Class 12 Physics Chapter 8: Waves in Pipes and Strings Notes (NEB New Syllabus) | Organ Pipes, Stretched String, Harmonics, Numerical Problems & Solutions | Mechanical Waves PDF
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1. Syllabus

Unit: Waves and Optics

Chapter 8: Waves in Pipes and Strings (6 hours)

8.1 Stationary waves in closed and open pipes
8.2 Harmonics and overtones in closed and open organ pipes
8.3 End correction in pipes
8.4 Velocity of transverse waves along a stretched string
8.5 Vibration of string and overtones
8.6 Laws of vibration of a fixed string

These notes follow the NEB Class 12 Physics syllabus (+2 Physics, Nepal). Practise the multiple choice questions first, then work through the numericals, and finish with the PDF notes for quick revision.

2. Multiple Choice Questions

Question 1. What is the fundamental frequency of a closed organ pipe of length \(l\) if the speed of sound in air is \(v\)?

  • a) \(\dfrac{v}{2l}\)
  • b) \(\dfrac{v}{l}\)
  • c) \(\dfrac{v}{4l}\)
  • d) \(\dfrac{2v}{l}\)

Correct Answer: c) \(\dfrac{v}{4l}\)

Reason: A closed pipe has a node at the closed end and an antinode at the open end. In the fundamental mode the pipe length is one quarter of a wavelength:

\[ l=\frac{\lambda}{4}\;\Rightarrow\;\lambda=4l,\qquad f=\frac v\lambda=\frac{v}{4l} \]

Question 2. Which of the following statements is true for an open organ pipe?

  • a) It produces only even harmonics.
  • b) It produces only odd harmonics.
  • c) It produces both even and odd harmonics.
  • d) It does not produce harmonics.

Correct Answer: c) It produces both even and odd harmonics.

Reason: An open pipe has antinodes at both ends. Its fundamental frequency is \(f_1=\dfrac{v}{2l}\) and the other modes are

\[ f_n=n\left(\frac{v}{2l}\right)=nf_1,\qquad n=1,2,3,\dots \]

Since \(n\) can be any integer, both even and odd harmonics are present.


Question 3. The end correction in an organ pipe is due to:

  • a) Friction of air molecules
  • b) Effect of temperature
  • c) Effect of pressure difference
  • d) Air bulging beyond the pipe’s end

Correct Answer: d) Air bulging beyond the pipe’s end

Reason: The antinode at the open end does not form exactly at the physical end of the tube. Because of the inertia of the air, the vibrating air column bulges slightly outward, so the acoustic length of the pipe is a little longer than its geometric length. This difference is the end correction.


Question 4. The third harmonic of a closed pipe corresponds to which overtone?

  • a) First overtone
  • b) Second overtone
  • c) Third overtone
  • d) Fourth overtone

Correct Answer: a) First overtone

Reason: A closed pipe produces only odd harmonics: \(f_1,\ 3f_1,\ 5f_1,\dots\) Overtones are the frequencies higher than the fundamental, so the first frequency above the fundamental, \(3f_1\), is the first overtone.


Question 5. If the length of a vibrating string is doubled while keeping tension and linear density constant, the frequency:

  • a) Doubles
  • b) Remains the same
  • c) Becomes half
  • d) Becomes four times

Correct Answer: c) Becomes half

Reason: The fundamental frequency of a stretched string (Mersenne’s law) is

\[ f=\frac1{2l}\sqrt{\frac T\mu}\;\Rightarrow\;f\propto\frac1l \]

If \(l\) becomes \(2l\), the frequency becomes \(\dfrac f2\).


Question 6. A string fixed at both ends vibrates in its fundamental mode. What is the nature of its displacement at the centre?

  • a) Zero displacement
  • b) Maximum displacement
  • c) Half of maximum displacement
  • d) Twice the maximum displacement

Correct Answer: b) Maximum displacement

Reason: In the fundamental mode a single loop forms. The fixed ends are nodes (zero displacement) and there is one antinode, with maximum displacement, exactly at the centre of the string.


Question 7. Two strings of the same material but different lengths and diameters are stretched under the same tension. Which factor most affects their vibrational frequency?

