Stationary Waves

Superposition of Waves · Maharashtra HSC

Theory

Formation of Stationary Waves

Imagine a string stretched between two fixed points. Pull it at the middle and release it: releasing the string produces two progressive waves travelling in opposite directions. These are reflected at the fixed ends, and the original waves combine with their reflections to produce a stationary wave.

When two identical waves travelling along the same path in opposite directions interfere with each other, the resultant wave is called a stationary wave.

Equation of a Stationary Wave on a Stretched String

Consider two simple harmonic progressive waves of equal amplitude $a$ and wavelength $\lambda$ propagating on a long uniform string in opposite directions. Recall $\dfrac{2\pi}{\lambda} = k$ and $2\pi n = \omega$ .

Wave travelling along the positive $x$ direction:

$y_1 = a\sin\left[2\pi\left(nt - \dfrac{x}{\lambda}\right)\right]$

Wave travelling along the negative $x$ direction:

$y_2 = a\sin\left[2\pi\left(nt + \dfrac{x}{\lambda}\right)\right]$

By the principle of superposition the resultant displacement is

$y = y_1 + y_2 = a\sin\left[2\pi\left(nt - \dfrac{x}{\lambda}\right)\right] + a\sin\left[2\pi\left(nt + \dfrac{x}{\lambda}\right)\right]$

Using

$\sin C + \sin D = 2\sin\left(\dfrac{C+D}{2}\right)\cos\left(\dfrac{C-D}{2}\right)$

we get

$y = 2a\sin(2\pi nt)\cos\dfrac{2\pi x}{\lambda}$

$\therefore\ y = 2a\cos\dfrac{2\pi x}{\lambda}\,\sin(2\pi nt) \qquad \ldots (1)$

Substituting $2a\cos\dfrac{2\pi x}{\lambda} = A$ in equation (1),

$y = A\sin(2\pi nt), \qquad \text{that is} \qquad y = A\sin\omega t \quad (\because\ \omega = 2\pi n)$

This is the equation of a stationary wave, giving the resultant displacement due to two simple harmonic progressive waves.

Why it is not a progressive wave

The terms in position $x$ and in time $t$ appear separately and not as the combination $2\pi\left(nt \pm \dfrac{x}{\lambda}\right)$ . The variable $x$ is present only in the expression for the amplitude, $A = 2a\cos\dfrac{2\pi x}{\lambda}$ , which is a periodic function of $x$ — the amplitude varies periodically in space. Different particles therefore have different amplitudes, but every point of the string oscillates with the same frequency $\omega$ , the same as that of each individual progressive wave. All particles of the string pass through their mean positions simultaneously, twice during each vibration. The wave moves neither to the left nor to the right, so we call it a stationary or standing wave.

Condition for Nodes

Nodes are the points of minimum displacement, so the amplitude must be zero:

$2a\cos\dfrac{2\pi x}{\lambda} = 0 \qquad \therefore\ \cos\dfrac{2\pi x}{\lambda} = 0$

$\therefore\ \dfrac{2\pi x}{\lambda} = \dfrac{\pi}{2},\ \dfrac{3\pi}{2},\ \dfrac{5\pi}{2},\ \ldots$

$\therefore\ x = \dfrac{\lambda}{4},\ \dfrac{3\lambda}{4},\ \dfrac{5\lambda}{4},\ \ldots$

$\text{i.e.}\quad x = (2p-1)\dfrac{\lambda}{4}, \qquad p = 1, 2, 3,\ldots$

Distance between two successive nodes:

$\dfrac{3\lambda}{4} - \dfrac{\lambda}{4} = \dfrac{\lambda}{2}; \qquad \dfrac{5\lambda}{4} - \dfrac{3\lambda}{4} = \dfrac{\lambda}{2}$

So the distance between two successive nodes is $\dfrac{\lambda}{2}$ .

