Stationary Waves in Air Column

Q: How is a stationary wave formed?

A stationary wave forms when two waves of the same frequency and amplitude travel in opposite directions and superpose, usually a wave and its reflection. The result, y = 2A cos kx sin omega t, does not travel: nodes stay at rest and antinodes oscillate with amplitude 2A.

Q: What is the distance between a node and an antinode?

Adjacent nodes, and adjacent antinodes, are half a wavelength apart. A node and the next antinode are a quarter wavelength apart. So a string or pipe length can always be counted in quarter wavelengths to find its allowed modes.

Q: Why does a closed organ pipe produce only odd harmonics?

The closed end must be a displacement node and the open end an antinode, so the pipe length must be an odd number of quarter wavelengths. This gives f = (2n minus 1) v over 4L: only odd multiples of the fundamental are possible.

Q: What is the difference between a harmonic and an overtone?

A harmonic is any whole-number multiple of the fundamental frequency; the fundamental is the first harmonic. Overtones are the higher frequencies actually produced, numbered from the first above the fundamental. In an open pipe the first overtone is the second harmonic; in a closed pipe it is the third harmonic.

Q: What is end correction in an organ pipe?

The displacement antinode at an open end lies slightly outside the pipe, about 0.6 r beyond the end, where r is the pipe's radius. So a closed pipe behaves as if its length were L + 0.6 r and an open pipe as if it were L + 1.2 r.

Q: How is the speed of sound found with a resonance tube?

A fork of known frequency is held over the tube and the water is lowered until the first and second resonances are heard at air-column lengths l1 and l2. They differ by half a wavelength, so v = 2 f (l2 minus l1); the end correction cancels out.

Q: Which stationary wave topics come in JEE Main?

JEE Main asks for frequencies of strings and closed and open pipes, counting harmonics below a limit, matching overtones of two systems, end correction, resonance tube calculations, and reading a standing-wave equation for nodes, wavelength and amplitude at a point.

Q: What should NEET students remember about organ pipes?

For NEET remember: closed pipe f = (2n minus 1) v over 4L with odd harmonics only, open pipe f = n v over 2L with all harmonics, the open pipe fundamental is twice that of a closed pipe of equal length, and the resonance tube gives v = 2 f (l2 minus l1).

PhysicsWavesFor JEE aspirants

Stationary waves form when two identical waves travel in opposite directions and superpose: $y = 2 A cos k x sin ω t$ . Points called nodes never move and antinodes swing the most. Stationary waves in an air column decide the notes of organ pipes and flutes: a pipe closed at one end gives only odd harmonics, $f = (2 n - 1) v /4 L$ , and an open pipe gives all harmonics, $f = n v /2 L$ . This page covers the theory, strings, closed and open pipes, end correction, the resonance tube and Kundt's tube. Asked every year in JEE Main and NEET.

On this page1Formation2Nodes and antinodes3Equations4Energy in a loop5Strings6Sound: pressure view7Closed pipe8Open pipe9End correction10Resonance tube11Kundt's tube

Key Formulas - Quick Reference

★ Must learnStationary wave: $y = 2 A cos k x sin ω t$ ; amplitude $R = 2 A ∣ cos k x ∣$
Antinodes ( $R = 2 A$ ): $x = 0, \frac{λ}{2}, λ, \dots$ ; nodes ( $R = 0$ ): $x = \frac{λ}{4}, \frac{3 λ}{4}, \dots$ ; $N$ to $N$ $= \frac{λ}{2}$
Energy in one loop (component amplitude $A$ ): $E = \frac{1}{2} μ ω^{2} A^{2} λ$
★ Must learnString fixed at both ends: $f_{n} = \frac{n v}{2 L} = \frac{n}{2 L} \frac{T}{μ}$ , all harmonics
String fixed at one end, free at the other: $f = \frac{( 2 n - 1 ) v}{4 L}$ , odd harmonics
★ Must learnClosed pipe: $f = \frac{( 2 n - 1 ) v}{4 L}$ (odd harmonics); open pipe: $f = \frac{n v}{2 L}$ (all harmonics)
Pressure node $=$ displacement antinode (open end); pressure antinode $=$ displacement node (closed end)
★ Must learnEnd correction $e = 0.6 r$ : closed $f = \frac{v}{4 ( L + 0.6 r )}$ , open $f = \frac{v}{2 ( L + 1.2 r )}$
★ Must learnResonance tube: $λ = 2 (l_{2} - l_{1})$ , $v = 2 f (l_{2} - l_{1})$ , $e = \frac{l _{2} - 3 l _{1}}{2}$
Kundt's tube: $\frac{v _{rod}}{v _{air}} = \frac{l _{0}}{l _{a}}$ ; rod clamped at middle: $f = \frac{2 n - 1}{2 l} \frac{Y}{ρ}$

