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Standing Waves and Normal Modes

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Standing Waves and Normal Modes - Lesson Summary

Let us consider a system which is bounded at both the ends such as a stretched string fixed at the ends.

Let us generate, a continuous sinusoidal wave of a certain frequency, moving to the right. This wave on reaching the right end gets reflected and travels to the left. The reflected wave on reaching the left end gets reflected again and travels to the right.

This process of reflection continues endlessly.

At any point x and at any time t there are always two overlapping waves, one moving to the right and the other moving to the left.

By superimposition waves 1 and 2, we get a combined wave which can be represented as the following.

y(x, t) = (2a sin kx) cos wt

The above equation does not represent a travelling wave, as the wave form does not move to either side. Therefore the above equation represents a standing wave.

Let us now find the positions of the elements of the stretched string where the amplitude of the standing wave is zero. For amplitude to be zero

2a sin kx = 0 ------- (1)

As a cannot be zero, sin kx = 0, where kx = np -------- (2)

n = 0, 1, 2, -----

Substituting in equation 2

kx = np

for n = 0, 1, 2, 3, -----

The positions of zero amplitude are called nodes. The distance between any two consecutive nodes is .

Now let us find the position of the elements of the stretched string where the amplitude of the standing wave is maximum.

Amplitude = 2a sin kx, The maximum possible amplitude is 2a when |sin kx| = 1

For for n = 0, 1, 2, ….

We know that

for n = 0, 1, 2,------ (1)

Equation 1 gives the position of maximum amplitude. These positions are called the antinodes.

The distance between two consecutive antinodes is . They are located midway between two consecutive nodes.


A stretched string is fixed at both the ends and plucked. We have nodes at the two ends.

If the left end of the string is taken as the origin then for the right end x = L, where L is the length of the string.

In order that the right end is a node, the length for n = 1, 2, 3, -----

for n = 1, 2, 3, ----- (1)

Equation 1 shows that the standing waves on a string of length L have restricted wavelength.

We know that


Velocity wavelength

----- (2)

Substituting 1 in 2

for n = 1,2,3 ----- (3)

Equation 3 gives natural frequencies or Modes of the oscillations of the system.

The natural frequencies of a string are integral multiples of the lowest frequency

Where n = 1

The oscillation mode with this lowest frequency is called the fundamental mode or the first harmonic.

By taking n = 2, 3, ---- we obtain the second harmonic, third harmonic and so on.

The frequencies associated with these modes are denoted as u1, u2, u3 etc. The collection of all possible modes is called the harmonic series and n is called the harmonic number.

A stretched string fixed at both the ends can vibrate simultaneously in more than one mode.

These two waves travelling in opposite directions are in phase. We get an antinode at the open end of the pipe.

Length of the air column = L

----- (1) For n = 0, 1, 2, 3, ------------- represents the position of antinodes.

Substituting x = L in equation 1

For n = 0, 1, 2, 3 -------

For n = 0, 1, 2, 3, ---------- (2)

Equation 2 gives the modes which are sustained in the air column.

Taking n = 0, Fundamental frequency

Taking n = 1, 2, 3, we get odd harmonics of the fundamental frequency, that is,


When the pipe in open at both the ends, there will be antinodes at both the ends and all harmonics will be generated.


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