  • a) Tension
  • b) Density
  • c) Length and diameter
  • d) Shape of the string

Correct Answer: c) Length and diameter

Reason: The linear density is \(\mu=\pi r^2\rho=\dfrac{\pi d^2}{4}\rho\). Substituting in \(f=\dfrac1{2l}\sqrt{T/\mu}\):

\[ f=\frac1{2l}\sqrt{\frac{4T}{\pi d^2\rho}}=\frac{1}{ld}\sqrt{\frac{T}{\pi\rho}} \]

Tension and material (density) are the same, so the frequency depends inversely on both the length \(l\) and the diameter \(d\).

Question 8. A flute and a guitar string produce the same frequency of sound. Which statement must be true?

  • a) They have the same mode of vibration.
  • b) They have the same amplitude.
  • c) They produce the same fundamental frequency.
  • d) They have the same length.

Correct Answer: c) They produce the same fundamental frequency.

Reason: Two instruments playing the same note have the same pitch, that is, the same fundamental frequency. They need not have the same length, mode of vibration or amplitude; they only have to be tuned to vibrate at the same base rate.


Question 9. A closed pipe has a fundamental frequency of \(220\ \mathrm{Hz}\). What will be the frequency of its first overtone?

  • a) \(220\ \mathrm{Hz}\)
  • b) \(440\ \mathrm{Hz}\)
  • c) \(660\ \mathrm{Hz}\)
  • d) \(880\ \mathrm{Hz}\)

Correct Answer: c) \(660\ \mathrm{Hz}\)

Reason: A closed pipe has only odd harmonics, so the first overtone is the third harmonic:

\[ f_3=3f_1=3\times220=660\ \mathrm{Hz} \]

Question 10. A string of length \(1.5\ \mathrm{m}\) is stretched under tension and vibrates with a fundamental frequency of \(150\ \mathrm{Hz}\). What is the speed of the wave along the string?

  • a) \(150\ \mathrm{m\,s^{-1}}\)
  • b) \(225\ \mathrm{m\,s^{-1}}\)
  • c) \(300\ \mathrm{m\,s^{-1}}\)
  • d) \(450\ \mathrm{m\,s^{-1}}\)

Correct Answer: d) \(450\ \mathrm{m\,s^{-1}}\)

Reason: In the fundamental mode \(l=\dfrac\lambda2\), so \(\lambda=2l=3.0\ \mathrm{m}\).

\[ v=f\lambda=150\times3.0=450\ \mathrm{m\,s^{-1}} \]

Question 11. The fundamental frequency of a stretched string is \(100\ \mathrm{Hz}\). If the tension is increased by a factor of 4, what will be the new frequency?

  • a) \(100\ \mathrm{Hz}\)
  • b) \(200\ \mathrm{Hz}\)
  • c) \(400\ \mathrm{Hz}\)
  • d) \(50\ \mathrm{Hz}\)

Correct Answer: b) \(200\ \mathrm{Hz}\)

Reason: By the law of tension, \(f\propto\sqrt T\):

\[ \frac{f_2}{f_1}=\sqrt{\frac{4T_1}{T_1}}=2\;\Rightarrow\;f_2=2\times100=200\ \mathrm{Hz} \]

Question 12. A closed pipe resonates at \(300\ \mathrm{Hz}\) and \(500\ \mathrm{Hz}\) but not at \(400\ \mathrm{Hz}\). What is the fundamental frequency of the pipe?

  • a) \(100\ \mathrm{Hz}\)
  • b) \(200\ \mathrm{Hz}\)
  • c) \(300\ \mathrm{Hz}\)
  • d) \(500\ \mathrm{Hz}\)

Correct Answer: a) \(100\ \mathrm{Hz}\)

Reason: A closed pipe has only odd harmonics, so the two resonances are consecutive odd harmonics, \(nf_1\) and \((n+2)f_1\). Their difference is \(2f_1\):

\[ 2f_1=500-300=200\ \mathrm{Hz}\;\Rightarrow\;f_1=100\ \mathrm{Hz} \]

Check: the harmonics are 100, 300, 500, \(\dots\) Hz.


