Wavelengths of two notes in air are `((90)/(175))` m and `((90)/(173))` m. Each note produces four beats per second with a third note of a fixed frequency. Calculate the velocity of sound in air.

Explanation

The lowest allowed frequency of vibration (fundamental ) of a bounded medium and all its integral multiples are called harmonics.
The stationary wave in the air column are subject to two boundary condition that there must be a node at the closed end and an antinode at the open end. In the following we shall ignore the end correction.

Let, v be speed of sound of sound in air, In the simplest mode of vibration as shown in the figure.
(a) There is a one node at the closed end and an antinode at the open end. The distance between a node and a constructive antinode is `lambda//4`, where `lambda` is the wavelength of sound. The corresponding wavelength `lambda` and frequency `eta` are
`lambda=4L and n=(v)/(lambda)=(v)/(4L) " " ` ...(i)
This gives the fundamental frequency of vibration, the first overtone, two nodes and two antinodes are formed as shown in the figure (b). The corresponding wavelength `lambda_(1)` and frequency `n_(1)` are
`lambda_(1)=(4L)/(3) and n_(1)=(v)/(lambda_(1))=(3v)/(4L)=3n " " `...(ii)
Therefore, the frequency in the first overtone is three times the fundamental frequency i.e., the first overtone is the third harmonic.
(c ) In the second ovetone, three nodes and three antinodes are formed as shown in the figure (c ).
The corresponding wavelength `lambda_(2)` and frequency `n_(2)` are
`lambda_(2)=(4L)/(5) and n_(2)=(v)/(lambda_(2))=(5v)/(4L)=5n " " `....(iii)
which is the fifth harmonic.
Therefore, in general the frequency of the `p^(th)` overtone (p = 1, 2, 3, ......) is
`n_(p)=(2p+1)n " " `...(iv)
i.e., the `p^(th)` overtone is the `(2p+1)^(th)`
harmonic.
Equation (i), (ii) and (iii) show that allowed frequencies in an air column is a pipe closed at one end are `n, 3n, 5n ....... .`
That is only odd harmonics are present as overtones.
Nomerical :
Given : `lambda_(1)=(81)/(173)m, lambda_(2)=(81)/(170)m.`
Let, `n_(1) and n_(2)` be the corresponding frequencies.
`therefore v=n_(1)lambda_(1)=n_(2)lambda_(2)`
where, v is the speed of sound in air.
`therefore n_(1)=(v)/(lambda_(1)) and n_(2)=(v)/(lambda_(2)) " " ` ...(i)
But `lambda_(2) gt lambda_(1)`
`therefore n_(2) lt n_(1)`
` therefore n_(1)-n_(2)=10`
`(v)/(lambda_(1))-(v)/(lambda_(2))=10`
`v[(1)/(lambda_(1))-(1)/(lambda_(2))=10`
`v[(173)/(81)-(170)/(81)]=10`
`v[(3)/(81)]=10`
`v=(10xx81)/(3)=270 m//s.`