Show that all harmonics are present on a stretched string between two rigid supports.

Explanation

Consider a string of linear density m stretched between two rigid supports a distance L apart. Let T be the temsion in the string
Stationary waves set up on the string are suhjected to two boundary conditions the displacement y=0 at x=0 and at x=L all times. That is. there must be a node at each fixed end. These two boundary conditions limit the possible modes of vibration to a discrete set of frequencies such that there are an integral number q of loops between the two fixed ends

Since the length of one loop (the distane between consecutive nodes) corresponds to half a wavelenth `(lamda)`
`(L)/(q)=(lamda)/(2)`
`therefore lamda =(2L)/q` ... (1)
The speed of a transverse wave on a stretched string is
`v=nlamda=sqrt(t//m)` ... (2) Therefore, from Eqs. (1) and (2), the allowed frequencies are given by
`n=(q)/(2L)sqrt((T)/(m))` (q 1, 2, 3, ..) ..(3)
In the simplest mode of vibration, only one loop (q 1) is formed [Fig. (a)]. The corresponding lowest allowed frequency,
`n=(1)/(2L)sqrt((T)/(m))` ...(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 (q=2) (Fig. (b) Its frequency, `n_1=(2)/(2L)sqrt((t)/(m))=2n` ...(5)
is twice the fundamental and is, therefore, the second harmonic.
In the second overtone, three loops are formed (q=3) [Fig. (c)]. Its frequency, `n_2=(3)/(2L)sqrt((t)/(m))=3n` ..(6)
is the third harmonic.
Therefore, in general, the frequency of the pth overtone (p =1, 2, 3, ..) is `n_p=(p+1)n` ...(7) i.e., the pth overtone corresponds to the (p + 1)th harmonic.
Equation (3) gives the set of discrete frequencles for the normal modes of vibration of a stretched string. Equation (7) shows that for a stretched string all the harmonics are present as overtones.