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Answer:

Consider a string of length \(L\), stretched with tension \(T\), of linear density \(m\), fixed at both ends. The speed of transverse waves on it is \(v = \sqrt{T/m}\). When it is plucked, stationary waves are formed with nodes at both fixed ends.

The distance between two successive nodes is \(\lambda/2\), so the length must contain a whole number \(p\) of loops:

\[L = p\frac{\lambda}{2} \Rightarrow \lambda = \frac{2L}{p}\qquad (p = 1, 2, 3, \dots)\]

Frequency of the \(p^{th}\) mode:

\[n_p = \frac{v}{\lambda} = \frac{p\,v}{2L} = \frac{p}{2L}\sqrt{\frac{T}{m}}\]
  • \(p = 1\) (one loop): fundamental frequency \(n_1 = \dfrac{1}{2L}\sqrt{\dfrac{T}{m}} = n\) — first harmonic.
  • \(p = 2\) (two loops): \(n_2 = 2n\) — second harmonic (first overtone).
  • \(p = 3\) (three loops): \(n_3 = 3n\) — third harmonic (second overtone), and so on.

Since \(p\) can take every integer value, the frequencies are \(n, 2n, 3n, 4n, \dots\): both odd and even multiples of the fundamental are possible. Hence all harmonics are present in a stretched string.

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report
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