1. Formation of Stationary Waves

When two coherent waves of equal amplitude and frequency travel in opposite directions in the same region, every particle there is under both at once. The result does not travel: it is a stationary (standing) wave. In practice the second wave is usually the reflection of the first from a boundary (the fixed end of a string or the end of a pipe).

Take $y_{1} = A sin (ω t - k x)$ (towards $+ x$ ) and $y_{2} = A sin (ω t + k x)$ (towards $- x$ ).
Add, using $sin (a - b) + sin (a + b) = 2 sin a cos b$ :
$y = y_{1} + y_{2} = 2 A cos k x sin ω t$
Write this as $y = R sin ω t$ with $R = 2 A cos k x$ : every particle performs SHM of the same frequency $ω$ , but its amplitude $R$ depends on its position $x$ .

Points	Condition	Positions	Amplitude
Antinodes	$cos k x = \pm 1$ , $k x = nπ$	$x = 0, \frac{λ}{2}, λ, \frac{3 λ}{2}, \dots$	$2 A$ (maximum)
Nodes	$cos k x = 0$ , $k x = (2 n + 1) \frac{π}{2}$	$x = \frac{λ}{4}, \frac{3 λ}{4}, \frac{5 λ}{4}, \dots$	$0$ (always at rest)

Figure 1:

y = 2 A cos k x sin ω t

at several instants. Nodes (

N

) never move; antinodes (

A

) swing between

\pm 2 A

N

N

= λ /2

A

N

= λ /4

. The shape does not travel.

If the two component waves carry equal energy, there is no net flow of energy through the region: energy is only redistributed, gathering at the antinodes (constructive interference) and vanishing at the nodes (destructive interference).

2. Motion of the Particles

Let all particles be at their extreme positions at $t = 0$ . They move towards the mean position, all cross it together at $t = T /4$ (the string is momentarily straight), reach the other extreme at $t = T /2$ , cross back at $t = 3 T /4$ , and return at $t = T$ . Nodes stay at rest throughout.

Figure 2: All particles reach their extremes together and cross the mean position together (the string is straight twice per period). Particles in one loop move in phase; adjacent loops move in opposite phase.

Some points are always at rest (nodes), some oscillate with the largest amplitude (antinodes); all others have amplitudes in between.
All particles between two successive nodes oscillate in the same phase; particles on opposite sides of a node oscillate in opposite phase. So the phase difference between any two particles is $0$ or $π$ .
In each period the medium becomes a straight line twice.
With equal component amplitudes, no energy is transported; energy only shifts between kinetic (all at the mean position) and potential (all at the extremes).

Travelling wave	Stationary wave
Advances through the medium with a definite speed	Stays between the boundaries; the pattern does not move
All particles have the same amplitude	Amplitude varies with position: zero at nodes, maximum at antinodes
Phase changes continuously from particle to particle (any value from $0$ to $2 π$ )	Same phase within a loop; opposite across a node (only $0$ or $π$ )
Particles never all pass the mean position together	All particles pass the mean position together, twice per period
Transmits energy	Does not transmit energy

Key idea

A stationary wave is an SHM of every particle with a position-dependent amplitude; nodes

λ /2

apart, antinodes midway.

JEE Advanced

Energy is trapped between nodes. Nodes never move, so no power crosses them and each loop keeps its energy. For $y = 2 A sin k x cos ω t$ on a string of linear density $μ$ , when the string is straight all the energy is kinetic:

E_{loop} = \int_{0}^{λ /2} \frac{1}{2} μ (2 A ω sin k x)^{2} d x = \frac{μ A ^{2} ω ^{2} λ}{2}

A quarter period later the string is at its extremes, at rest, and the same energy is all potential (stretch). The two component waves carry equal and opposite power $\frac{1}{2} μ ω^{2} A^{2} v$ , which cancel.

