📌 Snapshot
- Understand waves, wave speed, standing waves, organ pipes and beats with solved CUET numericals and traps.
- How a wave moves energy without moving matter, and why transverse waves cannot travel through air.
- Speed of waves on strings, in solids and in gases, including why Newton's value for sound was wrong.
- Open and closed pipes, harmonics and beats compared in tables, with six worked examples.
📖 Detailed Notes
Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XI, Chapter 14.
What actually travels in a wave?
In a cricket stadium, a crowd does "the wave". Each person stands up and sits down in the same spot, yet the wave runs all the way around the ground. Nobody moved around the stadium. Only the pattern did. A wave works the same way: the disturbance travels, the medium stays. A cork on a pond bobs up and down as ripples pass, and it ends up where it started.
What the wave does carry is energy and information. That is why you can hear a drum from across a field without any air travelling from the drum to your ear.
Kinds of waves
| Type | Needs a medium? | Examples |
|---|---|---|
| Mechanical | Yes, an elastic medium | Waves on a string, sound, water waves, seismic waves |
| Electromagnetic | No. They cross vacuum at $c = 3 \times 10^{8}$ m s⁻¹ | Light, radio waves, X-rays |
| Matter waves | No | Associated with electrons, protons and atoms |
This chapter deals with mechanical waves. They exist because the particles of a medium are coupled, like beads joined by springs. Pull one bead, and its neighbours feel a force and follow, then their neighbours do.
Transverse and longitudinal waves
If the particles of the medium vibrate perpendicular to the direction in which the wave travels, the wave is transverse (a wave on a rope). If they vibrate along the direction of travel, it is longitudinal (sound in air, with alternate compressions and rarefactions).
| Feature | Transverse | Longitudinal |
|---|---|---|
| Particle motion | At right angles to the wave direction | Along the wave direction |
| Pattern | Crests and troughs | Compressions and rarefactions |
| Medium needed | One that resists shear (a solid, or a stretched string) | Any medium that resists compression: solids, liquids, gases |
| Travels in air or the bulk of a liquid? | No | Yes |
| Travels in a solid like steel? | Yes | Yes |
Why can't a transverse wave travel in air? A transverse wave works by one layer dragging the next sideways. A solid pulls back on a sideways shift, which is called shear rigidity. A gas or a liquid has none. Layers just slide past each other, so there is nothing to carry the sideways disturbance forward. Compressing a gas, on the other hand, always pushes back, so a longitudinal wave passes without trouble.
Wind is not a sound wave. Wind carries air bodily from place to place. In a sound wave, every air molecule only vibrates a little about its mean position.
Beyond the textbook. For the exam: water waves are transverse, because they travel on the surface of a liquid. In reality: a particle of water on the surface moves in a small circle or ellipse as a wave passes, so surface waves are a mix of transverse and longitudinal motion. Why it matters: it explains why a floating cork moves forward and back a little as well as up and down.
Describing a travelling wave
A sinusoidal wave moving in the +x direction is described by
$$y(x, t) = a\sin(kx - \omega t + \phi)$$
For a longitudinal wave, y is replaced by the displacement of the particles along the wave. Every symbol answers one question about the wave:
| Symbol | What it tells you | Relations |
|---|---|---|
| a | Amplitude: the largest displacement of a particle | Energy carried grows with $a^{2}$ |
| k | Angular wave number: how many radians of phase per metre | $k = 2\pi/\lambda$ |
| $\omega$ | Angular frequency: how fast a particle oscillates | $\omega = 2\pi/T = 2\pi\nu$ |
| $\lambda$ | Wavelength: distance between two successive points in the same phase | $\lambda = vT$ |
| ($kx - \omega t + \phi$) | The phase, which fixes the displacement at any x and t | $\phi$ is the phase at $x = 0$, $t = 0$ |
The sign inside the bracket sets the direction. $kx - \omega t$ means the wave moves towards +x, and $kx + \omega t$ means towards −x. Any function of $(x - vt)$ is a wave moving at speed v along +x, and any function of $(x + vt)$ is one moving along −x.
