Physics by Lamhi: Not Your Boring Physics

CLASS 11 · CHAPTER 14 · WAVES

DETAILED NOTES

Waves

The complete chapter, section by section: how a disturbance travels without carrying matter along with it, the equation that describes it, why a stretched string has its own built-in speed limit, and how superposition quietly produces both standing waves and beats. For the interactive version of the beat phenomenon derived in §14.6, see the wave-beats simulation on the concept page.

14.1 Transverse and Longitudinal Waves

A wave is a way of moving energy (and information) from one place to another without moving matter along with it. Drop a stone in a still pond and the ripples travel outward across the surface, but a leaf floating nearby only bobs up and down in place; it never drifts toward the shore. Every particle of the medium does the disturbing locally, and passes it on to its neighbour, nobody actually makes the trip.

Waves that need a physical medium to exist in, string waves, water waves, sound, are called mechanical waves, and they need two properties from that medium: elasticity, so a displaced particle feels a restoring force, and inertia, so it overshoots and keeps the disturbance moving instead of simply snapping back. Light needs neither and travels through vacuum perfectly well, it is not a mechanical wave at all, and we leave it for a later chapter.

14.1.1 Transverse waves

In a transverse wave, each particle of the medium oscillates perpendicular to the direction the wave travels. A wave sent down a stretched string by flicking one end up and down is the clean textbook case: the disturbance races along the string’s length, while any given bit of string only ever moves up and down. Transverse waves rely on the medium resisting a shearing deformation, which is why they travel through solids and along surfaces, but not through the bulk of a fluid, a gas simply has no shear rigidity to restore a sideways displacement.

14.1.2 Longitudinal waves

In a longitudinal wave, particles oscillate along the same line the wave travels. Sound in air is the standard example: a loudspeaker cone pushes forward, briefly compressing the air in front of it into a compression, then pulls back, leaving a rarefaction, and this alternating pattern of compressions and rarefactions, not any one molecule, is what travels to your ear. Because this only needs the medium to resist being squeezed (bulk elasticity), longitudinal waves travel through solids, liquids, and gases alike, which is exactly why sound crosses all three.

14.2 The Displacement Relation in a Progressive Wave

A wave travelling steadily in the +x+x direction, without changing shape, is written as a function of both position and time:

Progressive wave
y(x,t)=Asin⁡(kx−ωt+ϕ)y(x,t) = A\sin(kx - \omega t + \phi)

Here AA is the amplitude, the maximum displacement of a particle from equilibrium, and ϕ\phi is a phase constant fixed by whatever is happening at x=0, t=0x=0,\,t=0. The two quantities worth knowing cold are the angular wave number kk, which packs the wavelength into radians per metre, and the angular frequency ω\omega, which does the same for time:

k=2πλ,ω=2πT=2πfk = \dfrac{2\pi}{\lambda}, \qquad \omega = \dfrac{2\pi}{T} = 2\pi f

A full cycle in space (one wavelength λ\lambda) and one cycle in time (one period TT) are linked by how fast the pattern itself glides along, the wave speed:

Wave speed
v=fλ=ωk\textcolor{#e08a1e}{v} = f\lambda = \dfrac{\omega}{k}

A wave of the form sin⁡(kx+ωt)\sin(kx+\omega t) (plus sign instead of minus) describes exactly the same kind of disturbance travelling in the −x-x direction instead. Note also that this wave speed is not the speed of any particle of the medium, it is the speed of the pattern; a single particle at fixed xx only oscillates up and down with velocity ∂y/∂t=−Aωcos⁡(kx−ωt+ϕ)\partial y/\partial t = -A\omega\cos(kx-\omega t+\phi), a completely different quantity that depends on amplitude and frequency, never on how “fast the wave travels”.

Worked example

Reading a wave equation

A wave on a string is described (SI units throughout) by y(x,t)=0.02sin⁡(4x−200t)y(x,t) = 0.02\sin(4x - 200t). Find its amplitude, wavelength, frequency, and speed, and state which way it travels.

Matching against Asin⁡(kx−ωt)A\sin(kx-\omega t) directly gives A=0.02 mA=0.02\text{ m}, k=4 rad/mk=4\text{ rad/m}, and ω=200 rad/s\omega=200\text{ rad/s}. From these:

λ=2πk=2π4≈1.57 m\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{4} \approx \textcolor{#e08a1e}{1.57\text{ m}}f=ω2π=2002π≈31.8 Hzf = \dfrac{\omega}{2\pi} = \dfrac{200}{2\pi} \approx \textcolor{#e08a1e}{31.8\text{ Hz}}v=ωk=2004=50 m/sv = \dfrac{\omega}{k} = \dfrac{200}{4} = \textcolor{#e08a1e}{50\text{ m/s}}

As a check, fλ=31.8×1.57≈50 m/sf\lambda = 31.8\times1.57 \approx 50\text{ m/s}, matching vv exactly, as it must. Since the phase is (kx−ωt)(kx-\omega t) and not (kx+ωt)(kx+\omega t), the wave travels in the +x direction.