3. Different Forms of the Equation

If the component waves have a phase difference, the standing wave shifts along $x$ . For example $y_{1} = A sin (ω t - k x)$ and $y_{2} = - A sin (ω t + k x)$ (the reflection from a fixed end at $x = 0$ ) give $y = - 2 A sin k x cos ω t$ , with a node at the origin. In general, with $A_{0} = 2 A$ the antinode amplitude:

At $x = 0$ there is	Use	Particle amplitude
a node	$y = A_{0} sin k x cos (ω t + ϕ)$ or $A_{0} sin k x sin (ω t + ϕ)$	$R = A_{0} ∣ sin k x ∣$
an antinode	$y = A_{0} cos k x sin (ω t + ϕ)$ or $A_{0} cos k x cos (ω t + ϕ)$	$R = A_{0} ∣ cos k x ∣$

Exam Trick

Read the pattern, then write the equation. Choose the $x$ part from what sits at the origin ( $sin k x$ for a node, $cos k x$ for an antinode); get $k$ from the loop length ( $loop = λ /2$ ); get $ϕ$ from where the particles are, and which way they move, at $t = 0$ . Solved Examples 1 and 2 do exactly this.

3.1 Energy in one loop

When every particle of a loop passes its mean position, all the energy of the loop is kinetic. A particle at $x$ then has speed $2 A ω sin k x$ (node at the origin). Summing over one loop of mass per length $μ$ :

E = \int_{0}^{λ /2} \frac{1}{2} μ d x (2 A ω sin k x)^{2} = 2 μ A^{2} ω^{2} \cdot \frac{λ}{4} = \frac{1}{2} μ ω^{2} A^{2} λ

The energy of a loop stays constant; it only changes form between kinetic and potential during each cycle.

4. Stationary Waves on Strings

4.1 Both ends fixed

A wave sent along a string stretched between two fixed points reflects at each end. The incident and reflected waves form a stationary wave, and both ends must be nodes. So the length holds a whole number of loops: $L = n \frac{λ}{2}$ .

Figure 3: Both ends fixed (nodes).

L = n \frac{λ}{2}

, so

f_{n} = \frac{n v}{2 L}

: all harmonics occur. The

n

-th harmonic is the

(n - 1)

-th overtone.

★ Must learn

f_{n} = \frac{n v}{2 L} = \frac{n}{2 L} \frac{T}{μ}, n = 1, 2, 3, \dots

n = 1

is the fundamental (first harmonic);

f_{2} = 2 f_{1}

is the second harmonic or first overtone;

f_{3} = 3 f_{1}

the third harmonic or second overtone. For a string fixed at both ends, every harmonic is an allowed overtone.

Laws of vibrating strings (from $f_{1} = \frac{1}{2 L} T / μ$ ): $f \propto \frac{1}{L}$ at fixed $T$ and $μ$ ; $f \propto T$ ; $f \propto \frac{1}{μ}$ . A sonometer verifies them, and a musician tunes a string by changing $T$ and plays notes by changing $L$ .

4.2 One end fixed, one end free

If one end is fixed (node) and the other is free (antinode, a ring sliding on a rod), the length holds an odd number of quarter wavelengths: $L = (2 n - 1) \frac{λ}{4}$ .

Figure 4: One end fixed (node), one free (antinode).

L = (2 n - 1) \frac{λ}{4}

, so

f = \frac{( 2 n - 1 ) v}{4 L}

: only odd harmonics.

f = \frac{( 2 n - 1 ) v}{4 L} = \frac{2 n - 1}{4 L} \frac{T}{μ}, n = 1, 2, 3, \dots

Only odd harmonics occur: the first overtone is the third harmonic.

5. Stationary Sound Waves: the Pressure Picture

Two sound waves of the same frequency travelling in opposite directions also form a stationary wave. Written for excess pressure, with $p_{1} = p_{0} sin (ω t - k x)$ and $p_{2} = p_{0} sin (ω t + k x + ϕ)$ :

p = 2 p_{0} cos (k x + \frac{ϕ}{2}) sin (ω t + \frac{ϕ}{2})

Pressure nodes (pressure always normal) are where $cos (k x + ϕ /2) = 0$ ; pressure antinodes (largest pressure swing, $\pm 2 p_{0}$ ) where $cos (k x + ϕ /2) = \pm 1$ . Because pressure and displacement differ by $π /2$ (Displacement and Pressure Wave concept), a pressure node is a displacement antinode, and a pressure antinode is a displacement node.