Wave speed. Follow one crest. Its phase $kx - \omega t$ stays constant, so it moves with
$$v = \omega/k = \lambda/T = \nu\lambda$$
In one period the pattern advances exactly one wavelength. Two points a distance $\Delta x$ apart along the wave differ in phase by $(2\pi/\lambda) \times \Delta x$.
Worked example: reading a wave equation. A wave on a string is $y = 0.005\sin(80x - 3t)$, with y and x in metres and t in seconds. Find the wavelength, period, frequency, wave speed, direction, and the greatest speed of a string particle.
| Step | What we do | Result |
|---|---|---|
| 1 | Compare with $a\sin(kx - \omega t)$ | $a = 0.005$ m, $k = 80$ m⁻¹, $\omega = 3$ rad s⁻¹ |
| 2 | $\lambda = 2\pi/k$ | $2\pi/80 = 0.0785$ m = 7.85 cm |
| 3 | $T = 2\pi/\omega$ and $\nu = 1/T$ | $T = 2.09$ s, $\nu = 0.48$ Hz |
| 4 | Wave speed $v = \omega/k$ | $3/80 = 0.0375$ m s⁻¹ |
| 5 | Direction: minus sign between $kx$ and $\omega t$ | Towards +x |
| 6 | Particle speed is a different thing: the maximum is $a\omega$ | 0.005 × 3 = 0.015 m s⁻¹ |
The common wrong step: mixing up the wave speed with the speed of the particle. The wave moves at 0.0375 m s⁻¹ along the string. Each particle moves up and down, at up to 0.015 m s⁻¹. They are different quantities, and in a transverse wave they point in different directions.
How fast does a wave travel?
Speed depends only on the medium. Every formula has the same shape: a restoring property on top, and inertia below. A tighter or stiffer medium pulls back faster, and a heavier medium responds slower.
| Medium | Speed | Elastic term | Inertia term |
|---|---|---|---|
| Stretched string | $v = \sqrt{T/\mu}$ | Tension T | Linear mass density $\mu = m/L$ |
| Solid rod (longitudinal) | $v = \sqrt{Y/\rho}$ | Young's modulus Y | Density $\rho$ |
| Fluid (longitudinal) | $v = \sqrt{B/\rho}$ | Bulk modulus B | Density $\rho$ |
| Ideal gas (sound) | $v = \sqrt{\gamma P/\rho}$ | Adiabatic bulk modulus $\gamma P$ | Density $\rho$ |
Because the speed is fixed by the medium, changing the frequency of the source changes only the wavelength ($\lambda = v/\nu$), never the speed. Sound moves at about 340 m s⁻¹ in air, about 1500 m s⁻¹ in water and over 5000 m s⁻¹ in steel, which is far stiffer for its density.
Worked example: speed on a wire. A steel wire 0.72 m long has a mass of $5.0 \times 10^{-3}$ kg and is stretched with a tension of 60 N. Find the speed of a transverse wave on it.
| Step | What we do | Result |
|---|---|---|
| 1 | Find the linear mass density $\mu = m/L$ | $5.0 \times 10^{-3} / 0.72 = 6.9 \times 10^{-3}$ kg m⁻¹ |
| 2 | Apply $v = \sqrt{T/\mu}$ | $\sqrt{60 / 6.9 \times 10^{-3}} = \sqrt{8700}$ |
| 3 | Take the root | $v \approx 93$ m s⁻¹ |
The common wrong step: putting the total mass $5.0 \times 10^{-3}$ kg in place of $\mu$. The formula needs mass per unit length. Also, if the tension is increased four times, the speed only doubles, because $v \propto \sqrt{T}$.
