Waves — Transverse, Longitudinal, Speed & Superposition for Class 11

What Are Waves, Really?

When you throw a stone into a still pond, ripples spread outward. The water itself doesn’t travel to the shore — the disturbance does. That’s the core idea behind waves: energy transfers through a medium (or vacuum) without bulk movement of matter.

Every wave has a source of disturbance and a medium that can be disturbed. The medium’s particles oscillate about their mean positions — they don’t go anywhere permanently. Your job in Class 11 is to deeply understand the mathematics of this oscillation: how fast it moves, what shape it takes, and what happens when two waves meet.

This chapter carries significant weightage in JEE Main — typically 2-3 questions per session — and CBSE boards regularly ask 5-mark derivations on standing waves and Doppler effect. Every concept here connects to the next chapter (Sound), so getting this right pays dividends twice.

Key Terms & Definitions

Wavelength (λ): The distance between two consecutive points in the same phase — crest-to-crest, trough-to-trough, or any matching pair. Unit: metres.

Frequency (f or ν): Number of complete oscillations per second. Unit: Hz. Don’t confuse this with angular frequency.

Angular frequency (ω): ω = 2πf. This is what appears naturally in the wave equation. Unit: rad/s.

Wave number (k): k = 2π/λ. Think of it as “spatial frequency” — how many radians of phase per metre.

Time period (T): Time for one complete oscillation. T = 1/f.

Amplitude (A): Maximum displacement of a particle from mean position. Energy of wave ∝ A².

Phase: The argument of the sine/cosine function, (kx − ωt + φ₀). Two points are “in phase” if their phase difference is 2nπ.

y(x, t) = A \sin(kx - \omega t + \phi_0)

Positive x-direction travel: $(kx - \omega t)$
Negative x-direction travel: $(kx + \omega t)$

Relation between parameters:

v = f\lambda = \frac{\omega}{k} = \frac{\lambda}{T}

Types of Waves

Transverse Waves

Particle displacement is perpendicular to the direction of wave propagation. Classic examples: light waves, waves on a stretched string.

For a transverse wave on a string, the particle at any x moves up and down while the wave pattern moves horizontally. If you’ve ever shaken one end of a long rope, you’ve made a transverse wave.

Longitudinal Waves

Particle displacement is parallel to the direction of propagation. Sound is the textbook example. Particles create compressions (high pressure) and rarefactions (low pressure) along the direction of travel.

Students often say “sound waves are transverse.” Sound in air is always longitudinal. Sound can be transverse only in solids (shear waves), and CBSE/JEE questions specifically test this distinction.

Wave Speed — The Most Asked Formula

Speed on a Stretched String

v = \sqrt{\frac{T}{\mu}}

$T$ = tension in the string (N), $\mu$ = linear mass density = mass/length (kg/m)

Why does this formula make sense? Higher tension → restoring force is stronger → wave moves faster. Higher μ → medium is heavier → harder to accelerate → wave moves slower. The formula captures both effects.

Speed in a Fluid (Bulk Modulus)

v = \sqrt{\frac{B}{\rho}}

where B = bulk modulus (resistance to compression), ρ = density.

Speed of Sound in Ideal Gas

v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma RT}{M}}

Here γ = ratio of specific heats (1.4 for diatomic gases like air), R = 8.314 J/mol·K, M = molar mass. Note: speed of sound increases with temperature — this appears in Doppler effect problems constantly.

JEE Main 2024 (Shift 2) asked about the effect of temperature on speed of sound. The key result: $v \propto \sqrt{T}$ where T is absolute temperature in Kelvin. A 1°C rise near room temperature increases sound speed by approximately 0.6 m/s.

The Superposition Principle

When two waves occupy the same medium simultaneously, the net displacement at any point is the algebraic sum of individual displacements:

y_{net}(x, t) = y_1(x, t) + y_2(x, t)

This is valid when the medium is linear (which it is for small amplitudes — an important assumption NCERT makes).

Constructive and Destructive Interference

Two waves: $y_1 = A\sin(kx - \omega t)$ and $y_2 = A\sin(kx - \omega t + \phi)$

Using the sum-to-product identity:

y_{net} = 2A\cos\left(\frac{\phi}{2}\right)\sin\left(kx - \omega t + \frac{\phi}{2}\right)

The resultant amplitude is $2A\cos(\phi/2)$ .

