Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers Waves for CBSE Class 11 Physics, bringing the mechanics of oscillations into continuous media.
Key Concepts
Wave Mechanics
How energy travels without transporting matter
1
Transverse & Longitudinal Waves
Definition
Transverse: Particle vibration is perpendicular to wave propagation (e.g., light, string waves). Consists of crests and troughs.
Longitudinal: Particle vibration is parallel to wave propagation (e.g., sound). Consists of compressions and rarefactions.
Longitudinal: Particle vibration is parallel to wave propagation (e.g., sound). Consists of compressions and rarefactions.
$$\text{Only transverse waves can be polarized.}$$
2
Equation of a Plane Progressive Wave
Formula
Describes the displacement $y$ of a particle at position $x$ and time $t$. A wave traveling in the positive x-direction uses a minus sign between $kx$ and $\omega t$.
$$y(x,t) = A \sin(kx – \omega t + \phi)$$
3
Wave Parameters ($k, \omega, v$)
Formula
Angular wave number ($k$), angular frequency ($\omega$), and wave speed ($v$) are deeply interconnected. Wave speed is the speed at which the pattern moves.
$$k = \frac{2\pi}{\lambda} \quad | \quad \omega = \frac{2\pi}{T} = 2\pi \nu \quad | \quad v = \frac{\omega}{k} = \nu \lambda$$
4
Speed of a Transverse Wave on a String
Formula
The speed of a mechanical wave depends only on the inertial and elastic properties of the medium. For a stretched string, it depends on Tension ($T$) and linear mass density ($\mu = \text{mass/length}$).
$$v = \sqrt{\frac{T}{\mu}}$$
5
Speed of Sound (Newton-Laplace Formula)
Formula
Newton assumed sound travels isothermally. Laplace corrected this, proving sound compression is adiabatic because it happens too fast for heat exchange. $\gamma$ is the ratio of specific heats ($C_p/C_v$).
$$v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma R T}{M}}$$
6
Principle of Superposition
Formula
When two or more waves overlap, the net displacement of the medium at any point is the algebraic sum of the individual displacements.
$$y_{\text{net}} = y_1 + y_2 + y_3 + \dots$$
7
Standing (Stationary) Waves
Formula
Formed when two identical waves traveling in opposite directions superimpose. The wave profile does not move. The equation separates position ($x$) and time ($t$) into independent trigonometric functions.
$$y(x,t) = (2A \sin kx) \cos \omega t$$
8
Harmonics in a Stretched String
Formula
A string fixed at both ends must have Nodes at both ends. It resonates at specific frequencies called harmonics. All harmonics (1st, 2nd, 3rd…) are present.
$$\nu_n = \frac{nv}{2L} = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \quad (n = 1, 2, 3\dots)$$
9
Organ Pipes (Open vs Closed)
Case Formulas
Open ends form Antinodes; closed ends form Nodes. An open pipe has all harmonics. A pipe closed at one end only has odd harmonics.
$$\text{Open Pipe: } \nu_n = \frac{nv}{2L} \quad (n = 1, 2, 3\dots)$$
$$\text{Closed Pipe: } \nu_n = \frac{nv}{4L} \quad (n = 1, 3, 5\dots)$$
10
Beats
Formula
Periodic variations in the intensity of sound caused by the superposition of two waves of slightly different frequencies. The beat frequency is the difference between the two source frequencies.
$$\nu_{\text{beat}} = |\nu_1 – \nu_2|$$
Concept Deep Dive
01
Phase Reversal on Reflection
Bouncing off wallsWhen a wave traveling on a string hits a solid wall (a Rigid Boundary), Newton’s Third Law dictates that the wall pushes back on the string in the opposite direction. This flips the wave upside down. A crest reflects as a trough. This introduces a phase change of $\pi$ ($180^\circ$).
If the string is attached to a frictionless sliding ring on a pole (a Free Boundary), the end is free to whip upwards, pulling the string up with it. The wave reflects right-side up. A crest reflects as a crest. There is a phase change of $0$.
If the string is attached to a frictionless sliding ring on a pole (a Free Boundary), the end is free to whip upwards, pulling the string up with it. The wave reflects right-side up. A crest reflects as a crest. There is a phase change of $0$.
02
Laplace’s Brilliant Correction
Why Newton was slightly wrong about soundNewton calculated the speed of sound assuming the compressions and rarefactions happened slowly enough that heat could escape, keeping the temperature constant (Isothermal, $P \propto 1/V$). His formula $v = \sqrt{P/\rho}$ gave $280 \text{ m/s}$, which was significantly lower than the experimental $331 \text{ m/s}$.
Laplace realized sound vibrations happen hundreds of times per second. There is no time for heat to flow between the compressed (hot) regions and rarefied (cold) regions! The process is Adiabatic ($P V^\gamma = \text{const}$). Multiplying by $\gamma$ ($\approx 1.4$ for air) perfectly corrected the formula to $v = \sqrt{\gamma P/\rho}$.
Laplace realized sound vibrations happen hundreds of times per second. There is no time for heat to flow between the compressed (hot) regions and rarefied (cold) regions! The process is Adiabatic ($P V^\gamma = \text{const}$). Multiplying by $\gamma$ ($\approx 1.4$ for air) perfectly corrected the formula to $v = \sqrt{\gamma P/\rho}$.
Compare & Contrast
✗ Progressive (Traveling) Waves
- The wave profile moves forward carrying energy through the medium.
- All particles execute SHM with the same amplitude.
- Phase changes continuously from particle to particle.
- No particle remains permanently at rest.
✓ Standing (Stationary) Waves
- The wave profile is stationary. Energy is trapped between boundaries, not transported.
