Is it correct to say: Treating every overtone as the next integer harmonic for a closed pipe (first overtone = 2n)?

A closed pipe has only odd harmonics. Its first overtone is 3n and its second overtone is 5n.

Is it correct to say: Using l + e for an open pipe?

An open pipe has two open ends, so add the correction twice: corrected length L = l + 2e = l + 0.6d.

Is it correct to say: Taking e = 0.3 × radius?

e = 0.3d where d is the inner diameter (equivalently 0.6r). Check which one the question gives.

Is it correct to say: Substituting linear density in g/m or g/cm in n = (1/2L)√(T/m)?

m must be in kg/m and L in metres. A wire of 5 g/m has m = 5×10⁻³ kg/m.

Is it correct to say: Thinking the distance between two successive nodes is λ?

Successive nodes (or antinodes) are λ/2 apart; a node and the next antinode are λ/4 apart.

Is it correct to say: Choosing n ± N at random in beat problems?

Use the clue in the question: wax loading lowers the frequency, filing the prongs raises it. Check which choice makes the beat count change the way the question says.

📄 Formula sheet (PDF)

Superposition of Waves

Progressive and stationary waves, vibrating strings, air columns, end correction, sonometer and beats — Maharashtra HSC Physics Ch 6

📐 16 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→Learn the analytical treatment of stationary waves: add y₁ = A sin 2π(nt − x/λ) and y₂ = A sin 2π(nt + x/λ), then find nodes (cosine term zero) and antinodes (cosine term ±1). It is a standard derivation question.
→For air columns, first write down which harmonics are present: a pipe closed at one end gives only odd harmonics, a pipe open at both ends gives all harmonics. Most numericals follow from this one line.
→If the question gives the diameter of the pipe, it expects end correction: e = 0.3d at each open end. The corrected length of the air column is L = l + e for a closed pipe and L = l + 2e for an open pipe, where l is the length of the pipe.
→Harmonic and overtone are not the same word. The first overtone is the next frequency actually present: the 2nd harmonic for a string or open pipe, but the 3rd harmonic for a closed pipe.
→In beats problems with an unknown fork, there are always two possible answers (n ± N). Use the extra clue — loading with wax lowers the frequency, filing raises it — to decide which one is correct.
→In sonometer numericals convert length to metres and linear density to kg/m (g/m ÷ 1000) before substituting in n = (1/2L)√(T/m).

📐 Formulas(16)

Equation of a Progressive Wave

y = A sin 2 π (n t - \frac{x}{λ}) = A sin (ω t - k x)

Symbol	Meaning
$y$	Displacement of the particle at position x and time t (m)
$A$	Amplitude (m)
$n$	Frequency (Hz)
$λ$	Wavelength (m)
$ω$	Angular frequency = 2πn (rad s⁻¹)
$k$	Propagation constant (wave number) = 2π/λ (rad m⁻¹)

Wave Speed

v = nλ = \frac{λ}{T} = \frac{ω}{k}

Symbol	Meaning
$v$	Wave speed (m s⁻¹)
$T$	Period = 1/n (s)

Phase Difference and Path Difference

Δ ϕ = \frac{2 π}{λ} Δ x

Symbol	Meaning
$Δ ϕ$	Phase difference (rad)
$Δ x$	Path difference (m)

Equation of a Stationary Wave★ Board fav

y = 2 A cos (\frac{2 π x}{λ}) sin (2 π n t)

Symbol	Meaning
$A$	Amplitude of each component wave (m)
$x$	Position along the medium (m)

Nodes and Antinodes

node to node = \frac{λ}{2}, node to adjacent antinode = \frac{λ}{4}

Fundamental Frequency of a Stretched String★ Board fav

n = \frac{1}{2 L} \frac{T}{m}

Symbol	Meaning
$L$	Vibrating length of the string (m)
$T$	Tension in the string (N)
$m$	Linear density — mass per unit length (kg m⁻¹)

Harmonics of a Stretched String

n_{p} = \frac{p}{2 L} \frac{T}{m} = p n, p = 1, 2, 3, \dots

Symbol	Meaning
$p$	Number of loops (harmonic number)

Laws of Vibrating Strings (Sonometer)

n \propto \frac{1}{L}, n \propto T, n \propto \frac{1}{m}

Pipe Closed at One End — Fundamental★ Board fav

n = \frac{v}{4 L} = \frac{v}{4 ( l + e )}

Symbol	Meaning
$v$	Speed of sound in air (m s⁻¹)
$L$	Corrected length of the air column, L = l + e (m)
$l$	Length of the pipe (m)
$e$	End correction (m)

Pipe Closed at One End — Harmonics

n_{p} = (2 p + 1) n = (2 p + 1) \frac{v}{4 L}, p = 0, 1, 2, \dots

Symbol	Meaning
$p$	Overtone number (p = 0 for the fundamental)

Pipe Open at Both Ends — Fundamental★ Board fav

n = \frac{v}{2 L} = \frac{v}{2 ( l + 2 e )}

Pipe Open at Both Ends — Harmonics

n_{p} = (p + 1) n = \frac{( p + 1 ) v}{2 L}, p = 0, 1, 2, \dots

Symbol	Meaning
$p$	Overtone number (p = 0 for the fundamental)

End Correction

e = 0.3 d

Symbol	Meaning
$e$	End correction (m)
$d$	Inner diameter of the pipe (m)

End Correction from Two Pipes (Closed at One End)

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

Symbol	Meaning
$n_{1}, n_{2}$	Fundamental frequencies of the two pipes (Hz)
$l_{1}, l_{2}$	Lengths of the two pipes (m)

Beat Frequency★ Board fav

N = n_{1} - n_{2}

Symbol	Meaning
$N$	Beats per second (Hz)
$n_{1}, n_{2}$	Frequencies of the two sources (Hz)

Speed of Sound from Beats

v = \frac{N λ _{1} λ _{2}}{λ _{2} - λ _{1}}

Symbol	Meaning
$λ_{1}, λ_{2}$	Wavelengths of the two waves, λ₁ < λ₂ (m)

✏️ Solved Examples

1Solved Exampleeasy3 steps

The equation of a progressive wave is y = 0.05 sin 2π(200t − x/1.5), where x and y are in metres and t in seconds. Find the amplitude, frequency, wavelength and speed of the wave.