Constructive (φ = 0, 2π, 4π…): amplitude = 2A (maximum reinforcement)
Destructive (φ = π, 3π, 5π…): amplitude = 0 (complete cancellation)

Standing Waves — The Heart of This Chapter

When two identical waves travel in opposite directions, they superpose to form a standing wave (stationary wave). No net energy transfer occurs.

y_1 = A\sin(kx - \omega t), \quad y_2 = A\sin(kx + \omega t)

y_{net} = 2A\sin(kx)\cos(\omega t)

This is crucial: the spatial part ( $\sin kx$ ) and time part ( $\cos \omega t$ ) are separated. Every point oscillates at the same frequency, but amplitude varies with position.

Nodes: Points where $\sin(kx) = 0$ , so displacement is always zero. Located at $x = 0, \lambda/2, \lambda, \ldots$

Antinodes: Points where $|\sin(kx)| = 1$ , maximum amplitude 2A. Located at $x = \lambda/4, 3\lambda/4, \ldots$

Modes in a String Fixed at Both Ends

Both ends are nodes (displacement = 0 always). This constrains allowed wavelengths.

f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}, \quad n = 1, 2, 3, \ldots

$n=1$ : Fundamental frequency (1st harmonic)
$n=2$ : 2nd harmonic (1st overtone)
All harmonics (odd and even) are present

Open Pipe vs. Closed Pipe

Property	Open Pipe (both ends open)	Closed Pipe (one end closed)
End conditions	Antinodes at both ends	Node at closed, antinode at open
Fundamental	$f_1 = v/2L$	$f_1 = v/4L$
Harmonics	All harmonics	Odd harmonics only

Closed pipes sound an octave lower than open pipes of the same length because $f_1^{closed} = f_1^{open}/2$ . This is why the NCERT figure shows the closed pipe with half the frequency. In CBSE, this comparison is a guaranteed 2-mark question.

Beats

When two waves of slightly different frequencies superpose, we hear periodic variations in loudness — these are beats.

f_{beat} = |f_1 - f_2|

The beat frequency equals the difference of the two frequencies. One “beat” per second means you hear one loud-soft cycle every second. The human ear can distinguish beats only up to about 7 Hz — beyond that, we just hear a rough tone.

A common PYQ: a tuning fork of unknown frequency when sounded with a 512 Hz fork gives 4 beats/second. The unknown frequency is either 508 Hz or 516 Hz. How do you determine which? Load the unknown fork with wax (this slightly decreases its frequency). If beats increase, the unknown was at 516 Hz. If beats decrease, it was at 508 Hz. This logic appears almost every year in CBSE boards.

Doppler Effect

The Doppler effect is the change in observed frequency when source and observer are in relative motion.

f_{observed} = f_0 \cdot \frac{v + v_o}{v - v_s}

Sign convention (NCERT): take positive direction as from observer to source.

$v_o$ = speed of observer (+ve if moving toward source)
$v_s$ = speed of source (+ve if moving toward observer)

The most common mistake here is with signs. Let’s be systematic.

Case 1 — Source moving toward stationary observer: $v_o = 0$ , $v_s =$ positive number.

f_{obs} = f_0 \cdot \frac{v}{v - v_s} > f_0 \quad \checkmark

Case 2 — Observer moving toward stationary source: $v_s = 0$ , $v_o =$ positive number.

f_{obs} = f_0 \cdot \frac{v + v_o}{v} > f_0 \quad \checkmark

Both cases give higher observed frequency, as expected when approach occurs.

Many students flip the signs when source moves away. If source moves away, $v_s$ is negative (moving away from observer, opposite to the observer-to-source direction). Plugging in a negative $v_s$ gives $f_{obs} = f_0 \cdot \frac{v}{v + |v_s|} < f_0$ , which correctly gives lower frequency. Always draw the diagram, mark the direction, then assign signs.

Solved Examples

Example 1 — Easy (CBSE Level)

A wave has frequency 500 Hz and speed 340 m/s. Find wavelength.

\lambda = \frac{v}{f} = \frac{340}{500} = 0.68 \text{ m}

That’s it. Direct formula application. In CBSE, write the formula first, substitute, box the answer.

Example 2 — Medium (JEE Main Level)

A string of length 2 m and mass 0.01 kg is under tension 100 N. Find the third harmonic frequency.