Speed of sound: Newton's mistake and Laplace's fix
Newton assumed that when sound squeezes and stretches a gas, the temperature stays constant (isothermal). The bulk modulus of a gas at constant temperature is just P, so he got
$$v = \sqrt{P/\rho}$$
For air at STP, with $P = 1.01 \times 10^{5}$ Pa and $\rho = 1.29$ kg m⁻³, this is about 280 m s⁻¹. The measured value is about 331 m s⁻¹, so Newton was nearly 15% too low.
Laplace saw the error. Sound vibrates the air so quickly, and air conducts heat so poorly, that no heat has time to flow between a compression and the next rarefaction. The changes are adiabatic, with $PV^{\gamma}$ constant, and the bulk modulus becomes $\gamma P$. The Newton–Laplace formula is
$$v = \sqrt{\gamma P/\rho}$$
For air $\gamma = 7/5 = 1.4$, so $v = 280 \times \sqrt{1.4} \approx 331$ m s⁻¹ at 0 °C, which matches experiment.
What decides the speed of sound in a gas? Using $P/\rho = RT/M$ for an ideal gas,
| Change | Effect on v | Why |
|---|---|---|
| Temperature rises | $v \propto \sqrt{T}$ (Kelvin). About +0.61 m s⁻¹ per °C | Molecules move faster and pass the disturbance on faster |
| Pressure changes at constant temperature | No change | P and $\rho$ rise together, so $P/\rho$ stays the same |
| Frequency or wavelength changes | No change | Speed belongs to the medium |
| Molar mass M larger | $v \propto 1/\sqrt{M}$, so slower | Heavier molecules respond more sluggishly |
Pause and predict: you pump air into a closed hall at the same temperature, doubling the pressure. Does a clap reach the back wall sooner?
Check your answer: no. Doubling the pressure doubles the density too (at constant temperature), so $P/\rho$ is unchanged and the speed of sound stays the same. Only a change in temperature, or in the gas itself, changes it.
Worked example: sound at a different temperature and in different gases. (a) The speed of sound in air at 0 °C is 331 m s⁻¹. Find it at 27 °C. (b) Compare the speed of sound in hydrogen and in oxygen at the same temperature, treating both as ideal diatomic gases.
| Step | What we do | Result |
|---|---|---|
| (a) 1 | Convert to Kelvin: 0 °C = 273 K, 27 °C = 300 K | $v \propto \sqrt{T}$ |
| (a) 2 | $v = 331 \times \sqrt{300/273} = 331 \times 1.048$ | $v \approx 347$ m s⁻¹ |
| (a) 3 | Quick check with 0.61 m s⁻¹ per °C | 331 + 0.61 × 27 = 347.5 m s⁻¹ |
| (b) 1 | Same T and same $\gamma$ (7/5), so $v \propto 1/\sqrt{M}$ | $v(\text{H}{2})/v(\text{O}{2}) = \sqrt{M(\text{O}{2})/M(\text{H}{2})}$ |
| (b) 2 | $\sqrt{32/2} = \sqrt{16}$ | Hydrogen is 4 times faster |
The common wrong step: using Celsius temperatures in the ratio, $\sqrt{27/0}$, which is meaningless. Always use Kelvin.
Superposition and interference
When two waves meet in the same medium, the displacement of each particle is simply the sum of the displacements the waves would cause separately: $y = y_{1} + y_{2}$. The waves then pass through each other unchanged. This is the principle of superposition.
Take two waves of equal amplitude a, equal $\omega$ and $k$, differing in phase by $\phi$. Their sum is
$$y = 2a\cos(\phi/2)\sin(kx - \omega t + \phi/2)$$
The resultant amplitude is $2a\cos(\phi/2)$. When $\phi = 0$ (in phase), the amplitude is 2a and the waves reinforce. This is constructive interference. When $\phi = \pi$ (opposite phase), the amplitude is zero and the waves cancel. This is destructive interference. Intensity goes as the square of amplitude, so two in-phase waves give four times the intensity of one, not twice. In destructive interference the energy is not destroyed. It reappears in the places where the waves reinforce.
