Consider a string of linear density m stretched between two rigid supports a distance L apart. Let T be the tension in the string.
Stationary waves set up on the string are subjected to two boundary conditions : the displacement y = 0 at x = 0 and at x = L at all times. That is, there must be a node at each fixed end. These conditions limit the possible modes of vibration to only a discrete set of frequencies such that there are an integral number of loops p between the two fixed ends.
Since, the length of one loop (the distance between consecutive nodes) corresponds to half a wavelength (λ),
\(\frac LP\) = \(\fracλ2\)
\(\therefore\) λ = \(\frac {2L}P\)........(1)
The speed of a transverse wave on a stretched string is
v = nλ = \(\sqrt{\frac Tm}\)......(2)
Therefore, from Eqs. (1) and (2), the allowed frequency are given by
n = \(\frac P{2L}\)\(\sqrt{\frac Tm}\),where p = 1,2,3,......(3)
Modes of vibration of a stretched string
In the simplest mode of vibration, only one loop (p = 1) is formed. The corresponding lowest allowed frequency, n, given by
n = \(\frac 1{2L}\)\(\sqrt{\frac Tm}\)......(4)
is called the fundamental frequency or the first harmonic. The possible modes of vibration with frequencies higher than the fundamental are called the overtones.
In the first overtone, two loops are formed (p = 2). Its frequency,
n1 = \(\frac 2{L}\)\(\sqrt{\frac Tm}\) = 2n…...... (5)
is twice the fundamental and is, therefore, the second harmonic.
In the second overtone, three loops are formed (p = 3). Its frequency,
n2 = \(\frac 3{2L}\)\(\sqrt{\frac Tm}\) = 3n.....… (6)
is the third harmonic.
