String

Resonance

 

When a body capable of vibration is displaced and then allowed to vibrate freely, it will vibrate with a frequency which is called the natural frequency. If an external periodic force is applied on the body and the body vibrates with the frequency of the force, the motion is called the forced vibration. Forced vibration leads to resonance. Resonance is the specific response of a system which is capable of vibrating with a certain frequency to an external force acting with the same frequency. For example, A suspension bridge has its own natural frequency. If the frequency of vibration coming out of marching soldier through the bridge is equal to the natural frequency of the bridge, then the bridge will vibrate violently with large amplitude and may collapse. That’s why the soldiers are ordered to break the steps while crossing the bridge.

The two conditions for occurring the resonance are

1. The frequency of applied force must be equal to the natural frequency of the system,
2. The applied force must be equal to the vibrating system.

The graph of the amplitude of vibration and frequency of forced vibration is shown below. The amplitude is large if the damping is small and vice versa.

Resonance air column tube: measurement of velocity of sound

The apparatus consists of a glass tube of length about one-meter and diameter 4 cm fitted on a vertical board with a meter scale attached to it. The tube is connected at its lower end by a rubber tube to reservoir which can be slided up and down. The tube and a part of the reservoir is filled with water. A tuning fork of known frequency is set into vibration and held horizontally above the mouth of the tube. The vibrating prongs of the tuning fork force the air in the tube to vibrate. As the vibration is forced, the frequency of tuning fork is same as the fundamental frequency of the pipe and the air inside the pipe is set into resonance by the periodic force. The length of air column in the tube is noted which is the first resonating length denoted by L₁. Let 'e' be the end correction and at the fundamental mode of vibration,

$$ \begin{align*} L_1 + e &= \frac {\lambda }{4} \\ \text {or,}\: \lambda &= 4(L_1 + e) \dots (i) \\ \end{align*} $$

Now, the length of air column is increased till another loud sound is heard with the same tuning fork.This is called the second resonance and this corresponds to the first overtone. Length of air column for is resonance is three times the length of the first resonance. So,

$$ \begin{align*} L_2 + e &= \frac {3\lambda }{4} \\ \text {or,} \: 3\lambda &= 4(L_2 + e) \dots (ii) \\ \text {Subtracting equation} \: (i)\: \text {from equation} \: (ii), \text {we have} \\ 2\lambda &= 4(L_2 – L_1) \\ \text {or,} \: \lambda &= 2(L_2 –L_1) \\ \text {Knowing the frequency of the tuning fork,} \\ \text { v can be calculated as} \\ v &= f\lambda = 2f(L_2 – L_1) \dots (iii) \\ \end{align*} $$

$$ \begin{align*} L_1 + e &= \frac {\lambda }{4} = \frac {v}{4f} \\ \text {or,} L_1 &= \frac {v}{4f} – e \end{align*} $$ The velocity of sound can be calculated at 0^oC using the relation as $$ \begin{align*} v_o &= v\sqrt {\frac {T_o}{T}} \\ &= v\sqrt {\frac {273}{273 + \theta }} \\ \end{align*} $$ At STP, velocity of aound can be calculated with correction of humidity as $$ \begin{align*} v_o &= \sqrt {\frac {P -0.35\times f}{p}}\sqrt {\frac {273}{273 + \theta }} \end{align*} $$ where f is the aqueous tension or saturated vapour pressure of water at lab temperature \(\theta \).

Velocity of Transverse Wave along a Stretched String

Suppose a stretched string. Consider a single symmetric pulse moving from left to right along the string with speed, v as shown in the figure. For convenience, we can consider a frame of reference in which the pulse remains stationary and then, we run along the pulse, keeping it constantly in view. In this frame, the string appears to move from right to left with the speed v.

Consider a small segment AB of the pulse of length DL forming an arc of a circle of radius R and subtending an angle 2θ at the centre, O of the circle. A tension T in the string pulls tangentially on this segment at each end. The horizontal component Tcos θ at two ends of the pulse cancel each other while the vertical components add to form a radial restoring force F given by

