Harmonics are the resonant frequencies of a stationary wave on a string

Wave Types, Polarisation & Stationary Waves — AQA A-Level Physics

A string fixed at both ends can only vibrate at specific resonant frequencies called harmonics.
The first harmonic (fundamental frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz).) has 2 nodes and 1 antinodeA point on a stationary wave where the displacement is a maximum. Located midway between adjacent nodes.. The wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). is 2L.
The second harmonic has 3 nodes and 2 antinodes. The wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). is L.
The third harmonic has 4 nodes and 3 antinodes. The wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). is 2L/3.
The nth harmonic has (n + 1) nodes and n antinodes.

Harmonic frequencies

f_1 = \frac{v}{2L}

\begin{aligned} f_2 &= \frac{v}{L} \\ &= 2f_1 \end{aligned}

\begin{aligned} f_3 &= \frac{3v}{2L} \\ &= 3f_1 \end{aligned}

In general: $f_n = nf_1$. Each harmonic is an integer multiple of the fundamental.
L is the length of the string between the two fixed ends.
v is the wave speedThe distance travelled by a wavefront per unit time. on the string.

Worked Example

A string vibrating in the third harmonic has a frequency of 150 Hz. Calculate the frequency of the fifth harmonic.

Show Solution

Find the fundamental frequency

$f_3 = 3f_1$, so $f_1 = 150 \div 3 = 50$ Hz

Calculate the fifth harmonic

$f_5 = 5f_1 = 5 \times 50 = 250$ Hz

Answer

$f_5 = 250$ Hz

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