MHT-CET Physics · Superposition of Waves

Stationary Waves and Vibrating Strings

Two equal waves running opposite ways make a stationary wave, y = 2A sin kx cos ωt, with fixed nodes λ/2 apart; a string fixed at both ends fits a whole number of loops, so it vibrates only at n = (p/2L)√(T/μ).

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Why this matters

36 PYQs, seven HARD — the largest page in the chapter. Two shapes: the pattern itself (node and antinode spacing, counting nodes and loops, reading λ off a standing-wave equation), and the string's frequencies — how they scale with length, tension, radius and density, which harmonic two consecutive resonances are, and matching overtones of two wires.

Concept 1 of 2: Nodes, Antinodes and the Standing-Wave Equation

In a stationary wave nothing travels: some points (nodes) never move and those halfway between (antinodes) swing hardest. Neighbouring nodes are half a wavelength apart, which is also the length of one loop.

Definition

$y = 2A\sin kx\cos\omega t$ : amplitude $2A\sin kx$ varies with position; $\lambda = \dfrac{2\pi}{k}$ .
Node to node (or antinode to antinode) $= \dfrac{\lambda}{2}$ ; node to next antinode $= \dfrac{\lambda}{4}$ .
String fixed at both ends: nodes at the ends. $x$ nodes means $x - 1$ loops, so $L = (x - 1)\dfrac{\lambda}{2}$ . The $p$ th overtone has $p + 1$ loops.
Waves on a stretched string are stationary TRANSVERSE waves; stationary waves form in solids, liquids and gases.
Sound reflected by a wall: node at the wall, first antinode $\dfrac{\lambda}{4}$ away.

Stationary wave

y = 2A\sin kx\cos\omega t, \qquad \text{node spacing} = \frac{\lambda}{2}

Worked example

y = 4\sin\left(\dfrac{\pi x}{5}\right)\cos(100\pi t)

, x in cm. Distance between consecutive nodes?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2022 · Shift 1 · Q5Easy

Example 1 · Superposition of Waves · Stationary Waves and Vibrating Strings

A stationary wave is represented by

y = 10\sin\left(\frac{\pi x}{4}\right)\cos(20\pi t)

where

x

and

y

are in cm and

t

in second. The distance between two consecutive nodes is

Taking the node spacing as λ

Nodes are HALF a wavelength apart. From

\sin\frac{\pi x}{4}

\lambda = 8

cm but the nodes are 4 cm apart.

Drill 14 more on nodes, antinodes and the standing-wave equation

Concept 2 of 2: Frequencies of a Stretched String

The fundamental fits half a wavelength into the string: n = v/2L with v = √(T/μ). Every other mode is a whole multiple. So frequency rises with tension (as √T), falls with length, and falls with thickness and density (μ = πr²ρ).

Definition

$n_p = \dfrac{p}{2L}\sqrt{\dfrac{T}{\mu}}$ , $\mu = \pi r^2\rho$ , so $n \propto \dfrac{1}{Lr}\sqrt{\dfrac{T}{\rho}}$ . First overtone = 2nd harmonic.
Two consecutive resonances $n_p, n_{p+1}$ : fundamental $= n_{p+1} - n_p$ (320 and 400 Hz ⇒ 80 Hz, $p = 4$ ).
Length −40% and tension +44%: $\dfrac{1.2}{0.6} = 2$ . Radius and length doubled: $\dfrac{n}{4}$ .
Segments of one wire: $\dfrac{1}{n} = \dfrac{1}{n_1} + \dfrac{1}{n_2} + \dfrac{1}{n_3}$ ; bridges for 1 : 2 : 3 divide it 6 : 3 : 2.
Hanging bob immersed: $\dfrac{n_1^2}{n_2^2} = \dfrac{W}{W - \text{upthrust}}$ , so relative density $= \dfrac{n_1^2}{n_1^2 - n_2^2}$ .

String harmonics

n_p = \frac{p}{2L}\sqrt{\frac{T}{\mu}}

Worked example

A 0.5 m string with

\mu = 0.01

kg/m is under 100 N. Fundamental frequency? And with four times the tension?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 1 · Q16Moderate

Example 2 · Superposition of Waves · Stationary Waves and Vibrating Strings

If length of stretched string is reduced by 40% and tension is increased by 44%, the ratio of final to initial frequency is

Frequency proportional to tension

n \propto \sqrt{T}

. A 44% rise in tension raises the frequency by 20% (

\sqrt{1.44} = 1.2

), not 44%.

Drill 20 more on frequencies of a stretched string

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Nodes, Antinodes and the Standing-Wave Equation
Stationary wave
$y = 2A\sin kx\cos\omega t, \qquad \text{node spacing} = \frac{\lambda}{2}$
Frequencies of a Stretched String
String harmonics
$n_p = \frac{p}{2L}\sqrt{\frac{T}{\mu}}$

Watch out for (2)

Taking the node spacing as λ
→ Nodes, Antinodes and the Standing-Wave Equation
Frequency proportional to tension
→ Frequencies of a Stretched String

Test yourself on Superposition of Waves

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Superposition of Waves chapter test Drill all 36 questions

Related notes

Beats — sonometers with tuning forks →