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IBPhysicsSLRevision NotesWave BehaviourStanding Waves & ResonanceHarmonics in Strings & Pipes

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Harmonics in Strings & Pipes (SL IB Physics)

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Katie M

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Katie M

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Harmonics

  • Stationary waves can have different wave patterns, known as harmonics
    • These depend on the frequency of the vibration and the boundary conditions (i.e. fixed and/or free ends)

  • The harmonics are the only frequencies and wavelengths that will form standing waves on strings or in pipes

Harmonics on Strings

  • The boundary condition is that both ends are fixed
  • The simplest wave pattern is a single loop made up of two nodes (i.e. the two fixed ends) and an antinode
    • This is called the first harmonic 
    • The wavelength of this harmonic is lambda subscript 1 space equals space 2 L
    • Using the wave equation, the frequency is f subscript 1 space equals space fraction numerator space v over denominator 2 L end fraction, where v is the wave speed of the travelling waves on the string (i.e. the incident wave and the reflected wave)

  • As the vibrating frequency increases, more complex patterns arise
    • The second harmonic has three nodes and two antinodes
    • The third harmonic has four nodes and three antinodes

Fixed end wavelengths and harmonics (1), downloadable AS & A Level Physics revision notesFixed end wavelengths and harmonics (2), downloadable AS & A Level Physics revision notes

Diagram showing the first three harmonics on a stretched string fixed at both ends

  • The nth harmonic will have (n + 1) nodes and n antinodes
  • The general expression for the wavelength of the nth harmonic on a string that is fixed at both ends is:

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

  • Where:
    • λn = wavelength in metres (m)
    • L = length of the string in metres (m)
    • n = integer number greater than zero - i.e. 1, 2, 3...

  • Knowing the wavelength λn of the standing wave and the speed v of the travelling waves (i.e. incident and reflected), the natural frequency fn of any harmonic can be calculated using the wave equation v = fλn, so that:

f subscript n space equals space fraction numerator n v over denominator 2 L end fraction

Harmonics in Pipes

  • The boundary conditions vary, since pipes can have:
    • two open ends
    • only one open end

  • For a pipe that is open at both ends:
    • The simplest wave pattern is one central node and two antinodes
    • The second harmonic consists of two nodes and three antinodes
    • The nth harmonic will have (n + 1) antinodes and n nodes
    • The expression for the wavelength of the nth harmonic in a pipe of length L is the same as that given above for nth harmonic on a string

4-5-4-harmonics-in-pipes-open-at-both-ends_sl-physics-rn

Diagram showing the first five harmonics in a pipe open at both ends

  

  • For a pipe that is open at one end:
    • The lowest harmonic is a "half-loop" with one node and one antinode
    • The next possible harmonic will have two nodes and two antinodes
      • This is the third harmonic, not the second one
      • Since only odd harmonics can exist under this boundary condition

 4-5-4-harmonics-in-pipes-open-at-one-end_sl-physics-rn

Diagram showing the first three possible harmonics in a pipe open at one end. Only the odd harmonics can form in this case

  • The expression for the wavelength of the nth harmonic in a pipe of length L is:

lambda subscript n space equals space fraction numerator 4 L over denominator n end fraction

  • Where this time, n is an odd number - i.e. 1, 3, 5...
  • Under both boundary conditions, the natural frequencies are once again calculated from the wavelength of the standing wave and the speed v of the travelling waves using the wave equation

Worked example

Transverse waves travel along a stretched wire 100 cm long. The speed of the waves is 250 m s–1.

Determine the maximum harmonic detectable by a person who can hear up to 15 kHz.

Answer:

Step 1: Write down the known quantities 

  • Length of the wire, L = 100 cm = 1.00 m
  • Speed of the waves, v = 250 m s–1
  • Maximum frequency of human hearing, fn = 15 kHz = 15 000 Hz

Step 2: Write down the equation for the frequency of the nth harmonic and rearrange for n 

f subscript n space equals space fraction numerator n v over denominator 2 L end fraction space space space space space rightwards double arrow space space space space space n space equals space fraction numerator 2 L f subscript n over denominator v end fraction

Step 3: Substitute the numbers into the above equation 

n space equals space fraction numerator 2 cross times 1.00 cross times 15 space 000 over denominator 250 end fraction space equals space 120

  • The person can hear up to the 120th harmonic

Examiner Tip

Before carrying out any calculation on standing waves, you should look carefully at the boundary conditions, since these will determine the wavelengths and natural frequencies of the harmonics.

The expressions for the wavelength of the nth harmonic on strings fixed at both ends (or in pipes open at both ends) and in pipes open at one end are not given in the data booklet and you must be able to recall them to make calculations easier.

Remember that n can take any integer value greater than zero in the case of standing waves on strings and in pipes open at both ends. For pipes open at one end, instead, n can only be an odd integer.

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    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.