Reflection, and why standing waves form
A pulse travelling on a string meets a boundary and returns. What happens depends on the boundary.
| Boundary | What the end does | Reflected wave |
|---|---|---|
| Rigid (string fixed to a wall) | The end cannot move, so the net displacement there must always be zero | Inverted, a phase change of $\pi$. A node forms at the end |
| Free (string end free to move) | The end moves freely | No phase change. An antinode forms at the end |
If a continuous wave is sent along and keeps reflecting, the incident and reflected waves of equal amplitude travel in opposite directions and superpose into a standing (stationary) wave:
$$y(x, t) = 2a\sin kx\cos\omega t$$
Every point oscillates with the same frequency, but the amplitude $2a\sin kx$ depends on position. Points of zero amplitude are nodes, at $kx = n\pi$, that is $x = n\lambda/2$. Points of maximum amplitude are antinodes, midway between nodes. So consecutive nodes (or consecutive antinodes) are $\lambda/2$ apart, and a node and the nearest antinode are $\lambda/4$ apart.
| Feature | Travelling wave | Standing wave |
|---|---|---|
| Amplitude | Same for every particle | Varies from zero at nodes to maximum at antinodes |
| Phase | Different for different particles | Same for all particles between two successive nodes, opposite across a node |
| Energy | Carried along | Not transported. It stays between the nodes |
| Pattern | Moves with speed v | Fixed in place |
Vibrating strings and air columns
A standing wave can only form if it fits the boundaries. This is why a string or a pipe has particular natural frequencies, the normal modes. The lowest is the fundamental (first harmonic).
String fixed at both ends. Both ends are nodes, so a whole number of half wavelengths fits in L: $L = n\lambda/2$. Hence
$$\nu_{n} = nv/(2L), \quad n = 1, 2, 3, \dots$$
All harmonics are present, in the ratio 1 : 2 : 3 : 4. Here $v = \sqrt{T/\mu}$, so a guitarist raises the pitch by tightening the string (larger v) or by pressing a finger to shorten the vibrating length L.
Air columns. At a closed end the air cannot move, which is a node. At an open end the air moves freely, which is an antinode.
| Feature | Open at both ends | Closed at one end |
|---|---|---|
| Ends | Antinode at both | Node at closed end, antinode at open end |
| Condition | L = $n\lambda/2$ | $L = (2n + 1)\lambda/4$ |
| Frequencies | $\nu_{n} = nv/(2L)$, $n = 1, 2, 3, \dots$ | $\nu_{n} = (2n + 1)v/(4L)$, $n = 0, 1, 2, \dots$ |
| Fundamental | $v/(2L)$ | $v/(4L)$ |
| Harmonics present | All: $\nu_{1}$, $2\nu_{1}$, $3\nu_{1}$, … | Only odd: $\nu_{1}$, $3\nu_{1}$, $5\nu_{1}$, … |
Why only odd harmonics in a closed pipe? The pipe must hold a node at one end and an antinode at the other. A node-to-antinode stretch is a quarter wavelength, so the length must be an odd number of quarter wavelengths. An even number would put an antinode at both ends or a node at both, which contradicts the boundary conditions.
Harmonic versus overtone. The harmonic number counts the fundamental as the first. The overtones are the frequencies above the fundamental. For a string or an open pipe the first overtone is the second harmonic, $2\nu_{1}$. For a closed pipe the first overtone is the third harmonic, $3\nu_{1}$.
Worked example: open or closed pipe? A pipe 30 cm long is excited by a source of frequency 1.1 kHz. Take the speed of sound as 330 m s⁻¹. (a) Which harmonic is excited if the pipe is open at both ends? (b) Does the source excite the pipe if one end is closed?
| Step | What we do | Result |
|---|---|---|
| (a) 1 | Open pipe fundamental: $\nu_{1} = v/(2L)$ | 330 / 0.60 = 550 Hz |
| (a) 2 | Compare 1100 Hz with $\nu_{n} = 550n$ | n = 2, the second harmonic |
| (b) 1 | Closed pipe fundamental: $\nu_{1} = v/(4L)$ | 330 / 1.20 = 275 Hz |
| (b) 2 | Allowed: 275, 825, 1375 Hz, and so on (odd multiples) | 1100 Hz = 4 × 275 is an even multiple, so no resonance |
The common wrong step: using $\nu_{n} = nv/(2L)$ for the closed pipe too, or allowing the closed pipe every multiple of 275 Hz instead of only the odd ones.