$$ \begin{align*} F &= 2(T\sin\theta ) 2T\theta = T2\theta \\ \text {where } \: \sin \theta = \theta \: \text {for small angle.} \\ \text {The small angle can be written as} \: 2\theta = \Delta L/R.\\ \text {If} \: \mu \: \text {is the mass per unit length or linear density of the string,} \\ \text { the mass of the segment is } \\ m &= \mu \:\Delta L \\ \text {Since the string segment is moving in an arc of a circle,} \\ \text { this force acts as a centripetal force } \\ \text {producing a centripetal acceleration towards the centre of the circle} \\ \text {given by} \\ a &= \frac {v^2}{R} \\ \text {So the centripetal force,} \\ F &= \frac {mv^2}{R} \\\text {or,} \: T2\theta &= \frac {\mu \Delta Lv^2}{R} \\ \text {or,} \: \frac {T\Delta L}{R} &= \frac {\mu \Delta Lv^2}{R} \\ \text {or,}\: T &= \mu \: v^2 \\ \text {or,} \: v &= \sqrt {\frac {T}{\mu }} \\ \end{align*} $$ The frequency of the wave is fixed entirely by whatever generates the wave as, $$ \begin{align*} v &= \lambda f \\ \text {or,} \: f &= \frac {v}{\lambda } = \frac {1}{\lambda } \sqrt {\frac {T}{\mu }} \\ \end{align*} $$

Waves and Laws of Transverse Vibration in Stretched String

Waves in Stretched Strings

A string is a tight wire. When it is plucked or bowed, progressive transverse waves travel along the wire and is reflect at the fixed ends. These waves superpose with the incident waves and produce a stationary wave in the wire. A progressive sound wave is produced in the surrounding air having a frequency equal to that of the stationary wave in the string.

Modes of Vibration

A stretched string can produce different frequencies. Since the ends are fixed, these are the position of nodes in the wire. When the string is plucked at the middle, an antinode is formed at the middle. This is the simplest mode of vibration and the distance between the consecutive nodes is λ/2 where λ is the wavelength of the transverse wave in the string.

$$ \begin{align*} L &= \lambda /2 \\ \text {or,} \: \lambda &= 2L \\ \text {The frequency of vibration is given by} \\ f &= \frac {v}{\lambda } = \frac {v}{2L} \\ \end{align*} $$

where v is the velocity of the transverse wave. This is fundamental frequency or frequency of first harmonic. It is the lowest frequency produced by the vibrating string.

Overtone in Stretched String

If the string is plucked at a point one-quarter of its length from one end, the string vibrates in two segments. This mode of vibration is called the first overtone. This vibration can be also be set when the vibrating antinodes are formed in the string as shown in the figure.

If λ₁ is wavelength and f₁ is the frequency of the resulting stationary wave, we have

$$ \begin{align*} L &= \frac {\lambda }{2} + \frac {\lambda }{2} = \lambda \\ \text {The frequency of the wave,} \: f_1 &= \frac {v}{\lambda } = \frac v L = 2f \\\end{align*} $$

Thus the frequency of the first overtone is two times the fundamental frequency. This is also called second harmonics.If the string is made to vibrate in three segments by touching it at one-third of the length from one end, additional nodes are produced in it.

$$ \begin{align*} \text {If } \: \lambda \: \text {is the wavelength and} \: f_2 \: \text {its frequency of the wave, then} \\ L &= \frac {\lambda }{2} + \frac {\lambda }{2} + \frac {\lambda }{2} = \frac {3\lambda }{2} \\ \text {or,} \: \lambda &= \frac {2L}{3} \\ \text {and the frequency is given by} \\ f_2 &= \frac {v}{\lambda } = \frac {3v}{2}L = 3f \\ \end{align*} $$

Hence, the frequency of second overtone is three times the fundamental frequency which is also called third harmonics. Similarly we can obtain other overtone in the same string with more segments. The ratio of the frequency of string is

$$ f: f_1 : f_2 : f_3:\dots = 1: 2: 3\dots $$

Laws of Transverse Vibration in Stretched String

The velocity of a transverse wave travelling in a stretched string is given by

$$ v = \sqrt {\frac {T}{\mu }} $$

where T is the tension in the stretched string and µ, the mass per unit length. Since the frequency, f = v/2L in fundamental mode, then

$$ f = \frac {1}{2L}\sqrt {\frac {T}{\mu }} $$

From this expression, it follows that there are three laws of transverse vibration of stretched string;

1. The length of length: The fundamental frequency is inversely proportional to the resonating length, L of the string.
   $$ f \propto \frac {1}{L} $$
2. The law of tension: The fundamental frequency is directly proportional to the square root of the stretching force or tension.
   $$ f \propto \sqrt {T} $$
3. The law of mass: The fundamental frequency is inversely proportional to the square root of the mass per unit length.

$$ f\propto \frac {1}{\sqrt {\mu }} $$

 

Verification of the Laws of Transverse Vibration

These laws can be verified experimentally using a sonometer. This device consists of a wire under tension which is arranged in a hollow wooden board as shown in the figure. The vibration of the wire are passed by the movable bridges to the box and then, to the air inside it.