14.3 The Speed of a Travelling Wave on a Stretched String

For a wave on a stretched string, the propagation speed is set entirely by two mechanical properties of the string itself: how hard it is pulled, and how heavy it is per unit length. If TT is the tension and μ=m/L\mu = m/L the linear mass density (mass per unit length), the full derivation from the string’s equation of motion lands on a remarkably clean result:

Wave speed on a string
v=Tμ\textcolor{#e08a1e}{v} = \sqrt{\dfrac{T}{\mu}}

A quick dimensional check is reassuring: [T]=MLT−2[T]=MLT^{-2} and [μ]=ML−1[\mu]=ML^{-1}, so [T/μ]=L2T−2[T/\mu] = L^2T^{-2}, exactly velocity squared. Physically, more tension stiffens the restoring force and speeds the wave up; more mass per length adds inertia and slows it down, precisely as you would guess by tightening or loosening a guitar string.

It is worth being deliberate here, because the chapter uses the letter vv for two genuinely different things. The v=T/μv=\sqrt{T/\mu} above is how fast the wave pattern glides along the string, fixed once and for all by TT and μ\mu. It has nothing to do with how fast an individual bit of string is moving up and down at some instant, that transverse particle speed depends on the wave’s amplitude and frequency (§14.2) and changes continuously as the particle oscillates. Mixing up these two speeds is one of the most common slips in this chapter.

Worked example

Wave speed and transit time on a wire

A wire of length 2.0 m and mass 20 g is stretched with a tension of 80 N. Find the speed of a transverse wave on it, and how long a pulse takes to travel from one end to the other.

The linear mass density first:

μ=mL=0.0202.0=0.01 kg/m\mu = \dfrac{m}{L} = \dfrac{0.020}{2.0} = 0.01\text{ kg/m}v=Tμ=800.01=8000≈89.4 m/sv = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{80}{0.01}} = \sqrt{8000} \approx \textcolor{#e08a1e}{89.4\text{ m/s}}

and the transit time is simply distance over speed:

t=Lv=2.089.4≈22.4 mst = \dfrac{L}{v} = \dfrac{2.0}{89.4} \approx \textcolor{#e08a1e}{22.4\text{ ms}}

14.4 The Principle of Superposition of Waves

What happens when two waves arrive at the same point in the medium at the same time? The medium does not have to “choose” between them. The principle of superposition says the net displacement is just the plain algebraic sum of what each wave would have produced on its own:

y(x,t)=y1(x,t)+y2(x,t)+⋯y(x,t) = y_1(x,t) + y_2(x,t) + \cdots

This holds because the wave equation governing small displacements is linear, double a displacement and you double the restoring force that produces it, with no cross-terms getting in the way. Two consequences follow immediately once the waves separate again: each one carries on completely unaffected by the encounter, as if the other had never been there, and while they overlap, the combined pattern can look very different from either wave alone, including constructive reinforcement, cancellation, or anything in between.

This one principle is doing all the work behind the next two sections: a standing wave (§14.5) is just the superposition of two identical waves travelling in opposite directions, and a beat (§14.6) is just the superposition of two waves of slightly different frequency at a single point. Nothing new is added mathematically, only the situation changes.

14.5 Reflection of Waves and Standing Waves

A wave reaching the end of its medium does not simply vanish, it reflects, and how it reflects depends on what kind of end it meets. At a fixed end (a string tied firmly to a wall), the support cannot move, so the reflected pulse comes back inverted, flipped through a phase of π\pi. At a free end (the string tied to a frictionless ring that can slide), the end overshoots and the reflected pulse comes back upright, with no phase change at all.

Now apply superposition (§14.4) to a string fixed at both ends, continuously fed with a travelling wave: the original wave and its reflections combine into two identical waves of the same amplitude and frequency moving in opposite directions, and the sum is a standing wave, a pattern that oscillates in place rather than travelling at all:

Asin⁡(kx−ωt)+Asin⁡(kx+ωt)=2Asin⁡(kx)cos⁡(ωt)A\sin(kx-\omega t) + A\sin(kx+\omega t) = 2A\sin(kx)\cos(\omega t)

14.5.1 Nodes and antinodes

In 2Asin⁡(kx)cos⁡(ωt)2A\sin(kx)\cos(\omega t), every particle still oscillates in time via the cos⁡(ωt)\cos(\omega t) factor, but the size of that oscillation is fixed by 2Asin⁡(kx)2A\sin(kx), which depends only on position. Where sin⁡(kx)=0\sin(kx)=0, the particle never moves at all, these points are nodes. Exactly halfway between consecutive nodes, sin⁡(kx)=±1\sin(kx)=\pm1 and the particle swings through the full 2A2A, these are antinodes. Consecutive nodes (and consecutive antinodes) sit λ/2\lambda/2 apart, and a node sits λ/4\lambda/4 from its nearest antinode. Because every point between two nodes moves in step, rising and falling together, a standing wave transports no energy down the string at all; it only stores it, sloshing locally between node-bound segments.