Figure 5: Third harmonic of a closed pipe. Displacement (solid) and pressure (dashed) envelopes are shifted by

λ /4

: every displacement node is a pressure antinode and vice versa.

Boundary	Displacement	Pressure	Reflected pressure wave
Rigid (closed end of a pipe)	Node: air cannot move	Antinode	Same phase: compression returns as compression
Open end of a pipe	Antinode: air moves freely	Node: pressure stays atmospheric	Phase change $π$ : compression returns as rarefaction

These are the sound versions of a fixed and a free string end (Superposition concept). A particle at a rigid wall cannot vibrate, so the displacement wave is inverted there, but the pressure there swings most, so the pressure wave reflects without inversion.

6. Vibration of Air in a Closed Organ Pipe

Hold a vibrating tuning fork near the open end of a pipe closed at the other end. The air column resonates when a stationary wave fits with a node at the closed end and an antinode at the open end. The lowest frequency (longest wavelength) has no other node or antinode in between: $L = \frac{λ}{4}$ .

Figure 6: Closed pipe (open at top). Closed end

N

, open end

A

L = \frac{λ}{4}, \frac{3 λ}{4}, \frac{5 λ}{4}

. Only odd harmonics:

f = \frac{( 2 n - 1 ) v}{4 L}

Figure 7: The air above the water is a closed pipe: node

N

at the water, antinode

A

at the mouth,

f = \frac{v}{4 L}

. With

v = 340 m s^{- 1}

L = 25

20

15 cm

give about

340

425

and

567 Hz

(end correction ignored). This is why the pitch rises as a vessel fills with water.

Fundamental: $L = \frac{λ}{4}$ , $f_{1} = \frac{v}{4 L}$ .
First overtone: $L = \frac{3 λ ^{'}}{4}$ , $f = \frac{3 v}{4 L} = 3 f_{1}$ (third harmonic).
Second overtone: $L = \frac{5 λ ^{''}}{4}$ , $f = \frac{5 v}{4 L} = 5 f_{1}$ (fifth harmonic).
In general
$f = \frac{( 2 n - 1 ) v}{4 L}$
the $(2 n - 1)$ -th harmonic and the $(n - 1)$ -th overtone.

A closed pipe resonates only at odd harmonics of its fundamental.

7. Vibration of Air in an Open Organ Pipe

In a pipe open at both ends, both ends are antinodes. The lowest mode has one node in the middle: $L = \frac{λ}{2}$ .

Figure 8: Open pipe. Both ends

A

L = \frac{λ}{2}, λ, \frac{3 λ}{2}

. All harmonics:

f = \frac{n v}{2 L}

(twice the fundamental of a closed pipe of the same length).

Fundamental: $f_{1} = \frac{v}{2 L}$ .
First overtone: $L = λ^{'}$ , $f = \frac{v}{L} = 2 f_{1}$ (second harmonic).
Second overtone: $L = \frac{3 λ ^{''}}{2}$ , $f = 3 f_{1}$ (third harmonic).
In general
$f = \frac{n v}{2 L}$
the $n$ -th harmonic and the $(n - 1)$ -th overtone.

Closed pipe (one end closed)

Closed end $N$ , open end $A$ . $f_{1} = \frac{v}{4 L}$ . Only odd harmonics: $f_{1}, 3 f_{1}, 5 f_{1}, \dots$ The $n$ -th overtone is the $(2 n + 1)$ -th harmonic.

Open pipe (both ends open)

Both ends $A$ . $f_{1} = \frac{v}{2 L}$ , twice a closed pipe of the same length. All harmonics: $f_{1}, 2 f_{1}, 3 f_{1}, \dots$ Richer sound, sweeter quality.

Exam Trick

Count quarter-wavelengths. Every mode is "number of $λ /4$ segments $= L$ ". Closed pipe and fixed-free string: odd numbers ( $1, 3, 5$ ). Open pipe and fixed-fixed string: even numbers ( $2, 4, 6$ ). Then $f = \frac{( number ) v}{4 L}$ . One formula for all four systems.

Natural oscillations of organ pipes. Which harmonics sound depends on how the air is excited. Blowing gently across a pipe gives mainly the fundamental; blowing harder (or raising the air pressure) brings in higher frequencies and the pipe jumps to an overtone.

