Question

Wavelength of two notes in air is $1\,m$ and $1\dfrac{1}{{164}}m$ . Each note produces $1\,beat/s$ with a third note of a fixed frequency. The speed of sound in air is
A. $330\,m/s$
B. $340\,m/s$
C. $350\,m/s$
D. $328\,m/s$

Answer 1

Hint: Start with the given information and then try to find the relation between the wavelength of the notes, frequency and speed of the sound. Then put all the given information from the question and then finally use the formula for finding the no of beats. As the value of no of beats is already given therefore we can get the speed of the sound.

Formula used:
The relation between velocity and wavelength is,
$v = f\lambda $
where, v is velocity or speed of sound and f is frequency.

Complete step by step solution:
From the question we know that:
Wavelength of first note is ${\lambda _1} = 1m$
Wavelength of second note is ${\lambda _2} = 1\dfrac{1}{{164}}m$
Solving; ${\lambda _2} = \dfrac{{165}}{{164}}m$
Each note produces one beat which means we have a total no of beats is 2.

Now we know that:
$v = f\lambda $
So, $f = \dfrac{v}{\lambda } \\ $
$\Rightarrow {f_1} = \dfrac{v}{{{\lambda _1}}} = \dfrac{v}{1} \\ $
$\Rightarrow {f_2} = \dfrac{v}{{\dfrac{{165}}{{164}}}} = \dfrac{{164v}}{{165}} \\ $
Now no of beats ${f_1} - {f_2} = 2$.......(given in question)
Putting values from above equations, we get;
$v - \dfrac{{164v}}{{165}} = 2$
further solving we get,
$\dfrac{v}{{165}} = 2$
$\therefore v = 330\,m/s$

Hence the correct answer is option A.

Note: Here the frequency of the third note is fixed it’s not the case all the time so if it is not fixed then the outcome will be different. Also it depends on the no of beats if it changes the answer will change too. Try to be careful about each and every value provided in the question and do not make mistakes in that.