Worked example

Locating nodes in a standing wave

A standing wave on a string is given (SI units) by y(x,t)=0.04sin⁡(5πx)cos⁡(200πt)y(x,t) = 0.04\sin(5\pi x)\cos(200\pi t). Find the wavelength, the spacing between nodes, and the speed of the two component travelling waves that combine to form it.

Comparing with 2Asin⁡(kx)cos⁡(ωt)2A\sin(kx)\cos(\omega t) gives k=5π rad/mk=5\pi\text{ rad/m} and ω=200π rad/s\omega=200\pi\text{ rad/s}:

λ=2πk=2π5π=0.4 m\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{5\pi} = \textcolor{#e08a1e}{0.4\text{ m}}

Nodes occur wherever sin⁡(5πx)=0\sin(5\pi x)=0, i.e. x=0, 0.2 m, 0.4 m,…x = 0,\ 0.2\text{ m},\ 0.4\text{ m}, \ldots, so consecutive nodes are λ/2=0.2 m\lambda/2 = \textcolor{#e08a1e}{0.2\text{ m}} apart, exactly as expected. The two travelling waves that superpose to give this pattern each move at:

v=ωk=200π5π=40 m/sv = \dfrac{\omega}{k} = \dfrac{200\pi}{5\pi} = \textcolor{#e08a1e}{40\text{ m/s}}

14.6 Beats

Superposition does not need opposite directions to produce something striking, two waves of nearly equal frequency travelling the same way and overlapping at one point in space produce beats: a tone whose loudness rhythmically swells and fades. Take two waves of equal amplitude AA at a fixed point, oscillating in time only:

y1=Asin⁡(2πf1t),y2=Asin⁡(2πf2t)y_1 = A\sin(2\pi f_1 t), \qquad y_2 = A\sin(2\pi f_2 t)

Superposing them and applying the sum-to-product identity, sin⁡α+sin⁡β=2sin⁡ ⁣(α+β2)cos⁡ ⁣(α−β2)\sin\alpha + \sin\beta = 2\sin\!\left(\frac{\alpha+\beta}{2}\right)\cos\!\left(\frac{\alpha-\beta}{2}\right), with α=2πf1t\alpha = 2\pi f_1 t and β=2πf2t\beta = 2\pi f_2 t, turns the sum into a product:

Superposed wave
y1+y2=[2Acos⁡(π(f1−f2)t)]sin⁡(π(f1+f2)t)y_1+y_2 = \Big[2A\cos\big(\pi(f_1-f_2)t\big)\Big]\sin\big(\pi(f_1+f_2)t\big)

Read this as a fast oscillation riding inside a slow envelope. The sin⁡(π(f1+f2)t)\sin\big(\pi(f_1+f_2)t\big) factor oscillates at the average frequency (f1+f2)/2(f_1+f_2)/2, that is the pitch you actually hear, close to either original tone since f1≈f2f_1\approx f_2. The bracketed 2Acos⁡(π(f1−f2)t)2A\cos\big(\pi(f_1-f_2)t\big) factor varies far more slowly (since f1−f2f_1-f_2 is small) and acts as a time-varying amplitude, the envelope, exactly the swelling-and-fading pattern traced by the orange sum curve in the wave-beats simulation, where it plots (y1+y2)/2(y_1+y_2)/2 for two waves of unit amplitude.

One subtlety is worth getting right: loudness depends only on the envelope’s size, ∣cos⁡(π(f1−f2)t)∣|\cos(\pi(f_1-f_2)t)|, not its sign, and a bare cosine reaches its peak size twice every cycle, once at +1+1 and once at −1-1. So a loud moment is heard twice as often as the envelope’s own mathematical frequency (f1−f2)/2(f_1-f_2)/2 would suggest, which exactly doubles it back to the familiar result:

Beat frequency
fbeat=∣f1−f2∣\textcolor{#e08a1e}{f_{\text{beat}}} = |f_1-f_2|

and correspondingly Tbeat=1/∣f1−f2∣T_{\text{beat}} = 1/|f_1-f_2| is the time between one loud moment and the next.