Quick Recall: tap to check

A closed pipe and an open pipe have the same length. Ratio of their fundamental frequencies?

\frac{v /4 L}{v /2 L} = 1 : 2

Which harmonics are missing in a closed pipe?

All even harmonics.

At the closed end of a pipe, is there a pressure node or a pressure antinode?

A pressure antinode (displacement node).

Key idea

Closed pipe: odd harmonics only,

f_{1} = \frac{v}{4 L}

. Open pipe: all harmonics,

f_{1} = \frac{v}{2 L}

, twice the closed pipe of the same length.

Figure 9: Look only at the two ends: same type gives

f_{n} = \frac{n v}{2 L}

(all harmonics); different types give

f = \frac{( 2 n - 1 ) v}{4 L}

(odd harmonics only). Then add

0.6 r

per open end of a pipe.

8. End Correction

The air just outside an open end still takes part in the vibration, so the displacement antinode sits slightly outside the open end. The distance of the antinode from the end is the end correction:

Figure 10: The antinode sits about

e = 0.6 r

beyond the open end. Closed pipe:

f = \frac{v}{4 ( L + 0.6 r )}

; open pipe (two ends):

f = \frac{v}{2 ( L + 1.2 r )}

e = 0.6 r, f_{closed} = \frac{v}{4 ( L + 0.6 r )}, f_{open} = \frac{v}{2 ( L + 1.2 r )}

r

is the radius of the pipe. An open pipe has two open ends, so it gets two end corrections.

9. Resonance Tube

The resonance tube measures the speed of sound in air (and compares the frequencies of two tuning forks). A long tube $T$ is filled with water whose level can be raised or lowered by moving a reservoir $R$ . The air column above the water acts as a closed pipe, with the water surface as the closed end.

Strike a fork of known frequency $f$ gently on a rubber pad and hold it over the open end.
Lower the water slowly. At a length $l_{1}$ of the air column the sound becomes loudest: the first resonance, with a node at the water surface.
Lower the water further. The next loud sound comes at length $l_{2}$ : the second resonance.
Including the end correction: $l_{1} + e = \frac{λ}{4}$ and $l_{2} + e = \frac{3 λ}{4}$ . Subtracting,
$l_{2} - l_{1} = \frac{λ}{2}, v = f λ = 2 f (l_{2} - l_{1})$
The end correction cancels, which is why the method is accurate. Also $e = \frac{l _{2} - 3 l _{1}}{2}$ .

Figure 11: Resonance tube.

l_{1} + e = \frac{λ}{4}

and

l_{2} + e = \frac{3 λ}{4}

, so the end correction cancels:

λ = 2 (l_{2} - l_{1})

v = 2 f (l_{2} - l_{1})

, and

e = \frac{l _{2} - 3 l _{1}}{2}

10. Kundt's Tube and the Clamped Rod

10.1 Rod clamped at its middle

A rod $A B$ clamped at its middle $C$ and struck at one end vibrates longitudinally. The clamp is a node and the free ends are antinodes. In the fundamental, $l = \frac{λ}{2}$ :

Figure 12: Rod clamped at the middle: clamp

N

, free ends

A

f = \frac{( 2 n - 1 )}{2 l} \frac{Y}{ρ}

: odd harmonics only (envelopes show displacement along the rod).

f_{0} = \frac{v}{2 l} = \frac{1}{2 l} \frac{Y}{ρ}, f = \frac{( 2 n - 1 )}{2 l} \frac{Y}{ρ}

Only odd harmonics occur, as the middle must always be a node.

10.2 Kundt's tube

Kundt's tube finds the speed of sound in a gas or in a solid. A glass tube holds a thin layer of lycopodium powder along its length. A rod $M$ of the material under test is clamped at its middle $C$ and carries a light disc $A$ inside the tube; a piston $B$ on a handle closes the other end. Stroking the rod with a resined cloth makes it vibrate in its fundamental mode, and the disc sets up a stationary wave in the air. Adjusting the piston for resonance, the powder collects in heaps at the displacement nodes, half an air wavelength apart.