1. To verify \( f \propto 1/L\)
   To verify this law, take a tuning fork of known frequency, such as 320 Hz. Taking a load of 1 kg on the string, find the resonating length of the wire between the bridges C and D. this is found by using a small paper on the wire. Let L₁ be the resonating length for this tuning fork. The same process is repeated for next tuning fork having a different frequency. Let L₂ resonating length for the second tuning fork. It will be found that the product \(f_1 \times L_1 = f_2 \times L_2 \) at constant tension on and mass per unit length of the string. This follows that
   $$F \propto \frac {1}{L} $$
   
   
2. To verify \(F \propto \sqrt {T} \)
   To verify this law, a length L of experimental wire AB is fixed between bridges C and D, and load W on it is varied to alter tension T. to measure the frequency of vibration in this wire, an auxiliary wire PQ is used which runs parallel to the experimental wire AB. The tension on this wire is kept constant which is not shown in a figure. The bridges M and N are moved until the note on length l of auxiliary wire between MN is same as that in experimental wire CD. Since the tension in the wire MN is constant, we have \(f \propto 1/l \). By varying W, the tension I in AB is varied. A graph between \(1/l\) and \(\sqrt {T}\) is a straight line passing through the origin, which shows that
   $$ F \propto \sqrt {T} $$
   at constant µ and resonating length.
   
   
3. To verify \( f \propto 1/\sqrt {\mu} \)
   To verify this law, the wires of different diameters and materials are used under same vibrating length and same tension. Parallel to the experimental wire AB, an auxiliary wire PQ with known tension is fixed on the sonometer as shown in a 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 M and N on the auxiliary string, let the resonating length of the auxiliary wire be L₁ and the frequency of vibration of the experimental wire is proportional to the length. The second experimental wire is taken with the same load and same vibrating length and the auxiliary wire is again made in resonance with the experimental wire. The resonating length is again noted as L₂. When a graph between \(1/L \) and \(\sqrt {\mu } \) is plotted, a straight line is obtained which passes through the origin.
   $$ f \propto 1/\sqrt {\mu} $$

Waves in Rods

The modes of vibration in a metal rod depend on the position at which the rod is clamped. When the rod is clamped tightly at the middle, and is struck along the length at a free end, stationary longitudinal waves are formed on the rod as the waves in an open pipe. Since the ends are free to vibrate, displacement antinodes are formed at the ends of the rod and a displacement node at the middle as all the atoms are at rest there. The wave in the length of the rod in the simplest mode of vibration have the length given by

$$ \begin{align*} L &= \lambda /2 \\ \text {or,} \: \lambda &= 2L \\\end{align*} $$

If v is the velocity of the wave, the frequency is given by

$$ \begin{align*}\\ f &= v/\lambda = v/2L \\ \end{align*} $$

Kundt’s Dust Tube Experiment

Kundt’s dust tube consists of a glass tube PQ. A cork B is a attached at one end of the tube with a wooden handle which can be moved inside the tube. A rod EA of given material for which the velocity of sound is to be measured is clamped at the middle. The front end of the rod has an aluminium disc attached to it. The tube contains air at room temperature and little lycopodium powder is sprinkled along the length of the tube as shown in the figure.

Applications

Velocity of Sound in Solids

When the rod is stroked at E by a resigned at E by a resigned cloth gently in the direction of EA, the longitudinal vibrations are set up in the rod. The free and E of the rod acts as a source of the same frequency, and sound wave travels along the tube through the air which is reflected at the fixed end B, the vibrations are transferred to the air column. The cork B in the tube can be adjusted such that the air column resonates with the vibrations of the rod and a stationary wave in the tube causes the licopodium powder to become violently agitated. The distance between the consecutive nodes is determined by measuring the distance between several of them and taking the average of them.