Condition for Antinodes

Antinodes are the points of maximum displacement, so $A = \pm 2a$ :

$2a\cos\dfrac{2\pi x}{\lambda} = \pm 2a \qquad \therefore\ \cos\dfrac{2\pi x}{\lambda} = \pm 1$

$\therefore\ \dfrac{2\pi x}{\lambda} = 0,\ \pi,\ 2\pi,\ 3\pi,\ \ldots$

$\therefore\ x = 0,\ \dfrac{\lambda}{2},\ \lambda,\ \dfrac{3\lambda}{2},\ \ldots$

$\text{i.e.}\quad x = \dfrac{p\lambda}{2}, \qquad p = 0, 1, 2, 3,\ldots$

Distance between two successive antinodes:

$\dfrac{\lambda}{2} - 0 = \dfrac{\lambda}{2}; \qquad \lambda - \dfrac{\lambda}{2} = \dfrac{\lambda}{2}$

So the distance between two successive antinodes is also $\dfrac{\lambda}{2}$ .

Nodes and antinodes are formed alternately and are equally spaced. Therefore the distance between a node and an adjacent antinode is $\dfrac{\lambda}{4}$ .

Motion of the Particles

When $\sin\omega t = 1$ , all the particles for which $\cos kx$ is positive have their maximum displacement in the positive direction, while at the same instant all those for which $\cos kx$ is negative have their maximum displacement in the negative direction. When $\sin\omega t = 0$ , all the particles cross their mean positions together, some moving in the positive and some in the negative direction.

Longitudinal Stationary Waves

Longitudinal waves such as sound in a tube or pipe of finite length are reflected at the ends in the same way as transverse waves on a string, and interference between the waves travelling in opposite directions gives standing waves. A longitudinal stationary wave is drawn as loops for convenience, but the actual motion of the particles is along the length of the loop and not perpendicular to it.

Diagram in Words

Draw the string as a straight horizontal line and superimpose on it a series of equal, symmetric loops (lens shapes) drawn above and below the line. The pinch points where consecutive loops meet are the nodes, marked N; the widest part of each loop is an antinode, marked A. Mark $\dfrac{\lambda}{2}$ between two adjacent N, and $\dfrac{\lambda}{4}$ between an N and the neighbouring A. Two-headed arrows inside each loop show the particles moving up and down, with adjacent loops arrowed in opposite senses. For a longitudinal stationary wave in a pipe closed at one end, draw the loop diagram alongside a row of dots showing air particles crowded at the nodes and spread out at the antinodes.

Properties of Stationary Waves

Produced by superposition of two identical waves, transverse or longitudinal, travelling through a medium along the same path in opposite directions.
Two identical transverse progressive waves give a transverse stationary wave; some points on the string are motionless and are called nodes, some oscillate with the greatest amplitude and are called antinodes, and points in between vibrate with amplitudes between $0$ and $A$ .
Two identical longitudinal progressive waves give a longitudinal stationary wave, with the same definitions of node and antinode in terms of the amplitude of the particles of the medium.
The distance between two consecutive nodes is $\dfrac{\lambda}{2}$ and between two consecutive antinodes is $\dfrac{\lambda}{2}$ .
Nodes and antinodes are produced alternately; the distance between a node and an adjacent antinode is $\dfrac{\lambda}{4}$ .
The amplitude of vibration varies periodically in space, but all points vibrate with the same frequency.
Though all the particles except those at the nodes possess energy, there is no propagation of energy; the wave is localised and its velocity is zero, which is why it is called stationary.
All the particles in one loop vibrate in phase, with no progressive change of phase from particle to particle; the phase reverses for the adjacent loop.

Comparison of Progressive and Stationary Waves

Progressive waves	Stationary waves
The disturbance travels from one region to the other with a definite velocity	The disturbance remains in the region where it is produced; the velocity of the wave is zero
Amplitudes of all particles are the same	Amplitudes of particles are different
The particles do not cross their mean positions simultaneously	All the particles cross their mean positions simultaneously
All the particles are moving	Particles at the position of nodes are always at rest
Energy is transmitted from one region to another	There is no transfer of energy
Phases of adjacent particles are different	All particles between two consecutive nodes move in the same direction and are in phase, while those in adjacent loops move in opposite directions and differ in phase by $180^{\circ}$