Question 13. Two strings of the same length but different tensions are vibrating. The first string has a tension of 4 times that of the second string. If the fundamental frequency of the second string is \(f\), what is the frequency of the first string?

  • a) \(f\)
  • b) \(2f\)
  • c) \(4f\)
  • d) \(8f\)

Correct Answer: b) \(2f\)

Reason: For the same length and material, \(f\propto\sqrt T\):

\[ \frac{f_1}{f_2}=\sqrt{\frac{4T}{T}}=2\;\Rightarrow\;f_1=2f \]

Question 14. A pipe closed at one end produces a sound of wavelength \(1.2\ \mathrm{m}\) in its fundamental mode. What is the length of the pipe if the velocity of sound is \(340\ \mathrm{m\,s^{-1}}\)?

  • a) \(0.15\ \mathrm{m}\)
  • b) \(0.30\ \mathrm{m}\)
  • c) \(0.60\ \mathrm{m}\)
  • d) \(1.20\ \mathrm{m}\)

Correct Answer: b) \(0.30\ \mathrm{m}\)

Reason: In the fundamental mode of a closed pipe, the length is one quarter of the wavelength:

\[ l=\frac\lambda4=\frac{1.2}{4}=0.30\ \mathrm{m} \]

The speed of sound is not needed because the wavelength is already given.

3. Numerical and Subjective Questions

Question 1. Harmonics and overtones in organ pipes, and the vibration of a stretched string.

a) Define harmonics and overtones.

Harmonics: integral multiples of the lowest (fundamental) frequency of an oscillating system. If the fundamental is \(f_1\), the harmonics are \(f_1,2f_1,3f_1,\dots,nf_1\). The fundamental is the first harmonic.

Overtones: all the frequencies present in the vibration that are higher than the fundamental. The first frequency above the fundamental is the first overtone, the next is the second overtone, and so on.

b) Differentiate between harmonics and overtones with respect to closed and open organ pipes.

The relation depends on whether the pipe produces all harmonics or only the odd ones.

PipeFundamentalFirst overtoneSecond overtone\(n\)th overtone
Open (all harmonics)\(f_1\) (first harmonic)\(2f_1\) (second harmonic)\(3f_1\) (third harmonic)\((n+1)\)th harmonic
Closed (odd harmonics only)\(f_1\) (first harmonic)\(3f_1\) (third harmonic)\(5f_1\) (fifth harmonic)\((2n+1)\)th harmonic

c) A closed organ pipe has a fundamental frequency of \(200\ \mathrm{Hz}\). What is the length of the pipe? (Speed of sound \(=340\ \mathrm{m\,s^{-1}}\))

Given: \(f_1=200\ \mathrm{Hz},\ v=340\ \mathrm{m\,s^{-1}}\)

For a closed pipe, \(f_1=\dfrac{v}{4l}\), so

\[ l=\frac{v}{4f_1}=\frac{340}{4\times200}=\frac{340}{800}=0.425\ \mathrm{m} \]

d) A string has linear density \(0.01\ \mathrm{kg\,m^{-1}}\) and is under a tension of \(144\ \mathrm{N}\). What will be its fundamental frequency if its length is \(0.6\ \mathrm{m}\)?

Given: \(\mu=0.01\ \mathrm{kg\,m^{-1}},\ T=144\ \mathrm{N},\ l=0.6\ \mathrm{m}\)
\[ v=\sqrt{\frac T\mu}=\sqrt{\frac{144}{0.01}}=\sqrt{14400}=120\ \mathrm{m\,s^{-1}} \]
\[ f_1=\frac{v}{2l}=\frac{120}{2\times0.6}=100\ \mathrm{Hz} \]

Question 2. Stationary waves in closed and open organ pipes, and the effect of pipe length on pitch.

a) Explain the formation of stationary waves in closed and open organ pipes.

Stationary (standing) waves form when two progressive waves of the same frequency and amplitude, travelling in opposite directions, superpose. In a pipe, a sound wave travelling along the pipe is reflected at the end and the incident and reflected waves interfere.

Closed pipe: the wave is reflected at the rigid closed end. Air particles cannot move there, so a node forms at the closed end, while an antinode forms at the open end where the particles vibrate freely.