Compare with the standard form y = A sin 2π(nt − x/λ).

A = 0.05 m, n = 200 Hz, λ = 1.5 m

2Solved Exampleboard4 steps

A sonometer wire 50 cm long has a linear density of 10 g/m and is stretched by a tension of 100 N. Find its fundamental frequency. To what length must the wire be reduced, at the same tension, so that it vibrates in unison with a fork of 250 Hz?

Convert to SI units.

L = 0.50 m, m = 10 \times 1 0^{- 3} = 0.01 kg m^{- 1}, T = 100 N

3Solved Exampleboard4 steps

A pipe closed at one end is 32 cm long and has an inner diameter of 4 cm. Taking the speed of sound as 340 m/s and including end correction, find the fundamental frequency and the frequency of the first overtone.

End correction for one open end.

e = 0.3 d = 0.3 \times 4 = 1.2 cm

4Solved ExampleHOTS3 steps

A tuning fork A produces 4 beats per second with a standard fork of 256 Hz. When A is loaded with a little wax, the number of beats falls to 2 per second. Find the original frequency of A.

4 beats/s means A differs from 256 Hz by 4 Hz.

n_{A} = 256 \pm 4 = 260 Hz or 252 Hz

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Treating every overtone as the next integer harmonic for a closed pipe (first overtone = 2n)
✓A closed pipe has only odd harmonics. Its first overtone is 3n and its second overtone is 5n.
2
Using l + e for an open pipe
✓An open pipe has two open ends, so add the correction twice: corrected length L = l + 2e = l + 0.6d.
3
Taking e = 0.3 × radius
✓e = 0.3d where d is the inner diameter (equivalently 0.6r). Check which one the question gives.
4
Substituting linear density in g/m or g/cm in n = (1/2L)√(T/m)
✓m must be in kg/m and L in metres. A wire of 5 g/m has m = 5×10⁻³ kg/m.
5
Thinking the distance between two successive nodes is λ
✓Successive nodes (or antinodes) are λ/2 apart; a node and the next antinode are λ/4 apart.
6
Choosing n ± N at random in beat problems
✓Use the clue in the question: wax loading lowers the frequency, filing the prongs raises it. Check which choice makes the beat count change the way the question says.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the fundamental frequency and the first overtone of a pipe 50 cm long, open at both ends. Speed of sound = 340 m/s; ignore end correction.
Q2
Two sound waves of wavelengths 1.00 m and 1.02 m in air produce 6 beats per second. Find the speed of sound in air.
Q3
In a resonance tube experiment with a 500 Hz fork, the first and second resonances occur at 16 cm and 50 cm. Find the speed of sound and the end correction.
Q4
In a stationary wave of frequency 680 Hz, the distance between two successive nodes is 25 cm. Find the speed of the wave.
Q5
Six tuning forks are arranged in increasing order of frequency. Each produces 5 beats per second with the next one, and the frequency of the last fork is 1.25 times that of the first. Find the frequencies of the first and last forks.
Q6
A pipe closed at one end and a pipe open at both ends have the same fundamental frequency. Find the ratio of their lengths (closed : open), ignoring end correction.

📝 Notes

Superposition of Waves — Maharashtra HSC Overview

Chapter 6 of the Maharashtra Board Std XII Physics textbook builds everything from one idea: when two waves meet, the resultant displacement is the algebraic sum of the individual displacements. Stationary waves, the modes of strings and air columns, and beats are all applications of this principle.

From progressive to stationary waves

A progressive wave $y = A\sin 2\pi(nt - x/\lambda)$ carries energy forward. Add an identical wave moving the other way and you get $y = 2A\cos(2\pi x/\lambda)\sin(2\pi nt)$ . Here the amplitude depends on position and nothing travels. Nodes are $\lambda/2$ apart, and energy stays trapped between them.

Strings and pipes: decide the end conditions first

Every vibration question starts with one question: what is at each end?

String fixed at both ends: node–node, $L = p\lambda/2$ , all harmonics present.
Pipe closed at one end: node–antinode, $L = (2p+1)\lambda/4$ with $p = 0, 1, 2, \ldots$ , odd harmonics only.
Pipe open at both ends: antinode–antinode, $L = (p+1)\lambda/2$ with $p = 0, 1, 2, \ldots$ , all harmonics present.

Draw the loop pattern, read off $\lambda$ in terms of $L$ , then use $n = v/\lambda$ . For pipes, $L$ is the corrected length of the air column: $L = l + e$ (closed) or $L = l + 2e$ (open), where $l$ is the length of the pipe and $e = 0.3d$ .

Beats

Two sounds of slightly different frequencies give a loudness that rises and falls $n_1 - n_2$ times per second. Beats are used to find an unknown frequency against a standard fork. When you write $n = n_0 \pm N$ , keep both possibilities until the loading or filing clue rules one out.

Approach tips

Write the given frequency symbol as $n$ , as the Maharashtra textbook does, and keep $T$ for tension clearly separate from period.
For the sonometer, state the three laws with the quantity held constant. Statements without the condition are incomplete.