First find μ: $\mu = 0.01/2 = 0.005$ kg/m

Wave speed: $v = \sqrt{T/\mu} = \sqrt{100/0.005} = \sqrt{20000} = 100\sqrt{2}$ m/s

Third harmonic (n=3, both ends fixed):

f_3 = \frac{3v}{2L} = \frac{3 \times 100\sqrt{2}}{2 \times 2} = \frac{300\sqrt{2}}{4} = 75\sqrt{2} \approx 106 \text{ Hz}

Example 3 — Hard (JEE Advanced Level)

Two speakers emit sound at 660 Hz. An observer walks at 2 m/s toward one speaker and away from the other along the line joining them. Speed of sound = 330 m/s. Find beat frequency heard.

From speaker 1 (observer approaching): $f_1 = 660 \times \frac{330 + 2}{330} = 660 \times \frac{332}{330}$

From speaker 2 (observer moving away): $f_2 = 660 \times \frac{330 - 2}{330} = 660 \times \frac{328}{330}$

Beat frequency:

f_{beat} = f_1 - f_2 = \frac{660}{330}(332 - 328) = 2 \times 4 = 8 \text{ Hz}

The observer hears 8 beats per second. This exact type — observer moving between two sources — appeared in JEE Main 2023 Session 1.

Exam-Specific Tips

CBSE Class 11 Boards: Standing waves and Doppler effect together typically account for 8-10 marks. Derivation of standing wave equation (superposition of two counter-propagating waves) is a full 5-mark question almost every year. Know it cold. Beats questions are usually 2-marks — direct formula.

JEE Main: Wave speed on string, harmonics in pipes, and Doppler effect are the three high-frequency areas. Expect one question that mixes two concepts — e.g., a resonating pipe whose resonant frequency you must then use in a Doppler calculation. The paper typically has 1-2 questions from this chapter. Score both.

For CBSE marking scheme: Always state the principle before the formula. “According to the superposition principle…” earns the concept mark even if your arithmetic slips later.

For JEE: The wave equation $y = A\sin(kx - \omega t)$ is tested in sneaky ways — questions about the velocity of a particle on the string ( $\partial y/\partial t$ ) versus wave speed ( $\omega/k$ ). These are completely different quantities.

Common Mistakes to Avoid

Mistake 1 — Confusing particle velocity with wave speed. Wave speed $v = \omega/k$ is constant for a given medium. Particle velocity $v_p = \partial y/\partial t = -A\omega\cos(kx - \omega t)$ varies with position and time. JEE loves asking for “maximum particle velocity” which is Aω, not $\omega/k$ .

Mistake 2 — Applying speed of sound formula without checking temperature. $v = 331 + 0.6t$ m/s (where t is in °C) is the empirical formula. The theoretical formula uses absolute temperature T (Kelvin). When a problem says “temperature doubles,” use $v \propto \sqrt{T}$ with Kelvin values — doubling °C temperature does NOT double the wave speed.

Mistake 3 — Forgetting closed pipes only have odd harmonics. Open pipes: $f_n = nv/2L$ for n = 1, 2, 3, … Closed pipes: $f_n = nv/4L$ for n = 1, 3, 5, … only. A question that asks “which frequency is NOT a resonant frequency of a closed pipe?” is testing exactly this.

Mistake 4 — Wrong node/antinode count. A string of length L fixed at both ends in its nth harmonic has (n+1) nodes and n antinodes. Students often get this off by one. Draw the picture every time.

Mistake 5 — Doppler effect applied to light without checking. The formula $f_{obs} = f_0(v \pm v_o)/(v \mp v_s)$ is for mechanical waves only. For light (EM waves), you need the relativistic Doppler formula. NCERT Class 11 only asks about sound, so don’t apply this formula to light questions.

Practice Questions

Q1. The equation of a wave is $y = 0.05\sin(4\pi x - 20\pi t)$ metres. Find the amplitude, wavelength, frequency, and wave speed.

From the equation $y = A\sin(kx - \omega t)$ :

A = 0.05 m
k = 4π rad/m → λ = 2π/k = 2π/4π = 0.5 m
ω = 20π rad/s → f = ω/2π = 20π/2π = 10 Hz
v = ω/k = 20π/4π = 5 m/s (or v = fλ = 10 × 0.5 = 5 m/s ✓)

Q2. A wire of length 1 m and mass 4 g is under tension 25 N. What is the frequency of its second overtone?

μ = 0.004/1 = 0.004 kg/m

v = √(25/0.004) = √6250 = 25√10 m/s

Second overtone = 3rd harmonic (n=3):

f₃ = 3v/2L = 3 × 25√10 / (2 × 1) = 37.5√10 ≈ 118.6 Hz

Q3. A closed organ pipe of length 35 cm resonates in its first overtone. If speed of sound is 350 m/s, find the frequency.