Open pipe: at the open end a compression is reflected as a rarefaction. Particles at both open ends have maximum displacement amplitude, so antinodes form at both ends.

b) How do boundary conditions affect the formation of nodes and antinodes?

Rigid or closed boundary: air molecules cannot move past a solid barrier, so the displacement must be zero and a node is formed.

Free or open boundary: air molecules are free to oscillate against the open atmosphere, so the displacement amplitude is maximum and an antinode is formed.

c) How does changing the length of an organ pipe affect the pitch of the sound produced? Explain using wave theory.

The pipe length restricts the allowed wavelengths of the stationary waves. For an open pipe \(f=\dfrac{v}{2l}\) and for a closed pipe \(f=\dfrac{v}{4l}\), so in both cases \(f\propto\dfrac1l\).

Pitch is the perception of frequency. When the length increases, the allowed wavelength increases; since \(v=f\lambda\) and \(v\) is constant in air, the frequency decreases. A longer pipe therefore gives a lower pitch and a shorter pipe gives a higher pitch.

d) An open organ pipe is \(85\ \mathrm{cm}\) long. Calculate its fundamental frequency. (Speed of sound \(=340\ \mathrm{m\,s^{-1}}\))

Given: \(l=85\ \mathrm{cm}=0.85\ \mathrm{m},\ v=340\ \mathrm{m\,s^{-1}}\)
\[ f_1=\frac{v}{2l}=\frac{340}{2\times0.85}=\frac{340}{1.7}=200\ \mathrm{Hz} \]
Question 3. Fundamental frequency and overtones of closed and open organ pipes.

a) Derive the expressions for the fundamental frequency and overtones in (i) a closed organ pipe and (ii) an open organ pipe.

Let \(l\) be the length of the pipe and \(v\) the speed of sound.

(i) Closed organ pipe (node at the closed end, antinode at the open end)

Fundamental (first harmonic): a quarter of a wave fits in the pipe.

\[ l=\frac{\lambda_1}{4}\;\Rightarrow\;\lambda_1=4l,\qquad f_1=\frac{v}{\lambda_1}=\frac{v}{4l} \]

First overtone (third harmonic): three quarters of a wave fit.

\[ l=\frac{3\lambda_3}{4}\;\Rightarrow\;\lambda_3=\frac{4l}{3},\qquad f_3=\frac{3v}{4l}=3f_1 \]

Second overtone (fifth harmonic): five quarters of a wave fit.

\[ l=\frac{5\lambda_5}{4}\;\Rightarrow\;\lambda_5=\frac{4l}{5},\qquad f_5=\frac{5v}{4l}=5f_1 \]

Only odd harmonics are present: \(f_1:f_3:f_5=1:3:5\).

(ii) Open organ pipe (antinodes at both ends)

Fundamental (first harmonic): half a wave fits (node in the middle).

\[ l=\frac{\lambda_1}{2}\;\Rightarrow\;\lambda_1=2l,\qquad f_1=\frac{v}{2l} \]

First overtone (second harmonic): one full wave fits.

\[ l=\lambda_2,\qquad f_2=\frac vl=2f_1 \]

Second overtone (third harmonic): one and a half waves fit.

\[ l=\frac{3\lambda_3}{2}\;\Rightarrow\;\lambda_3=\frac{2l}{3},\qquad f_3=\frac{3v}{2l}=3f_1 \]

All harmonics are present: \(f_1:f_2:f_3=1:2:3\).

b) A closed pipe resonates with a tuning fork of frequency \(256\ \mathrm{Hz}\) in its first harmonic. What is the length of the pipe? (\(v=343\ \mathrm{m\,s^{-1}}\))

Given: \(f_1=256\ \mathrm{Hz},\ v=343\ \mathrm{m\,s^{-1}}\)
\[ l=\frac{v}{4f_1}=\frac{343}{4\times256}=\frac{343}{1024}\approx0.335\ \mathrm{m} \]

Question 4. Harmonic content of closed and open organ pipes.

a) Why are only odd harmonics present in closed pipes, while all harmonics are present in open pipes?

