Question

Wavelength of two notes in air are $\left(\frac{90}{175}\right)mand\left(\frac{90}{173}\right)m$, 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 in m/s.

• A. 320
• B. 360
• C. 420
• D. 460

Answer 1

The correct option is B 360

Given ${\lambda }_1=\frac{90}{175}mand{\lambda }_1=\frac{90}{173}m$
If $f_{1}and{f}_2$ are the corresponding frequencies and v is the velocity of sound in air, we have
$v=f_1{\lambda }_{1}andv=f_2{\lambda }_2$
$f_1=\frac{v}{{\lambda }_1}and{f}_2=\frac{v}{{\lambda }_2}$
Since, ${\lambda }_1<{\lambda }_2$ we must have $f_1>f_2$
If f is the frequency of the third note, then $f_1-f=4andf-f_2=4$
$\Rightarrow f_1-f_2=8or\frac{v}{{\lambda }_1}-\frac{v}{{\lambda }_2}=8$
$v[\frac{175}{9}-\frac{173}{90}]=8$ v = 360m/s