Standard Results

$y = 2a\cos\dfrac{2\pi x}{\lambda}\sin(2\pi nt) = A\sin\omega t$ , with $A = 2a\cos\dfrac{2\pi x}{\lambda}$ .
Nodes at $x = (2p-1)\dfrac{\lambda}{4}$ , $p = 1, 2, 3,\ldots$ ; antinodes at $x = \dfrac{p\lambda}{2}$ , $p = 0, 1, 2,\ldots$
Node to node $= \dfrac{\lambda}{2}$ ; antinode to antinode $= \dfrac{\lambda}{2}$ ; node to adjacent antinode $= \dfrac{\lambda}{4}$ .
Musical instruments such as the violin and the tanpura work on the formation of stationary waves.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
State the points of comparison between progressive waves and stationary waves.
Maharashtra HSC board question — asked in March 2009Answer 1 ↓
2
Explain analytically how the stationary waves are formed. Hence show that the distance between node and adjacent antinode is $\dfrac{\lambda}{4}$ .
Maharashtra HSC board question — asked in March 2016Answer 2 ↓
3
Explain the formation of stationary waves by analytical method. Show the formation of stationary wave diagrammatically.
Maharashtra HSC board question — asked in March 2017Answer 3 ↓
4
Explain the formation of stationary waves by analytical method. Show that nodes and antinodes are equally spaced in a stationary wave. OR Derive an expression for equation of stationary wave on a stretched string. Show that the distance between two successive nodes or antinodes is λ/2.
Maharashtra HSC board question — asked in February 2024; July 2016; March 2019Answer 4 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question

Sr. No.	Progressive waves	Stationary waves
i.	The disturbance travels from one region to the other with definite velocity.	Disturbance remains in the region where it is produced, velocity of the wave is zero.
ii.	Amplitudes of all particles are same.	Amplitudes of particles are different.
iii.	Particles do not cross each other.	All the particles cross their mean positions simultaneously.
iv.	All the particles are moving.	Particles at the position of nodes are always at rest.
v.	There is no transfer of energy.	Energy is transmitted from one region to another.
vi.	Phases of adjacent particles are different.	All particles between two consecutive nodes are moving in the same direction and are in phase while those in adjacent loops are moving in opposite directions and differ in phase by 180°.