Figure 13: Kundt's tube. Rod and air vibrate at the same

f

\frac{v _{rod}}{v _{air}} = \frac{λ _{rod}}{λ _{air}} = \frac{l _{0}}{l _{a}}

The rod and the air vibrate at the same frequency: $f = \frac{v _{rod}}{2 l _{0}} = \frac{v _{air}}{2 l _{a}}$ , where $l_{0}$ is the rod length and $l_{a}$ the distance between heaps. So $\frac{v _{rod}}{v _{air}} = \frac{l _{0}}{l _{a}}$ , and with $v_{rod} = Y / ρ$ the Young's modulus of the rod can be found.

Quick Recall: tap to check

A closed pipe and an open pipe have the same length. What is the ratio of their fundamentals?

1 : 2

(closed

\frac{v}{4 L}

, open

\frac{v}{2 L}

What is the phase difference between particles in adjacent loops of a standing wave?

π

: they always move in opposite directions.

Why does a resonance tube use

l_{2} - l_{1}

The end correction cancels:

l_{2} - l_{1} = \frac{λ}{2}

exactly.

Figure 14: Revision map: standing waves, strings, closed and open pipes, end correction and Kundt's tube.

11. Solved Examples

Solved Example 1

Find the equation of the standing wave on a string between

x = 0

and

x = L

with nodes at both ends and at the middle (the second harmonic in Figure 3).
(A)

A sin (2 π x / L) cos ω t

(B)

A sin (π x / L) cos ω t

(C)

A cos (π x /2 L) cos ω t

(D)

A cos (π x / L) cos ω t

Solution:

Two loops fill $L$ , so $λ = L$ and $k = \frac{2 π}{L}$ . General form: $y = A sin (k x + θ) cos ω t$ . A node at $x = 0$ needs $sin θ = 0$ , so $θ = 0$ .

Answer: (A) $y = A sin \frac{2 π x}{L} cos ω t$ .

Solved Example 2

A standing wave on a string has antinode amplitude

2 A

and a node at

x = 0

. At

t = 0

the particle at the first antinode is at

y = A

and moving towards its mean position. Find the equation of the standing wave.

Solution:

Let $y = R sin (ω t + ϕ)$ with $R = 2 A sin (k x + θ)$ . Node at $x = 0$ : $2 A sin θ = 0$ , so $θ = 0$ .

At the antinode, $2 A sin ϕ = A$ gives $ϕ = \frac{π}{6}$ or $\frac{5 π}{6}$ . Moving towards the mean position means $∣ y ∣$ is decreasing: $cos ϕ < 0$ , so $ϕ = \frac{5 π}{6}$ ( $= \frac{π}{2} + \frac{π}{3}$ ).

Answer: $y = 2 A sin k x sin (ω t + \frac{5 π}{6})$ .

Solved Example 3

A string

120 cm

long sustains a standing wave. The points where the displacement amplitude is

3.5 mm

are equally spaced,

15.0 cm

apart. Find the maximum displacement amplitude and the wavelength.

Solution:

Points of equal amplitude $a ∣ sin k x ∣ = c$ lie at $k x = \pm θ + nπ$ ; their spacings alternate between $\frac{2 θ}{k}$ and $\frac{π - 2 θ}{k}$ . These are equal only if $θ = \frac{π}{4}$ , and then the spacing is $\frac{λ}{4}$ . (Measuring from an antinode, $a ∣ cos k x ∣$ , gives the same condition.)

So $a sin \frac{π}{4} = 3.5 mm$ : $a = 3.5 2 \approx 4.95 mm$ . And $\frac{λ}{4} = 15 cm$ : $λ = 60 cm$ (four loops in $120 cm$ ).

Answer: maximum amplitude $\approx 4.95 mm$ ; $λ = 60 cm$ .

Solved Example 4

A tube of length

48 cm

is open at both ends. Its fundamental frequency is

320 Hz

and the speed of sound is

320 m s^{- 1}

. Estimate the diameter of the tube. One end is now closed. Find the lowest resonant frequency.

Solution:

$λ = \frac{v}{f} = \frac{32000}{320} = 100 cm$ . For an open tube the antinodes lie $0.6 r$ outside each end: $\frac{λ}{2} = 48 + 2 (0.6 r)$ , so $50 = 48 + 1.2 r$ and $r \approx 1.67 cm$ . Diameter $D = 2 r \approx 3.33 cm$ .