Pause and predict: a closed pipe of length L and an open pipe of length 2L are sounded at their fundamental. Which note is higher?
Check your answer: neither. The closed pipe gives $v/(4L)$. The open pipe of length 2L gives $v/(2 \times 2L) = v/(4L)$. They are identical.
Beyond the textbook. For the exam: take the antinode of an open pipe exactly at its open end, so $\nu_{1} = v/(2L)$. In reality: the air vibrates a little beyond the open mouth, so the antinode sits outside the end. The pipe behaves as if it were longer by about 0.6 times its radius at each open end. Why it matters: for a narrow pipe this is negligible, but for a wide tube it shifts the frequencies noticeably, and it is the reason resonance tube experiments need an end correction.
Worked example: a string and its harmonics. A string 0.5 m long, fixed at both ends, carries waves at 200 m s⁻¹. Find the fundamental and the third harmonic. How far apart are the nodes in the third harmonic? What happens to the fundamental if the tension is made four times larger?
| Step | What we do | Result |
|---|---|---|
| 1 | Fundamental $\nu_{1} = v/(2L)$ | 200 / 1.0 = 200 Hz |
| 2 | Third harmonic $\nu_{3} = 3\nu_{1}$ | 600 Hz |
| 3 | Wavelength in this mode: $\lambda = 2L/3$ | 0.333 m |
| 4 | Adjacent nodes are $\lambda/2$ apart | 0.167 m |
| 5 | Tension × 4 gives v × 2, so $\nu_{1} \times 2$ | 400 Hz |
The common wrong step: writing the wavelength as L/3 for the third harmonic. A fixed string holds n half wavelengths, so $\lambda = 2L/n$.
Beats
Strike two tuning forks of nearly equal frequencies together, say 256 Hz and 260 Hz. You do not hear two separate notes. You hear one note whose loudness rises and falls, four times each second. These are beats.
Why? The two waves sometimes arrive in phase, so the sound is loud, and sometimes out of phase, so it fades. The phase relationship slips by one full cycle each time one wave gains a complete oscillation on the other, which happens $\nu_{1} - \nu_{2}$ times per second. So
beat frequency = $\nu_{1} - \nu_{2}$ (the positive difference)
The pitch heard is the average frequency $(\nu_{1} + \nu_{2})/2$, while the loudness swells at the difference. Beats are only noticeable when the difference is small, below about 10 Hz.
Worked example: finding an unknown frequency from beats. Two sitar strings A and B are playing the same note. A has frequency 427 Hz. They produce 5 beats per second. When the tension of B is slightly increased, the beat frequency falls to 3 per second. Find the original frequency of B.
| Step | What we do | Result |
|---|---|---|
| 1 | Beats of 5 Hz means $\nu(B) = 427 \pm 5$ | Either 432 Hz or 422 Hz |
| 2 | Increasing tension raises the frequency of B ($v \propto \sqrt{T}$) | $\nu(B)$ goes up |
| 3 | Beats decreased, so $\nu(B)$ moved closer to 427 Hz | It must have been below 427 Hz |
| 4 | Choose the lower value | $\nu(B) = 422$ Hz |
| 5 | Confirm: slightly above 422 Hz gives a difference of 3 Hz at 424 Hz | Consistent with the observation |
The common wrong step: picking 432 Hz. If B started at 432 Hz, raising its tension would push it farther from 427 Hz and the beats would increase, which contradicts the observation.