2.↑ back to the question
i. Consider two simple harmonic progressive waves of equal amplitudes (a) and wavelength (λ) propagating on a long uniform string in opposite directions. ii. The equation of wave travelling along the X-axis in the positive direction is given by, $y_1=a\sin\left[2\pi\left(nt-\dfrac{x}{\lambda}\right)\right]$ The equation of wave travelling along the X-axis in the negative direction is given by, $y_2=a\sin\left[2\pi\left(nt+\dfrac{x}{\lambda}\right)\right]$ iii. When these waves interfere, the resultant displacement of particles of string is given by the principle of superposition of waves as y = $y_1+y_2$ $\therefore y=a\sin\left[2\pi\left(nt-\dfrac{x}{\lambda}\right)\right]+a\sin\left[2\pi\left(nt+\dfrac{x}{\lambda}\right)\right]$ iv. By using trigonometry formula, $\sin C+\sin D=2\sin\left(\dfrac{C+D}{2}\right)\cos\left(\dfrac{C-D}{2}\right)$ $\therefore y=2a\sin(2\pi nt)\cos\dfrac{2\pi x}{\lambda}$ $y=2a\cos\dfrac{2\pi x}{\lambda}\sin(2\pi nt)\qquad....(1)$ v. Substituting $2a\cos\dfrac{2\pi x}{\lambda}=A$ in equation (1), y = A sin(2πnt) y = A sin ωt $\qquad(\because\omega=2\pi n)$ This is the equation of a stationary wave which gives resultant displacement due to two simple harmonic progressive waves. vi. Condition for node: Nodes are the points of minimum displacement. This is possible if the amplitude is minimum, i.e., A = 0. $\therefore 2a\cos\dfrac{2\pi x}{\lambda}=0$ $\therefore \cos\dfrac{2\pi x}{\lambda}=0$ $\therefore \dfrac{2\pi x}{\lambda}=\dfrac{\pi}{2},\dfrac{3\pi}{2},\dfrac{5\pi}{2},....$ $\therefore x=\dfrac{\lambda}{4},\dfrac{3\lambda}{4},\dfrac{5\lambda}{4},....$ i.e., $x=(2p-1)\dfrac{\lambda}{4}$ where p = 1, 2, 3,.... Distance between two successive nodes: $\dfrac{3\lambda}{4}-\dfrac{\lambda}{4}=\dfrac{\lambda}{2}$ ; $\dfrac{5\lambda}{4}-\dfrac{3\lambda}{4}=\dfrac{\lambda}{2}$ $\therefore$ Distance between two successive nodes is $\dfrac{\lambda}{2}$ . vii. Condition for antinode: Antinodes are the points of maximum displacement, i.e., A = ± 2a $\therefore 2a\cos\dfrac{2\pi x}{\lambda}=\pm2a$ $\therefore \cos\dfrac{2\pi x}{\lambda}=\pm1$ $\therefore \dfrac{2\pi x}{\lambda}=0,\pi,2\pi,3\pi,....$ $\therefore x=0,\dfrac{\lambda}{2},\lambda,\dfrac{3\lambda}{2},....$ i.e., $x=\dfrac{\lambda p}{2}$ where p = 0, 1, 2, 3,... $\therefore$ Distance between two successive antinodes: $\dfrac{\lambda}{2}-0=\dfrac{\lambda}{2}$ ; $\lambda-\dfrac{\lambda}{2}=\dfrac{\lambda}{2}$ $\therefore$ Distance between two successive antinodes is $\dfrac{\lambda}{2}$ viii. Nodes and antinodes are formed alternately and are equally spaced. Therefore, the distance between a node and an adjacent antinode is $\dfrac{\lambda}{4}$ .
3.↑ back to the question
i. [From 6.5, Q2]
i. Consider two simple harmonic progressive waves of equal amplitudes (a) and wavelength (λ) propagating on a long uniform string in opposite directions. ii. The equation of wave travelling along the X-axis in the positive direction is given by, $y_1=a\sin\left[2\pi\left(nt-\dfrac{x}{\lambda}\right)\right]$ The equation of wave travelling along the X-axis in the negative direction is given by, $y_2=a\sin\left[2\pi\left(nt+\dfrac{x}{\lambda}\right)\right]$ iii. When these waves interfere, the resultant displacement of particles of string is given by the principle of superposition of waves as y = $y_1+y_2$ $\therefore y=a\sin\left[2\pi\left(nt-\dfrac{x}{\lambda}\right)\right]+a\sin\left[2\pi\left(nt+\dfrac{x}{\lambda}\right)\right]$ iv. By using trigonometry formula, $\sin C+\sin D=2\sin\left(\dfrac{C+D}{2}\right)\cos\left(\dfrac{C-D}{2}\right)$ $\therefore y=2a\sin(2\pi nt)\cos\dfrac{2\pi x}{\lambda}$ $y=2a\cos\dfrac{2\pi x}{\lambda}\sin(2\pi nt)\qquad....(1)$ v. Substituting $2a\cos\dfrac{2\pi x}{\lambda}=A$ in equation (1), y = A sin(2πnt) y = A sin ωt $\qquad(\because\omega=2\pi n)$ This is the equation of a stationary wave which gives resultant displacement due to two simple harmonic progressive waves. vi. Condition for node: Nodes are the points of minimum displacement. This is possible if the amplitude is minimum, i.e., A = 0. $\therefore 2a\cos\dfrac{2\pi x}{\lambda}=0$ $\therefore \cos\dfrac{2\pi x}{\lambda}=0$ $\therefore \dfrac{2\pi x}{\lambda}=\dfrac{\pi}{2},\dfrac{3\pi}{2},\dfrac{5\pi}{2},....