- Particles execute SHM with different amplitudes (zero at Nodes, max at Antinodes).
- All particles between two nodes vibrate in the exact same phase.
- Particles at Nodes are permanently at rest.
Remember
In a standing wave, the distance between a Node (N) and the very next Antinode (A) is exactly $\lambda/4$. The distance from Node to Node (N to N) is $\lambda/2$.
Common Mistakes to Avoid
Mistake 1
Confusing Particle Velocity with Wave Velocity: Wave velocity ($v = \omega/k$) is constant and is the speed at which the crest moves forward. Particle velocity ($v_p = \frac{\partial y}{\partial t}$) is the speed of the actual string/medium moving up and down. Particle velocity is zero at the extremes and maximum at the mean position, completely independent of the wave’s forward speed.
Mistake 2
Misreading the Direction of Propagation: If the equation is $y = A \sin(kx – \omega t)$, the wave moves in the Positive X direction. If it is $y = A \sin(kx + \omega t)$, it moves in the Negative X direction. A difference in signs means positive motion; same signs mean negative motion.
Mistake 3
Using the wrong Pipe formula: If an organ pipe is closed at one end, its fundamental frequency is $\nu = v / 4L$. If it is open at both ends, it is $\nu = v / 2L$. A pipe open at both ends produces a note exactly one octave higher (double the frequency) than a closed pipe of the exact same length.
Exam Tips
Tip 1
Filing and Loading a Tuning Fork: This is a favorite exam question for Beats.
1. If you attach a tiny bit of wax (load) to a tuning fork, it gets heavier, and its frequency decreases.
2. If you file the prongs of a tuning fork, it gets lighter, and its frequency increases.
Use this logic to deduce unknown frequencies when beat counts change.
1. If you attach a tiny bit of wax (load) to a tuning fork, it gets heavier, and its frequency decreases.
2. If you file the prongs of a tuning fork, it gets lighter, and its frequency increases.
Use this logic to deduce unknown frequencies when beat counts change.
Tip 2
Extracting Data from the Wave Equation: Given $y = 5 \sin(10\pi t – 2\pi x)$. The coefficient of $t$ is ALWAYS $\omega$ ($\omega = 10\pi$). The coefficient of $x$ is ALWAYS $k$ ($k = 2\pi$). Wave speed is instantly found by $v = \omega / k = 10\pi / 2\pi = 5 \text{ m/s}$.
Expected Exam Questions
SQ
Board Pattern Questions
Class 11 · Waves · CBSE Exam1
A tuning fork produces 4 beats per second with another tuning fork of frequency $256 \text{ Hz}$. When the first tuning fork is loaded with a little wax, the beat frequency reduces to 2 beats per second. What is the original frequency of the first tuning fork? [2 marks]
Answer
$260 \text{ Hz}$
📝
Explanation
Let unknown frequency be $\nu_1$. The known frequency $\nu_2 = 256 \text{ Hz}$.
Since beats = 4, $\nu_1$ could be $256 + 4 = 260 \text{ Hz}$, OR $256 – 4 = 252 \text{ Hz}$.
When loaded with wax, $\nu_1$ must decrease.
If it was 252, decreasing it (e.g., to 250) would increase the gap with 256, producing MORE beats (e.g., 6 beats).
If it was 260, decreasing it (e.g., to 258) would close the gap with 256, reducing the beats to 2.
Since the question states beats reduced to 2, the original frequency must be $260 \text{ Hz}$.
2
The equation of a wave traveling on a string is $y(x,t) = 0.05 \sin(80x – 3t)$, where $x$ and $y$ are in meters and $t$ in seconds. Find the amplitude, wavelength, and velocity of the wave. [3 marks]
Answer
$A = 0.05 \text{ m}$, $\lambda = \pi/40 \text{ m}$, $v = 3/80 \text{ m/s}$
📝
Explanation
Comparing with $y = A \sin(kx – \omega t)$:
1. Amplitude ($A$): $0.05 \text{ m}$.
2. Wavelength ($\lambda$): We see $k = 80$. Since $k = 2\pi / \lambda$, we have $\lambda = 2\pi / 80 = \pi / 40 \text{ meters}$.
3. Velocity ($v$): We see $\omega = 3$. Velocity $v = \omega / k = 3 / 80 \text{ m/s}$.
(The wave is traveling in the positive x-direction due to the negative sign).
3
Show that an organ pipe open at both ends produces both odd and even harmonics, whereas a pipe closed at one end produces only odd harmonics. [3 marks]
Answer
Derivation of Open ($\nu = nv/2L$) vs Closed ($\nu = nv/4L$ for odd n).
📝
Explanation
Open Pipe: Antinodes form at both ends.
Fundamental mode (1st harmonic): length $L = \lambda/2 \implies \lambda = 2L \implies \nu_1 = v/2L$.
Second harmonic: $L = \lambda \implies \lambda = L \implies \nu_2 = v/L = 2\nu_1$.
Thus, frequencies are $\nu_1, 2\nu_1, 3\nu_1 \dots$ (All harmonics present).
Closed Pipe: Node at the closed end, Antinode at the open end.
Fundamental mode (1st harmonic): $L = \lambda/4 \implies \lambda = 4L \implies \nu_1 = v/4L$.
Next mode (3rd harmonic): $L = 3\lambda/4 \implies \lambda = 4L/3 \implies \nu_3 = 3v/4L = 3\nu_1$.
Thus, frequencies are $\nu_1, 3\nu_1, 5\nu_1 \dots$ (Only odd harmonics present).
Concept Map
Waves connects to →
Macroscopic Physics
Oscillations (The source of waves)
Optics (Light as an EM wave)
Wave Optics (Class 12 interference)
Quantum Mechanics (Matter waves)
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