A closed pipe needs a node at one end and an antinode at the other. This is possible only when the pipe length contains an odd number of quarter wavelengths (\(\lambda/4,\ 3\lambda/4,\ 5\lambda/4,\dots\)), so only the odd harmonics \(1,3,5,\dots\) times the fundamental occur.

An open pipe needs antinodes at both ends. This is possible whenever the length contains a whole number of half wavelengths (\(\lambda/2,\ 2\lambda/2,\ 3\lambda/2,\dots\)), so all harmonics \(1,2,3,\dots\) occur.

b) Compare the harmonic content in a pipe open at both ends and in a pipe closed at one end, and explain how this affects the quality of sound.

Harmonic content: an open pipe produces \(f_1,2f_1,3f_1,4f_1,\dots\) (all harmonics); a closed pipe produces \(f_1,3f_1,5f_1,7f_1,\dots\) (odd harmonics only).

Quality (timbre): the quality of a sound depends on the number and relative intensities of the overtones present. An open pipe has a richer mixture of overtones, so its sound is fuller and richer. A closed pipe lacks the even harmonics, so its tone is hollower and plainer than that of an open pipe of the same fundamental frequency.

c) An open pipe resonates in its second harmonic at \(680\ \mathrm{Hz}\). What is its length? (\(v=340\ \mathrm{m\,s^{-1}}\))

Given: \(f_2=680\ \mathrm{Hz},\ v=340\ \mathrm{m\,s^{-1}}\)

For an open pipe \(f_n=n\dfrac{v}{2l}\), so \(f_2=\dfrac vl\):

\[ l=\frac{v}{f_2}=\frac{340}{680}=0.5\ \mathrm{m} \]

d) Calculate the frequency of the third harmonic of a closed pipe of length \(0.4\ \mathrm{m}\). (\(v=340\ \mathrm{m\,s^{-1}}\))

Given: \(l=0.4\ \mathrm{m},\ v=340\ \mathrm{m\,s^{-1}}\)
\[ f_3=3f_1=\frac{3v}{4l}=\frac{3\times340}{4\times0.4}=\frac{1020}{1.6}=637.5\ \mathrm{Hz} \]

Question 5. End correction in resonance tubes.

a) Explain the concept of end correction in resonance tubes.

The antinode at the open end of a pipe does not form exactly at the geometric opening. The air just outside the pipe is partly restrained by the surrounding air, so the point of maximum displacement (the acoustic antinode) lies slightly beyond the end of the tube. This small extra distance is the end correction \(e\). It depends on the radius \(r\) of the pipe and is approximately

\[ e\approx0.6\,r \]

b) Derive the expression for the effective length of a pipe including end correction.

Let \(l\) be the geometric length of the pipe and \(e\) the end correction.

Closed pipe: there is one open end, so one correction applies and the air column behaves as if it were longer by \(e\):

\[ L=l+e \]

Open pipe: there are two open ends, and an antinode forms slightly outside each, so the correction is applied twice:

\[ L=l+2e \]

c) A resonance tube has an actual length of \(0.75\ \mathrm{m}\). If the end correction is \(0.03\ \mathrm{m}\), what is the effective length of the pipe?

Given: \(l=0.75\ \mathrm{m},\ e=0.03\ \mathrm{m}\)

A resonance tube is closed at one end, so

\[ L=l+e=0.75+0.03=0.78\ \mathrm{m} \]

(If the tube were open at both ends, \(L=l+2e=0.81\ \mathrm{m}\).)

Question 6. Velocity of transverse waves and the laws of vibration of a stretched string.

a) Mention the factors affecting the velocity of transverse waves along a stretched string.

The velocity depends on two factors:

(i) Tension \(T\): the restoring force on the string. A larger tension gives a faster wave.
(ii) Linear density \(\mu\): the mass per unit length (the inertia). A heavier string gives a slower wave.
\[ v=\sqrt{\frac T\mu} \]

b) Explain the laws of vibration of a stretched string.