| Action on a fork | Effect on its frequency |
|---|---|
| Adding a little wax | Lowers it |
| Filing the prong | Raises it |
| Increasing tension of a string | Raises it |
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| "Sound is a transverse wave in air." | Pictures of waves show crests and troughs | Air resists compression but not shear, so sound in air is longitudinal |
| "Wave speed is the same as particle speed." | Both are called speed | Wave speed is $v = \omega/k$. A particle's speed varies, with maximum $a\omega$ |
| "Newton's $v = \sqrt{P/\rho}$ is correct." | It looks like the natural formula | It gives 280 m s⁻¹. The adiabatic correction $\sqrt{\gamma}$ gives 331 m s⁻¹ |
| "Sound is faster at higher pressure." | Gas feels stiffer when compressed | Density rises equally, so v is unchanged at constant temperature |
| "A closed pipe has all harmonics." | An open pipe does | Only odd ones: $\nu_{1}$, $3\nu_{1}$, $5\nu_{1}$ |
| "Nodes in a standing wave are $\lambda$ apart." | Crest to crest is $\lambda$ | Nodes are $\lambda/2$ apart. Node to antinode is $\lambda/4$ |
| "Beat frequency is the average of the two." | The average sets the pitch | Beat frequency is the difference |
| "Frequency decides the wave speed on a string." | $v = \nu\lambda$ shows $\nu$ | $v = \sqrt{T/\mu}$. Changing $\nu$ changes $\lambda$ only |
Formula sheet
| Quantity | Formula | SI unit | When to use |
|---|---|---|---|
| Travelling wave | $y = a\sin(kx - \omega t + \phi)$ | m | Read a, k, $\omega$ from the equation |
| Wave number, frequency | $k = 2\pi/\lambda$, $\omega = 2\pi/T = 2\pi\nu$ | rad m⁻¹, rad s⁻¹ | Converting between parameters |
| Wave speed | $v = \omega/k = \nu\lambda$ | m s⁻¹ | Any wave |
| String | $v = \sqrt{T/\mu}$, $\mu = m/L$ | m s⁻¹ | Stretched wire or rope |
| Solid, fluid | $v = \sqrt{Y/\rho}$, $v = \sqrt{B/\rho}$ | m s⁻¹ | Longitudinal waves |
| Gas | $v = \sqrt{\gamma P/\rho}$ | m s⁻¹ | Sound in a gas |
| Interference | Resultant amplitude $2a\cos(\phi/2)$ | m | Two equal waves, phase difference $\phi$ |
| Standing wave | $y = 2a\sin kx\cos\omega t$ | m | Nodes at $x = n\lambda/2$ |
| String, open pipe | $\nu_{n} = nv/(2L)$, $n = 1, 2, 3, \dots$ | Hz | Both ends nodes or both antinodes |
| Closed pipe | $\nu_{n} = (2n + 1)v/(4L)$, $n = 0, 1, 2, \dots$ | Hz | Only odd harmonics |
| Beats | $\nu_{1} - \nu_{2}$ | Hz | Two close frequencies |
| Power on a string | Proportional to $a^{2}\omega^{2}$ | W | Doubling a or $\omega$ gives four times the power |
60-second revision
| Topic | Lock this in |
|---|---|
| Wave | Carries energy, not matter |
| Transverse vs longitudinal | Perpendicular vs along. Fluids support only longitudinal |
| Wave equation | $y = a\sin(kx - \omega t)$. Minus sign means towards +x. $v = \omega/k = \nu\lambda$ |
| Speeds | $\sqrt{T/\mu}$ for strings, $\sqrt{Y/\rho}$ for solids, $\sqrt{B/\rho}$ for fluids, $\sqrt{\gamma P/\rho}$ for gases |
| Sound in air | 331 m s⁻¹ at 0 °C, $v \propto \sqrt{T}$, independent of pressure and frequency |
| Reflection | Rigid: phase change $\pi$, node. Free: no change, antinode |
| Standing wave | Nodes $\lambda/2$ apart. Node to antinode $\lambda/4$ |
| String or open pipe | $\nu = nv/2L$, all harmonics |
| Closed pipe | $\nu = (2n + 1)v/4L$, odd harmonics only. First overtone is $3\nu_{1}$ |
| Beats | Difference of the two frequencies. Pitch is their average |
Where this chapter connects
- Before this: Oscillations supplies the sine curves and the idea of a restoring force that every particle of the medium obeys.