Let the length of the rod be l_rand the mean distance between the consecutive heaps be l_a. Then

$$ l_r= \frac {\lambda _r}{2} \: \text {and} \: l_a = \frac {\lambda _a}{2}$$

Let v_r and v_a be the velocity of sound in rod and air respectively. So for the rod, frequency $$ \begin{align*}\\ f &= \frac {v_r}{\lambda _r} = \frac {v_r}{2l_r} \dots (i) \\ \text {and for the air column,} \\ f &= \frac {v_a}{\lambda _a} = \frac {v_a}{2l_a} \dots (ii)\\ \text {Equation} \: (i) \: \text {and} \: (ii), \: \text {we have} \\ \frac {v_r}{2l_r} &= \frac {v_a}{2l_a} \\ \therefore v_r &= v_a \left [\frac {l_r}{l_a}\right ] \dots {iii} \\\end{align*} $$

Thus, knowing v_a, l_rand l_a, the velocity of sound in rod can be calculated.

Determination of Young’s modulus of a rod

Velocity of sound in a medium is given by

$$ \begin{align*} v &= \sqrt {\frac {Y}{\rho }} \\ \text {where Y is Young’s modulus of the medium and} \: \rho , \\ \text {its density. If } \: v_r \: \text { is the velocity of sound in the rod, then} \\ v_r &= \sqrt {\frac {Y}{\rho }} \\ \text {or,} Y &= v_r^2 \times \rho \dots (iv) \\ \text {Thus, knowing} \: v_r \: \text {and} \rho \: \text {Y can be calculated.} \\ \end{align*} $$

Velocity of Sound in Gases

The experiment is performed first with air and then with the required gas, using same rod. Suppose the distance between two consecutive heaps in the air column is l_a and the mean distance between two consecutive heaps in the gas column is l_g.

$$ \begin{align*} \text {From equation} \: (iii), \: \text {we have} \\ v_r &= v_a \left [\frac {l_r}{l_a}\right ]\dots (v) \\ v_r &= v_g \left [\frac {l_r}{l_g}\right ]\dots (vi) \\ \text {Equation} \: (v) \: \text {and} \: (vi), \: \text {we have} \\ v_g \left [\frac {l_r}{l_g}\right ]\ &= v_r \left [\frac {l_r}{l_a}\right ]\\ \text {or,} \: v_g &= v_a \left [\frac {l_g}{l_a}\right ]\dots (vii)\\ \end{align*} $$

$$\text{Thus, knowing} \: v_a, l_g\: \text {and} \: l_a, \: \text {the velocity of sound in a gas can be calculated.}$$

1. Determination of ratio of molar heat capacities of a gas and its molecular structure:
   Velocity of sound in a gas is given by
   $$ \begin{align*} v_g &= \sqrt {\frac {\gamma P}{\rho }} \\ \text {where} \: \gamma \text {is the ratio of two molar heat capacities} \: (C_p/C_v) \: \text {of a gas,} \\ \text {P is pressure and} \: \rho \: \text {is its density.} \\ v_g^2 &= \frac {\gamma P}{\rho } \\ \therefore \gamma &= \frac {v_g^2 \rho }{P} \\ \text {Thus, knowing} \: v_g, \rho\: \text {and P}, \: \gamma \text { can be calculated.} \end{align*} $$
2. To find Bulk modulus (K)
   $$ \begin{align*} v_g &= \sqrt {\frac {K}{\rho }} \\ \text {or,} \: \left ( \frac {l_g}{l_a} \right ) v_a &= \sqrt {\frac {K}{\rho }} \\ \text {or,} \: \left ( \frac {l_g}{l_a} \right )^2 \times v_a^2 &= \frac {K}{\rho } \\ \therefore K &= \left ( \frac {l_g}{l_a} \right )^2 \times v_a^2 \times \rho \\ \end{align*}$$ \text{Thus, knowing} \: l_g, l_a, v_a\: \text {and }, \: \rho , \:\text {K can be calculated.}$$

Velocity of Sound in Liquid

Velocity of sound in a liquid can be determined by filling the tube PQ with the liquid and fine sand or metal fillings spread along the length of the tube instead of lycopodium powder. The experiment for velocity of sound in solids is repeated and the distance between two consecutive heaps of sand or metal fillings is determined. Let l₁ be the distance between two consecutive heaps in liquid and l_a be the distance between two consecutive heaps in air, then we get

$$ V_l = \left ( \frac {l_g}{l_a} \right ) v_a $$

Thus knowing the values of v_a , l_l and l_a, v_l can be calculated.

Stationary Waves in a Closed Organ Pipe

Musical sounds can be produced by oscillating strings, air columns, membrane, steel bars and many other oscillating bodies. In most of the instruments, more than a single part take part in oscillation such as strings and body f violin vibrate producing musical sound.