The laws of transverse vibration (Mersenne’s laws) follow from the fundamental frequency \(f=\dfrac1{2l}\sqrt{\dfrac T\mu}\):

Law of length: \(f\propto\dfrac1l\) when \(T\) and \(\mu\) are constant.
Law of tension: \(f\propto\sqrt T\) when \(l\) and \(\mu\) are constant.
Law of linear density: \(f\propto\dfrac1{\sqrt\mu}\) when \(l\) and \(T\) are constant.

c) How do length, tension and mass per unit length affect the fundamental frequency of transverse vibration of a string?

From \(f=\dfrac1{2l}\sqrt{\dfrac T\mu}\):

Length: a longer string has a proportionally lower frequency (lower pitch).
Tension: the frequency rises in proportion to the square root of the tension (a tighter string has a higher pitch).
Mass per unit length: the frequency falls in proportion to the inverse square root of \(\mu\) (a thicker, heavier string has a lower pitch).

d) A string of length \(1.2\ \mathrm{m}\) has a mass of \(0.01\ \mathrm{kg}\) and a tension of \(100\ \mathrm{N}\). Find the velocity of the transverse wave.

Given: \(l=1.2\ \mathrm{m},\ m=0.01\ \mathrm{kg},\ T=100\ \mathrm{N}\)
\[ \mu=\frac{m}{l}=\frac{0.01}{1.2}=\frac1{120}\approx0.00833\ \mathrm{kg\,m^{-1}} \]
\[ v=\sqrt{\frac T\mu}=\sqrt{100\times120}=\sqrt{12000}\approx109.5\ \mathrm{m\,s^{-1}} \]

Question 7. Stationary waves and harmonics on a string fixed at both ends.
NNNNAAA

Figure: Third harmonic of a string fixed at both ends (N = node, A = antinode)

a) A string is vibrating in its third harmonic. Sketch the wave pattern and explain the number of nodes and antinodes formed.

The sketch is shown in the figure above. In the third harmonic the string vibrates in 3 segments (loops).

For a string fixed at both ends and vibrating in its \(n\)th harmonic, there are \(n\) loops. The fixed ends are always nodes.

Nodes: the two fixed ends plus two points inside the string, so there are 4 nodes.

Antinodes: one at the centre of each of the 3 loops, so there are 3 antinodes.

In general, the \(n\)th harmonic has \(n\) antinodes and \((n+1)\) nodes.

b) Find the frequency of the fundamental mode of a stretched string \(1\ \mathrm{m}\) long, under a tension of \(150\ \mathrm{N}\) and mass per unit length \(0.005\ \mathrm{kg\,m^{-1}}\).

Given: \(l=1\ \mathrm{m},\ T=150\ \mathrm{N},\ \mu=0.005\ \mathrm{kg\,m^{-1}}\)
\[ v=\sqrt{\frac T\mu}=\sqrt{\frac{150}{0.005}}=\sqrt{30000}=100\sqrt3\approx173.2\ \mathrm{m\,s^{-1}} \]
\[ f_1=\frac{v}{2l}=\frac{173.2}{2\times1}=86.6\ \mathrm{Hz} \]

c) A stretched string vibrates in its second overtone at \(300\ \mathrm{Hz}\). What is the fundamental frequency?

A string fixed at both ends has all harmonics, so: fundamental \(=f_1\), first overtone \(=2f_1\), second overtone \(=3f_1\).

\[ 3f_1=300\ \mathrm{Hz}\;\Rightarrow\;f_1=\frac{300}{3}=100\ \mathrm{Hz} \]

