Question

Wavelength of two notes in air are $\frac{85}{141}\text{m}$ and $\frac{85}{143}\text{m}$ respectively. Each of these notes produces $4$ beats per second with a third note of a fixed frequency. Calculate the velocity of sound in air.

• A. $360\text{m/s}$
• B. $330\text{m/s}$
• C. $340\text{m/s}$
• D. $332\text{m/s}$

Answer 1

The correct option is C $340\text{m/s}$

Given that
${\lambda }_1=\frac{85}{141}\text{m}$ and ${\lambda }_2=\frac{85}{143}\text{m}$
Let the frequency of the first note be $f_1$ and that of second be $f_2$ and $\upsilon$ be the velocity of sound in air.
So,
$\upsilon =f_1{\lambda }_1$ and $\upsilon =f_2{\lambda }_2$
Since, ${\lambda }_1>{\lambda }_2,$ so we must have $f_1<f_2.$
If $f$ is the frequency of the third note, then from question
$f_2-f=4$ ...(i)
and
$f-f_1=4$ ...(ii)
adding eq (i) and (ii)
$f_2-f_1=8$
$\Rightarrow \frac{\upsilon }{{\lambda }_2}-\frac{\upsilon }{{\lambda }_1}=8\Rightarrow \upsilon [\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}]=8$
$\Rightarrow \upsilon [\frac{143}{85}-\frac{141}{85}]=8$
$\Rightarrow \upsilon =\frac{8\times 85}{2}=340\text{m/s}$