- Where the formulas come from: Mechanical Properties of Solids defines Young's modulus, Mechanical Properties of Fluids the density and pressure used here, and Thermodynamics the adiabatic change behind Laplace's correction.
- Class XII link: Wave Optics reuses superposition and interference for light, and Electromagnetic Waves gives the waves that need no medium.
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. In which of the following can both transverse and longitudinal mechanical waves propagate?
▸ Show answer & explanation
Answer: B
A medium that supports a transverse wave must resist shear, and only a solid does that. Steel also resists compression, so it carries both kinds. Air and the interior of water carry only longitudinal waves. Vacuum carries no mechanical wave at all.
Q2. A travelling wave on a string is $y = 0.005\sin(80.0x - 3.0t)$, with all quantities in SI units. The wavelength is closest to
▸ Show answer & explanation
Answer: B
$k = 80$ rad per metre, so $\lambda = 2\pi/k = 6.283/80 = 0.0785$ m = 7.85 cm. (A) is half of that. (C) and (D) correspond to $k = 50$ and $k = 25$.
Q3. Assertion (A): Increasing the frequency of the vibrator that drives a stretched string does not change the speed of the waves on the string. Reason (R): The speed of a transverse wave on a string is $v = \sqrt{T/\mu}$, which depends only on the tension and the mass per unit length.
▸ Show answer & explanation
Answer: A
The speed is set by the string, through T and $\mu$, and not by the source. A higher frequency only shortens the wavelength, because $v = \nu\lambda$ is fixed. So R explains A.
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Q4. A steel wire 0.72 m long has mass $5.0 \times 10^{-3}$ kg and is under a tension of 60 N. The speed of a transverse wave on it is about
▸ Show answer & explanation
Answer: C
$\mu = m/L = 5.0 \times 10^{-3}/0.72 = 6.9 \times 10^{-3}$ kg per metre. $v = \sqrt{T/\mu} = \sqrt{60/6.9 \times 10^{-3}} = \sqrt{8700} \approx 93$ m per second. (A) is half the right value, (D) is twice it, and (B) is 93 divided by $\sqrt{2}$. None of them follows from the formula.
Q5. Newton's formula for the speed of sound in air at STP gives about 280 m s⁻¹, while the measured value is about 331 m s⁻¹. If Laplace's correction for air ($\gamma = 1.4$) is applied to Newton's value, the result is
▸ Show answer & explanation
Answer: B
The corrected formula is $v = \sqrt{\gamma P/\rho} = \sqrt{\gamma} \times \sqrt{P/\rho} = 1.18 \times 280 \approx 331$ m per second. (A) divides by $\sqrt{\gamma}$. (C) multiplies by $\gamma$ instead of $\sqrt{\gamma}$. (D) overshoots by multiplying by a further factor.
Q6. Two waves of equal amplitude 4 mm travel in the same direction on a string with a phase difference of 120°. The amplitude of the resultant wave is
▸ Show answer & explanation
Answer: B
The resultant amplitude is $2a\cos(\phi/2) = 2 \times 4 \times \cos 60^{\circ} = 8 \times 0.5 = 4$ mm. (D) is the amplitude for $\phi = 0$. (C) uses $\cos 30^{\circ}$ by treating $\phi/2$ as $30^{\circ}$. (A) uses $a\cos\phi$ in magnitude.
Q7. Consider the statements about a pulse reflected on a string. I. At a rigid boundary the reflected pulse is inverted, with a phase change of $\pi$. II. At a free end the reflected pulse has no phase change. III. A standing wave on a string fixed at both ends has an antinode at each end. Which statements are correct?