4. View PDF Notes

i. Handwritten Notes

ii. Typed PDF Notes

5. Frequently Asked Questions

What are waves in pipes and strings in Class 12 Physics?
Waves in pipes and strings is the NEB Class 12 Physics chapter on stationary (standing) waves. It covers longitudinal stationary waves in air columns of closed and open organ pipes and transverse stationary waves on a stretched string, including harmonics, overtones, end correction, wave speed on a string and the laws of vibration.
What topics are included in Chapter 8 Waves in Pipes and Strings of NEB Class 12 Physics?
The chapter (6 hours, part of the unit Waves and Optics) covers: 8.1 stationary waves in closed and open pipes; 8.2 harmonics and overtones in closed and open organ pipes; 8.3 end correction in pipes; 8.4 velocity of transverse waves along a stretched string; 8.5 vibration of a string and overtones; 8.6 laws of vibration of a fixed string.
What are mechanical waves and what are the types of mechanical waves?
Mechanical waves are disturbances that transfer energy through a material medium such as air, water, a string or the ground, and they cannot travel through a vacuum. The two types of mechanical waves are transverse waves, where particles vibrate perpendicular to the direction of travel, and longitudinal waves, where particles vibrate parallel to it. The basic formula is v = fλ.
What are some examples of mechanical waves?
Ten examples of mechanical waves are sound waves in air, water surface waves, waves on a guitar string, waves on a rope, seismic P-waves, seismic S-waves, ultrasound, the vibrations of a tuning fork, compression waves in a slinky spring and ocean swells.
What is the difference between mechanical waves and electromagnetic (non-mechanical) waves?
Mechanical waves need a material medium and can be transverse or longitudinal, for example sound and string waves. Electromagnetic waves, which are non-mechanical, need no medium and can travel through a vacuum at 3 × 10⁸ m s⁻¹; they are always transverse, for example light, radio waves and X-rays.
What are stationary waves, nodes and antinodes?
A stationary (standing) wave is formed when two waves of the same frequency and amplitude travelling in opposite directions superpose. Nodes are points of zero displacement and antinodes are points of maximum displacement. A closed end of a pipe and a fixed end of a string are nodes, while an open end of a pipe is an antinode.
What are the frequencies of a closed organ pipe?
A closed pipe has a node at the closed end and an antinode at the open end. Its fundamental frequency is f₁ = v/4l and it produces only odd harmonics: f₁, 3f₁, 5f₁ and so on. The third harmonic is the first overtone and the fifth harmonic is the second overtone.
What are the frequencies of an open organ pipe?
An open pipe has antinodes at both ends. Its fundamental frequency is f₁ = v/2l and it produces all harmonics: f₁, 2f₁, 3f₁ and so on. The second harmonic is the first overtone and the third harmonic is the second overtone.
What is the difference between harmonics and overtones?
Harmonics are integral multiples of the fundamental frequency, with the fundamental being the first harmonic. Overtones are all the frequencies higher than the fundamental. In an open pipe the nth overtone is the (n + 1)th harmonic, and in a closed pipe the nth overtone is the (2n + 1)th harmonic.
What is end correction in an organ pipe?
End correction is the small extra length by which the antinode at an open end forms outside the pipe, because the air just outside is not free to vibrate completely. It is approximately e = 0.6r, where r is the radius of the pipe. The effective length is l + e for a closed pipe and l + 2e for an open pipe.
What is the velocity of transverse waves along a stretched string?
The speed of a transverse wave on a stretched string is v = √(T/μ), where T is the tension and μ is the mass per unit length (linear density). The speed increases with tension and decreases with linear density. For a string fixed at both ends, the fundamental frequency is f = (1/2l)√(T/μ).
What are the laws of vibration of a stretched string?
Mersenne’s laws state that the fundamental frequency of a stretched string is inversely proportional to its length (f ∝ 1/l) when T and μ are constant, directly proportional to the square root of the tension (f ∝ √T) when l and μ are constant, and inversely proportional to the square root of the linear density (f ∝ 1/√μ) when l and T are constant.
Why does a closed pipe have only odd harmonics while an open pipe has all harmonics?
A closed pipe needs a node at one end and an antinode at the other, so its length must contain an odd number of quarter wavelengths (λ/4, 3λ/4, 5λ/4 and so on), giving only odd harmonics. An open pipe needs antinodes at both ends, so its length contains any whole number of half wavelengths, giving all harmonics.
How does the length of an organ pipe change the pitch?
The fundamental frequency is inversely proportional to the length of the pipe (f = v/2l for open and f = v/4l for closed). A longer pipe allows a longer wavelength and so produces a lower frequency (lower pitch), while a shorter pipe produces a higher pitch.
Where can I get Class 12 Physics Chapter 8 Waves in Pipes and Strings notes in PDF?
This page includes handwritten notes and typed PDF notes for Waves in Pipes and Strings that you can view online, along with solved multiple choice questions and numericals.

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