First overtone of closed pipe = 3rd harmonic (n=3):

f = 3v/4L = 3 × 350 / (4 × 0.35) = 1050/1.4 = 750 Hz

Q4. Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. How many beats are heard per minute?

Beat frequency = |260 - 256| = 4 beats/second

Beats per minute = 4 × 60 = 240 beats per minute

Q5. A train moves toward a stationary observer at 20 m/s blowing a whistle of 500 Hz. Speed of sound = 340 m/s. Find the frequency heard by the observer.

Source moving toward stationary observer: v_o = 0, v_s = 20 m/s (toward observer, positive)

f_obs = 500 × 340/(340 - 20) = 500 × 340/320 = 531.25 Hz

Q6. The maximum particle velocity in a wave is 4 times the wave speed. Find the relation between amplitude and wavelength.

Maximum particle velocity = Aω = 4 × wave speed = 4v

Also, v = fλ, ω = 2πf

So: Aω = 4v → A(2πf) = 4fλ → A = 4λ/2π = 2λ/π

Q7. A standing wave is formed on a string: $y = 0.04\sin(3\pi x)\cos(200\pi t)$ m. Find the distance between adjacent nodes.

From the standing wave equation, k = 3π rad/m

λ = 2π/k = 2π/3π = 2/3 m

Distance between adjacent nodes = λ/2 = 1/3 m ≈ 0.33 m

Q8. An open pipe of length 20 cm and a closed pipe of length 15 cm are sounded simultaneously. Do they produce beats? (v = 340 m/s)

Open pipe fundamental: f_open = v/2L = 340/(2 × 0.20) = 850 Hz

Closed pipe fundamental: f_closed = v/4L = 340/(4 × 0.15) = 566.7 Hz

These are far apart, but let’s check harmonics:

Open pipe harmonics: 850, 1700, 2550 Hz…
Closed pipe harmonics (odd only): 566.7, 1700, 2833 Hz…

Both have 1700 Hz! If sounded at their 2nd harmonic (open) and 3rd harmonic (closed) simultaneously, they produce the same frequency — no beats at 1700 Hz.

At fundamental frequencies, beat frequency = 850 - 566.7 = 283.3 Hz (too fast to perceive as beats).

FAQs

Why do standing waves not transfer energy?

In a standing wave, energy oscillates between kinetic and potential form locally at each point. The two counter-propagating waves carry equal energy in opposite directions, exactly canceling the net energy flux. The Poynting vector (or acoustic intensity) sums to zero everywhere.

What is the difference between a node and a point of zero displacement in a progressive wave?

In a progressive wave, every point eventually oscillates — there are no permanently stationary points. A node in a standing wave is permanently at zero displacement because the two waves always cancel there. One is a momentary zero crossing; the other is a structural feature of the standing wave.

Why is it called the “fundamental frequency” and not just the “lowest frequency”?

“Fundamental” because all other resonant frequencies (harmonics) are integer multiples of it. The fundamental sets the base frequency; every overtone is built on it. In music, the fundamental determines the perceived pitch of a note.

Does the Doppler effect change the speed of sound received, or just the frequency?

Only the frequency changes. The speed of sound depends on the medium (temperature, pressure), not on the motion of source or observer. A faster-moving ambulance doesn’t send sound waves at higher speed — it compresses the wavefronts (shorter wavelength, higher frequency) but the wave still travels at 340 m/s in still air.

Why do closed pipes produce only odd harmonics?

The boundary conditions force a node at the closed end and an antinode at the open end. For this to work, the pipe length must be an odd multiple of λ/4: L = λ/4, 3λ/4, 5λ/4… Translating to frequency: f = v/4L, 3v/4L, 5v/4L… — only odd multiples of the fundamental. Even harmonics would require a node at the open end, which is physically impossible for a free surface.

Is the principle of superposition always valid?

No — only for linear media where disturbances are small. For large-amplitude waves (shock waves, violent ocean waves), the medium response is non-linear and superposition breaks down. NCERT Class 11 assumes small amplitudes throughout.

What happens at the resonant frequency of a pipe?

The driving frequency matches a natural frequency of the air column. The wave reflects back and forth inside the pipe and interferes constructively with itself, building up a large-amplitude standing wave. The amplitude grows until energy losses (viscosity, sound radiation) balance the energy input. That large amplitude is what you hear as a loud resonant tone.