▸ Show answer & explanation
Answer: A
The fixed end cannot move, so the reflected wave must cancel the incident displacement there. That needs a phase change of $\pi$ and creates a node, which makes I true and III false. A free end can move and reflects without inversion, so II is true.
Q8. A string of length 1.2 m is fixed at both ends and vibrates in its third harmonic. The wavelength of the standing wave is
▸ Show answer & explanation
Answer: B
$L = n\lambda/2$ gives $\lambda = 2L/n = 2 \times 1.2/3 = 0.8$ m. (A) results from $\lambda = L/n$. (C) is the wavelength of the second harmonic. (D) is the fundamental wavelength $2L$.
Q9. A pipe closed at one end has a fundamental frequency of 120 Hz. Which one of the following frequencies cannot be produced by it?
▸ Show answer & explanation
Answer: B
A closed pipe produces only odd multiples of its fundamental: 120, 360, 600, 840 and so on. 480 Hz is 4 × 120, an even multiple, so it is absent. The others are 3, 5 and 7 times 120 Hz.
Q10. A pipe 30 cm long is open at both ends. Taking the speed of sound as 330 m s⁻¹, which harmonic is resonantly excited by a source of 1.1 kHz?
▸ Show answer & explanation
Answer: B
$\nu_{1} = v/(2L) = 330/0.60 = 550$ Hz. Then 1100 Hz = 2 × 550 Hz, which is the second harmonic. If one end were closed, the allowed frequencies would be 275, 825, 1375 Hz, and 1100 Hz would not be a mode.
Q11. Two sitar strings A and B give 5 beats per second. The frequency of A is 427 Hz. When the tension of B is slightly increased, the beat frequency becomes 3 per second. The original frequency of B was
▸ Show answer & explanation
Answer: D
B is either 422 Hz or 432 Hz. Increasing the tension raises its frequency. If it had been 432 Hz, the beats would have risen above 5. Since they fell to 3, B was 422 Hz and became 424 Hz. (B) and (C) do not give 5 beats with 427 Hz.
Q12. Match List I with List II. List I: (P) String fixed at both ends (Q) Pipe closed at one end (R) Pipe open at both ends (S) Two sources of nearly equal frequency. List II: (1) Only odd harmonics are present (2) Nodes at both ends, all harmonics present (3) Antinodes at both ends, all harmonics present (4) Waxing and waning of loudness
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Answer: A
A fixed string has a node at each end, so P-2. A closed pipe needs a node and an antinode, which allows only odd harmonics, so Q-1. An open pipe has an antinode at each end, so R-3. Nearly equal frequencies produce beats, which is S-4.
Q13. The speed of sound in air at 0 °C is 331 m s⁻¹. Treating air as an ideal gas with v proportional to the square root of the absolute temperature, its speed at 27 °C is closest to
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Answer: B
$v = 331 \times \sqrt{300/273} = 331 \times 1.048 \approx 347$ m per second. (A) ignores the temperature rise. (C) and (D) overstate the rise and do not follow from the square root rule.
Q14. In a standing wave on a string, a node and the adjacent antinode are 20 cm apart. If the speed of the waves on the string is 320 m s⁻¹, the frequency is
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Answer: B
A node and the nearest antinode are $\lambda/4$ apart, so $\lambda = 4 \times 0.20 = 0.80$ m. Then $\nu = v/\lambda = 320/0.80 = 400$ Hz. (C) treats 20 cm as $\lambda/2$, giving $\lambda = 0.4$ m and 800 Hz. (A) would need $\lambda = 1.6$ m, and (D) would need $\lambda = 0.2$ m.
Q15. Assertion (A): An open organ pipe sounds an octave higher than a closed pipe of the same length when both are sounded in their fundamental modes. Reason (R): The fundamental frequency of an open pipe is $v/(2L)$ while that of a closed pipe is $v/(4L)$.
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Answer: A
The ratio of the open to the closed fundamental is $[v/(2L)] / [v/(4L)] = 2$, which is one octave. R states exactly this, so it explains A.
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