Worked example

Beats from two tuning forks

Two tuning forks ring at 326 Hz and 320 Hz. Find the beat frequency and beat period, and write the resultant displacement as a product of a carrier and an envelope (amplitude AA each, SI units).

fbeat=∣326−320∣=6 Hz,Tbeat=16≈0.167 sf_{\text{beat}} = |326-320| = \textcolor{#e08a1e}{6\text{ Hz}}, \qquad T_{\text{beat}} = \dfrac{1}{6} \approx \textcolor{#e08a1e}{0.167\text{ s}}

Substituting directly into the boxed result above:

y1+y2=2Acos⁡(6πt) sin⁡(646πt)y_1+y_2 = 2A\cos(6\pi t)\,\sin(646\pi t)

The carrier term sin⁡(646πt)=sin⁡(2π⋅323 t)\sin(646\pi t) = \sin(2\pi\cdot323\,t) oscillates at the average, 323 Hz, close to either fork, while the envelope cos⁡(6πt)\cos(6\pi t) swells and fades, and because its size peaks twice per cycle, a listener hears exactly 6 pulses of loudness every second, matching fbeatf_{\text{beat}} above.

14.7 The Doppler Effect

The Doppler effect is the shift in a wave’s observed frequency caused by relative motion between the source and the observer, even though the source itself never changes what frequency it is emitting. The everyday giveaway is a train sounding a fixed, steady whistle: its pitch sounds distinctly higher while it approaches a platform and drops to distinctly lower the instant it passes and recedes, despite the driver hearing the exact same note the whole time.

A source moving toward a stationary observer effectively “crowds” each successive wavefront closer to the one before it (it has moved forward slightly before emitting the next crest), shortening the wavelength reaching the observer and raising the frequency heard; moving away does the reverse. For a source moving at speed vsv_s and sound speed vv in the medium:

fapproach=f0vv−vs,frecede=f0vv+vsf_{\text{approach}} = f_0\dfrac{v}{v-v_s}, \qquad f_{\text{recede}} = f_0\dfrac{v}{v+v_s}

A moving observer instead changes how often wavefronts are intercepted, not their spacing: approaching the source at speed vov_o sweeps up extra wavefronts per second, while receding misses some, giving f′=f0(v±vo)/vf' = f_0(v\pm v_o)/v. One detail trips up almost everyone at first: the actual speed of sound through the medium, vv, is a property of the medium alone (its elasticity and density) and is completely unaffected by how fast the source or observer happens to be moving.

— From the NCERT exercises

Two of the chapter’s own classic exercise questions, with original worked solutions.

NCERT Exercise 14.4

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10⁻² kg and its linear mass density is 4.0 × 10⁻² kg/m. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

Solution

In the fundamental mode, a string fixed at both ends fits exactly half a wavelength, so first find the string’s length from its mass and linear density, then double it:

L=mμ=3.5×10−24.0×10−2=0.875 m,λ=2L=1.75 mL = \dfrac{m}{\mu} = \dfrac{3.5\times10^{-2}}{4.0\times10^{-2}} = 0.875\text{ m}, \qquad \lambda = 2L = 1.75\text{ m}

(a) The wave speed follows from the standard relation:

v=fλ=45×1.75=78.75 m/s≈78.8 m/sv = f\lambda = 45\times1.75 = \textcolor{#e08a1e}{78.75\text{ m/s} \approx 78.8\text{ m/s}}

(b) and the tension comes from inverting §14.3’s formula, v=T/μv=\sqrt{T/\mu}:

T=μv2=4.0×10−2×(78.75)2≈248 NT = \mu v^2 = 4.0\times10^{-2}\times(78.75)^2 \approx \textcolor{#e08a1e}{248\text{ N}}

NCERT Exercise 14.14

A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. The speed of sound in still air is 340 m/s. (a) What is the frequency of the whistle for a platform observer when the train (i) approaches the platform with a speed of 10 m/s, (ii) recedes from the platform with a speed of 10 m/s? (b) What is the speed of sound in each of these cases?

Solution

This is §14.7’s moving-source formula with f0=400 Hzf_0=400\text{ Hz}, v=340 m/sv=340\text{ m/s}, vs=10 m/sv_s=10\text{ m/s}:

(i) approaching: f′=f0vv−vs=400×340330≈412.1 Hz\text{(i) approaching: } f' = f_0\dfrac{v}{v-v_s} = 400\times\dfrac{340}{330} \approx \textcolor{#e08a1e}{412.1\text{ Hz}}(ii) receding: f′=f0vv+vs=400×340350≈388.6 Hz\text{(ii) receding: } f' = f_0\dfrac{v}{v+v_s} = 400\times\dfrac{340}{350} \approx \textcolor{#e08a1e}{388.6\text{ Hz}}

(b) In both cases the speed of sound stays exactly 340 m/s\textcolor{#e08a1e}{340\text{ m/s}}. Sound speed is fixed by the medium’s own elastic and inertial properties, not by whatever is making the sound, the train’s motion changes the wavelength the platform hears, never the speed at which that wave crosses the platform.

Want the interactive version instead? Try the wave-beats simulation → or take a practice paper →