Closed at one end: $\frac{λ ^{'}}{4} = 48 + 0.6 r = 49 cm$ , so $λ^{'} = 196 cm$ and $f = \frac{32000}{196} \approx 163.3 Hz$ .

Answer: $D \approx 3.33 cm$ ; lowest frequency $\approx 163 Hz$ .

Solved Example 5

A standing wave on a string is

y = (4 cm) sin (\frac{π x}{15}) cos (96 π t)

, with

x

in cm and

t

in s. Find (a) the amplitude and speed of the component waves, (b) the positions of the nodes, (c) the amplitude of the particle at

x = 5 cm

Solution:

(a) Component amplitude $= 4/2 = 2 cm$ . $k = \frac{π}{15} cm^{- 1}$ gives $λ = 30 cm$ ; $ω = 96 π$ gives $f = 48 Hz$ ; $v = f λ = 1440 cm s^{- 1} = 14.4 m s^{- 1}$ .

(b) Nodes where $sin (π x /15) = 0$ : $x = 0, 15, 30, \dots cm$ .

(c) $R = 4 sin \frac{5 π}{15} = 4 sin \frac{π}{3} \approx 3.46 cm$ .

Solved Example 6

1.0 m

string of linear density

0.01 kg m^{- 1}

is fixed at both ends under a tension of

100 N

. Find the fundamental frequency and the frequency of the second overtone.

Solution:

$v = T / μ = 100/0.01 = 100 m s^{- 1}$ ; $f_{1} = \frac{v}{2 L} = 50 Hz$ .

Second overtone $=$ third harmonic $= 3 f_{1} = 150 Hz$ .

Answer: $50 Hz$ and $150 Hz$ .

Solved Example 7

A pipe

85 cm

long is closed at one end. How many of its natural frequencies lie below

1250 Hz

? (

v = 340 m s^{- 1}

)

Solution:

$f_{1} = \frac{v}{4 L} = \frac{340}{4 \times 0.85} = 100 Hz$ . Allowed: $100, 300, 500, 700, 900, 1100 Hz$ (odd harmonics); the next, $1300 Hz$ , is above the limit.

Answer: 6.

Solved Example 8

The first overtone of a closed pipe of length

L_{c}

has the same frequency as the first overtone of an open pipe. Find the length of the open pipe.

Solution:

Closed pipe first overtone: $\frac{3 v}{4 L _{c}}$ . Open pipe first overtone: $\frac{2 v}{2 L _{o}} = \frac{v}{L _{o}}$ .

Equate: $\frac{3}{4 L _{c}} = \frac{1}{L _{o}}$ , so $L_{o} = \frac{4 L _{c}}{3}$ .

Answer: $L_{o} = \frac{4}{3} L_{c}$ .

Solved Example 9

In a resonance tube experiment with a

512 Hz

fork, the first and second resonances occur at

16.0 cm

and

49.5 cm

. Find the speed of sound and the end correction.

Solution:

$v = 2 f (l_{2} - l_{1}) = 2 \times 512 \times 0.335 \approx 343 m s^{- 1}$ .

$e = \frac{l _{2} - 3 l _{1}}{2} = \frac{49.5 - 48.0}{2} = 0.75 cm$ (so the tube's radius is about $0.75/0.6 \approx 1.25 cm$ ).

Answer: $343 m s^{- 1}$ ; $e = 0.75 cm$ .

Solved Example 10

In Kundt's tube, a

1.0 m

rod clamped at its middle gives powder heaps

6.5 cm

apart in air (

v = 340 m s^{- 1}

). Find the frequency and the speed of sound in the rod.

Solution:

$λ_{air} = 2 \times 6.5 = 13 cm$ ; $f = \frac{340}{0.13} \approx 2615 Hz$ .

$v_{rod} = v_{air} \frac{l _{0}}{l _{a}} = 340 \times \frac{100}{6.5} \approx 5.23 \times 1 0^{3} m s^{- 1}$ .

Answer: $f \approx 2.6 kHz$ ; $v_{rod} \approx 5.2 km s^{- 1}$ .

Solved Example 11

The tension in a sonometer wire is made four times and its length is doubled. The fundamental frequency
(A) doubles
(B) halves
(C) stays the same
(D) becomes four times

Solution:

$f \propto \frac{T}{L}$ : $\frac{4}{2} = 1$ .

Answer: (C).

