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Wavelength of Stationary Waves (CIE AS Physics)

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Wavelength of stationary waves

  • Stationary waves have different wave patterns depending on the frequency of the vibration and the situation in which they are created

 

Two fixed ends

  • When a stationary wave, such as a vibrating string, is fixed at both ends, the simplest wave pattern is a single loop made up of two nodes and an antinode
    • This is called the fundamental mode of vibration or the first harmonic
  • The particular frequencies (i.e. resonant frequencies) of standing waves possible in the string depend on its length L and its speed v
  • As you increase the frequency, the higher harmonics begin to appear
    • The frequencies can be calculated from the string length and wave equation
  • For a string of length L, the wavelength of the lowest harmonic is 2L
    • This is because there is only one loop of the stationary wave, which is a half wavelength

    lambda subscript n space equals space fraction numerator 2 L over denominator n end fraction

lambda subscript 1 space equals space fraction numerator 2 L over denominator 1 end fraction space equals space 2 L

  • The second harmonic has three nodes and two antinodes
    • So the wavelength is equal to the length of the string

lambda subscript 2 space equals space fraction numerator 2 L over denominator 2 end fraction space equals space L

  • The third harmonic has four nodes and three antinodes

lambda subscript 3 space equals space fraction numerator 2 L over denominator 3 end fraction

  • The nth harmonic has n antinodes and n + 1 nodes
  • The nth wavelength has a wavelength of fraction numerator bold 2 bold italic L over denominator bold n end fraction
  • The wavelengths and frequencies of the first three harmonics can be summarised as follows:

Wavelength and frequencies of different harmonics

Fixed end wavelengths and harmonics, downloadable AS & A Level Physics revision notes

Diagram showing the first three modes of vibration of a stretched string with corresponding frequencies

One or two open ends in an air column

  • When a stationary wave is formed in an air column with one or two open ends,  slightly different wave patterns are seen in each

Harmonics in an air column

3-2-3-closed-and-open-ends

Diagram showing modes of vibration in pipes with one end closed and the other open or both ends open

  • Image 1 shows stationary waves in a column which is closed at one end
    • At the closed end, a node forms
    • At the open end, an antinode forms
  • Therefore, the fundamental mode is made up of a quarter wavelength with one node and one antinode
    • Every harmonic after that adds on an extra node or antinode
    • Hence, only odd harmonics form
  • Image 2 shows stationary waves in a column which is open at both ends
    • An antinode forms at each open end
  • Therefore, the fundamental mode is made up of a half wavelength with one node and two antinodes
    • Every harmonic after that adds on an extra node and an antinode
    • Hence, odd and even harmonics can form
  • In summary, a column length L for a wave with wavelength λ and resonant frequency f for stationary waves to appear is as follows:

Air Column Length & Frequencies Summary Table

new-8-1-4-table-of-length

Worked example

A standing wave is set up in a column of length L when a loudspeaker placed at one end emits a sound wave of frequency f. The column is closed at the other end. The speed of sound is 340 m s−1.

For a column of length 7.5 m, what is the wavelength of the second lowest note produced?

Answer:

Step 1: Determine the positions of the nodes and antinodes

  • One end of the column is closed, and the loudspeaker represents an open end
  • Hence, an antinode forms at the loudspeaker (open end) and a node forms at the closed end
  • The fundamental frequency represents the lowest note - this would be 1 node and 1 antinode
  • So, the second-lowest note must have 2 nodes and 2 antinodes

closed-and-open-ends-ma

Step 2: Write an expression for the length of the sound wave in the column

  • In the column, there is a quarter wavelength and a half wavelength, or 3 over 4 lambda
  • Therefore the length of the column is:

L space equals fraction numerator space 3 lambda over denominator 4 end fraction

  • Note: for a column with an open and closed end, L space equals fraction numerator space n lambda over denominator 4 end fraction, this would represent the third harmonic (n = 3)

Step 3: Determine the wavelength of the second lowest note

lambda space equals space fraction numerator 4 L over denominator 3 end fraction space equals space fraction numerator 4 cross times 7.5 space over denominator 3 end fraction space equals space 10 space straight m

Examiner Tip

The fundamental counts as the first harmonic or n = 1 and is the lowest frequency with half or quarter of a wavelength. A full wavelength with both ends open or both ends closed is the second harmonic. Make sure to match the correct wavelength with the harmonic asked for in the question!

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Ashika

Author: Ashika

Expertise: Physics Project Lead

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.