Organ Pipe

A hollow wooden or metallic tube used to produce sound is called an organ pipe. It is wind instrument such as a flute, whistle, violin, clarinet etc. their air column in it set into vibrations by blowing air into it from one end. If both ends of the pipe are open, it is called an open organ pipe; flute is an example pipe.

 

Figure (a) is a closed organ pipe. When air is blown from the mouthpiece M, a jet of air strikes a sharp edge L oscillating the air there and progressive sound waves travel along the pipe towards the closed end. The waves are reflected from the end and the reflected waves interfere with incident waves producing stationary waves in the pipe. Sound waves of other frequencies die out, so organ pipe can produce notes of definite frequencies depending on its length. Stationary waves are formed in an open pipe as shown in figure (b) due to reflection at the open end.

Stationary Waves in a Closed Organ Pipe

Consider a closed organ pipe of length L as shown in the figure (a). A blast of air is blown into it at the open end and a wave thus travels through the pipe and is reflected at the next end. Due to a superposition of incident and reflected waves, stationary waves are produced. In the simplest mode of vibration, there is a displacement node, N at the closed end air is at rest there and a displacement antinode, An at the open end as the air can vibrate freely.

Fundamental Mode

In this mode of vibration, the pipe has one node at its closed end and one antinode at its open end. As observed in figure (a), the length, L of the pipe is equal to distance between a node and an antinode which is λ/4 where λ is the wavelength of the stationary wave. Then

$$ \begin{align*} L &= \frac {\lambda }{4} \\ \text {or,} \: \lambda &= 4L \\ \text {If v is the velocity of sound and} \: f_1 \: \text {is the frequency of vibration,} \\ \text {we have} \\ v &= \lambda f_1 \\ \text {or,} \: f_1 &= \frac {v}{\lambda } = \frac {v}{4L} \\ \therefore f_1 &= \frac {v}{4L} \dots (i)\\ \end{align*} $$

It is the lowest frequency produced in the pipe. It is called fundamental frequency or first harmonic.

Overtones in Closed Pipe

If a stronger blast of air is blown into the pipe, the notes of higher frequencies are obtained. The two nodes of vibration in the same pipe as shown in the figure which is called the first overtone and second overtone.

Second Mode (first overtone)

In this mode of vibration, there are two antinodes and two nodes within the pipe. If λ is the wavelength of the wave, then we have

$$ \begin{align*} L &= \frac {3\lambda }{4} \\ \text {or,} \: \lambda &= \frac {4L}{3} \\ \text {The frequency of vibration,} \: f_2 \: \text {is given by} \\ v &= \lambda f_2 \\ \text {or,} \: f_2 &= \frac {v}{\lambda } = \frac {v}{4L/3} = \frac {3v}{4L} = 3f_1 \\ \therefore f_2 &= 3f_1 \dots (ii) \\ \end{align*} $$

This is the frequency of first overtone or third harmonic. The frequency of the first over tone is three times the fundamental frequency.

Third Mode (Second Overtone)

In this node of vibration, there are three nodes and three antinodes within the pipe. If λ is wavelength of the wave, then

$$ \begin{align*} L &= \frac {5\lambda }{4} \\ \text {or,} \: \lambda &= \frac {4L}{5} \\ \text {If} \: f_3 \: \text {is frequency of second overtone, then} \\ v &= \lambda f_3 \\ \text {or,} \: f_3 &= \frac {v}{\lambda } = \frac {v}{4L/5} = 5\frac {v}{4L} = 5f_1 \\ \therefore f_3 &= 5f_1 \dots (iii) \\ \end{align*} $$

This is the frequency of second overtone or fifth harmonic which is five times the fundamental frequency.

In this way, other higher modes of vibration can be obtained. From equations (i), (ii) and (iii), it is observed that the frequency of higher modes of vibration is odd integral multiples of fundamental frequency f₁. That is,

$$ \begin{align*} \: f:f_1, \: 3f_1, \: 5f_1, \: 7f_1, \dots \dots \end{align*} $$

Conclusions

1. The frequencies of various harmonics are odd integral multiples of fundamental frequency.
2. The frequency of the first overtone is three times the fundamental frequency, frequency of the second overtone is five times the fundamental frequency. So the frequency of n^th overtone is (2n + 1) times the fundamental frequency.
3. The even harmonics are missing.
4. Since only odd harmonics are present, the sound quality is poor.

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