Practice Questions

Find the fundamental and first overtone of an open pipe $50 cm$ long ( $v = 340 m s^{- 1}$ ).Answer: $340 Hz$ and $680 Hz$ .
A closed pipe $25 cm$ long ( $v = 340 m s^{- 1}$ ): find its first two resonant frequencies.Answer: $340 Hz$ and $1020 Hz$ .
A $60 cm$ string vibrates in 3 loops at $300 Hz$ . Find $λ$ and the wave speed.Answer: $λ = 40 cm$ , $v = 120 m s^{- 1}$ .
First resonance in a resonance tube with a $500 Hz$ fork is at $17 cm$ (ignore end correction). Find $v$ .Answer: $v = 4 f l_{1} = 340 m s^{- 1}$ .
Where are the nodes of $y = 2 sin (\frac{π x}{10}) cos (20 π t)$ ( $x$ in cm)?Answer: $x = 0, 10, 20, \dots cm$ .
Two open pipes of lengths $L$ and $2 L$ : ratio of fundamental frequencies?Answer: $2 : 1$ .

Common Mistakes to Avoid

Watch out

Taking node-to-node distance as $λ$ . It is $λ /2$ ; node to antinode is $λ /4$ .
Giving a closed pipe even harmonics. A closed pipe (and a fixed-free string) has only odd harmonics.
Calling $2 f_{1}$ the second overtone. The $n$ -th harmonic is the $(n - 1)$ -th overtone for all-harmonic systems; for a closed pipe the first overtone is $3 f_{1}$ .
Putting a pressure node at the closed end. The closed end is a displacement node and a pressure antinode.
Using one end correction for an open pipe. It has two: $L + 1.2 r$ .
Forgetting that the end correction cancels in $l_{2} - l_{1}$ for the resonance tube.
Using the component amplitude as the antinode amplitude. The antinode amplitude is $2 A$ .
Expecting a stationary wave to carry energy. With equal components it does not; energy only redistributes.

Frequently Asked Questions

How is a stationary wave formed?

A stationary wave forms when two waves of the same frequency and amplitude travel in opposite directions and superpose, usually a wave and its reflection. The result, $y = 2 A cos k x sin ω t$ , does not travel: nodes stay at rest and antinodes oscillate with amplitude $2 A$ .

What is the distance between a node and an antinode?

Adjacent nodes, and adjacent antinodes, are half a wavelength apart. A node and the next antinode are a quarter wavelength apart. So a string or pipe length can always be counted in quarter wavelengths to find its allowed modes.

Why does a closed organ pipe produce only odd harmonics?

The closed end must be a displacement node and the open end an antinode, so the pipe length must be an odd number of quarter wavelengths, $L = (2 n - 1) λ /4$ . This gives $f = (2 n - 1) v /4 L$ : only odd multiples of the fundamental are possible.

What is the difference between a harmonic and an overtone?

A harmonic is any whole-number multiple of the fundamental frequency; the fundamental is the first harmonic. Overtones are the higher frequencies actually produced, numbered from the first above the fundamental. In an open pipe the first overtone is the second harmonic; in a closed pipe it is the third harmonic.

What is end correction in an organ pipe?

The displacement antinode at an open end lies slightly outside the pipe, about $0.6 r$ beyond the end, where $r$ is the pipe's radius. So a closed pipe behaves as if its length were $L + 0.6 r$ and an open pipe as if it were $L + 1.2 r$ .

How is the speed of sound found with a resonance tube?

A fork of known frequency is held over the tube and the water is lowered until the first and second resonances are heard at air-column lengths $l_{1}$ and $l_{2}$ . They differ by half a wavelength, so $v = 2 f (l_{2} - l_{1})$ ; the end correction cancels out.

Which stationary wave topics come in JEE Main?

JEE Main asks for frequencies of strings and closed and open pipes, counting harmonics below a limit, matching overtones of two systems, end correction, resonance tube calculations, and reading the equation $y = A sin k x cos ω t$ for nodes, wavelength and amplitude at a point.

What should NEET students remember about organ pipes?

For NEET remember: closed pipe $f = (2 n - 1) v /4 L$ with odd harmonics only, open pipe $f = n v /2 L$ with all harmonics, the open pipe fundamental is twice that of a closed pipe of equal length, and the resonance tube gives $v = 2 f (l_{2} - l_{1})$ .

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