Question

The wavelength of two sound notes in the air are $\dfrac{{40}}{{195}}m$ and $\dfrac{{40}}{{193}}m$. Each note produces $9$beats per second separately with a third note of fixed frequency. The velocity of sound in $m/s$is:
(A) $360$
(B) $320$
(C) $300$
(D) $340$

Answer 1

Hint: By definition, a beat is the periodic and repeating fluctuations heard in the intensity of a sound when two sound waves of very similar frequencies interfere with one another. From the given data in question, we know that the difference in the frequencies of the two sound notes with the third note frequency separately gives us $9beats/\sec $. First, we find the frequencies of both the notes by dividing the velocity of sound in air with the wavelength given and subtract with the frequency of the third note to give 2 equations. These 2 equations are added to get the velocity of sound.

Formula used:
Frequency is equal to the velocity of sound in air divided by wavelength. $f = \dfrac{v}{\lambda }$

Complete step by step answer:
We know that the two notes produce $9\text{beats/sec} $.
So we first find frequencies, let $v,{f_1},{f_2}$ be the velocity of sound in air, frequency of sound note 1, and sound note 2 respectively.
${f_1} = \dfrac{v}{{\dfrac{{40}}{{195}}}}$
$ \Rightarrow {f_1} = \dfrac{{195v}}{{40}} $
${f_2} = \dfrac{v}{{\dfrac{{40}}{{193}}}} $
$ \Rightarrow {f_2} = \dfrac{{193v}}{{40}} $
From the question with the third note combined with the first two notes separately produces $9beats/\sec $, let $f$ be the frequency of the third note, we get
$ \dfrac{{195v}}{{40}} - f = 9..........(1) $
$f - \dfrac{{193v}}{{40}} = 9...........(2) $
Adding both the equations, we get :
$ \Rightarrow 18 = \dfrac{\nu }{{40}}(195 - 193) $
$ \Rightarrow 18 = \dfrac{v}{{20}} $
$ \Rightarrow v = 360m/\operatorname{s}$
The units of the velocity of sound In the air is $\text{meter/second}$

Hence, the correct answer is option (A).

Note: We subtract the frequency of the first note with the frequency of the third note whereas in the case of the second note we subtract the frequency of the third note with the second note (${\nu _1} - \nu = 9, \nu - {\nu _2} = 9 $). This is because the frequency of the third note always lies between the frequency of the two notes. Frequency and wavelength are inversely proportional. In the question the wavelength of the first note is shorter than that of the second note, therefore the frequency of the first note is the highest and the second note is the lowest and the third note frequency is in between.