$ $\therefore x=\dfrac{\lambda}{4},\dfrac{3\lambda}{4},\dfrac{5\lambda}{4},....$ i.e., $x=(2p-1)\dfrac{\lambda}{4}$ where p = 1, 2, 3,.... Distance between two successive nodes: $\dfrac{3\lambda}{4}-\dfrac{\lambda}{4}=\dfrac{\lambda}{2}$ ; $\dfrac{5\lambda}{4}-\dfrac{3\lambda}{4}=\dfrac{\lambda}{2}$ $\therefore$ Distance between two successive nodes is $\dfrac{\lambda}{2}$ . vii. Condition for antinode: Antinodes are the points of maximum displacement, i.e., A = ± 2a $\therefore 2a\cos\dfrac{2\pi x}{\lambda}=\pm2a$ $\therefore \cos\dfrac{2\pi x}{\lambda}=\pm1$ $\therefore \dfrac{2\pi x}{\lambda}=0,\pi,2\pi,3\pi,....$ $\therefore x=0,\dfrac{\lambda}{2},\lambda,\dfrac{3\lambda}{2},....$ i.e., $x=\dfrac{\lambda p}{2}$ where p = 0, 1, 2, 3,... $\therefore$ Distance between two successive antinodes: $\dfrac{\lambda}{2}-0=\dfrac{\lambda}{2}$ ; $\lambda-\dfrac{\lambda}{2}=\dfrac{\lambda}{2}$ $\therefore$ Distance between two successive antinodes is $\dfrac{\lambda}{2}$ viii. Nodes and antinodes are formed alternately and are equally spaced. Therefore, the distance between a node and an adjacent antinode is $\dfrac{\lambda}{4}$ ..(i,ii,iii,iv,v). ii.
4.↑ back to the question
[From 6.5, Q2]
i. Consider two simple harmonic progressive waves of equal amplitudes (a) and wavelength (λ) propagating on a long uniform string in opposite directions. ii. The equation of wave travelling along the X-axis in the positive direction is given by, $y_1=a\sin\left[2\pi\left(nt-\dfrac{x}{\lambda}\right)\right]$ The equation of wave travelling along the X-axis in the negative direction is given by, $y_2=a\sin\left[2\pi\left(nt+\dfrac{x}{\lambda}\right)\right]$ iii. When these waves interfere, the resultant displacement of particles of string is given by the principle of superposition of waves as y = $y_1+y_2$ $\therefore y=a\sin\left[2\pi\left(nt-\dfrac{x}{\lambda}\right)\right]+a\sin\left[2\pi\left(nt+\dfrac{x}{\lambda}\right)\right]$ iv. By using trigonometry formula, $\sin C+\sin D=2\sin\left(\dfrac{C+D}{2}\right)\cos\left(\dfrac{C-D}{2}\right)$ $\therefore y=2a\sin(2\pi nt)\cos\dfrac{2\pi x}{\lambda}$ $y=2a\cos\dfrac{2\pi x}{\lambda}\sin(2\pi nt)\qquad....(1)$ v. Substituting $2a\cos\dfrac{2\pi x}{\lambda}=A$ in equation (1), y = A sin(2πnt) y = A sin ωt $\qquad(\because\omega=2\pi n)$ This is the equation of a stationary wave which gives resultant displacement due to two simple harmonic progressive waves. vi. Condition for node: Nodes are the points of minimum displacement. This is possible if the amplitude is minimum, i.e., A = 0. $\therefore 2a\cos\dfrac{2\pi x}{\lambda}=0$ $\therefore \cos\dfrac{2\pi x}{\lambda}=0$ $\therefore \dfrac{2\pi x}{\lambda}=\dfrac{\pi}{2},\dfrac{3\pi}{2},\dfrac{5\pi}{2},....$ $\therefore x=\dfrac{\lambda}{4},\dfrac{3\lambda}{4},\dfrac{5\lambda}{4},....$ i.e., $x=(2p-1)\dfrac{\lambda}{4}$ where p = 1, 2, 3,.... Distance between two successive nodes: $\dfrac{3\lambda}{4}-\dfrac{\lambda}{4}=\dfrac{\lambda}{2}$ ; $\dfrac{5\lambda}{4}-\dfrac{3\lambda}{4}=\dfrac{\lambda}{2}$ $\therefore$ Distance between two successive nodes is $\dfrac{\lambda}{2}$ . vii. Condition for antinode: Antinodes are the points of maximum displacement, i.e., A = ± 2a $\therefore 2a\cos\dfrac{2\pi x}{\lambda}=\pm2a$ $\therefore \cos\dfrac{2\pi x}{\lambda}=\pm1$ $\therefore \dfrac{2\pi x}{\lambda}=0,\pi,2\pi,3\pi,....$ $\therefore x=0,\dfrac{\lambda}{2},\lambda,\dfrac{3\lambda}{2},....$ i.e., $x=\dfrac{\lambda p}{2}$ where p = 0, 1, 2, 3,... $\therefore$ Distance between two successive antinodes: $\dfrac{\lambda}{2}-0=\dfrac{\lambda}{2}$ ; $\lambda-\dfrac{\lambda}{2}=\dfrac{\lambda}{2}$ $\therefore$ Distance between two successive antinodes is $\dfrac{\lambda}{2}$ viii. Nodes and antinodes are formed alternately and are equally spaced. Therefore, the distance between a node and an adjacent antinode is $\dfrac{\lambda}{4}$ .[(i) to (vii)]