The wavelengths of two sound notes in air are | vedclass

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(A) Let the velocity of sound be $v$ and the frequency of the third note be $f_0$. The frequencies of the two given notes are $f_1 = \frac{v}{\lambda_

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$360$

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(A) Let the velocity of sound be $v$ and the frequency of the third note be $f_0$. The frequencies of the two given notes are $f_1 = \frac{v}{\lambda_1} = \frac{v}{40/195} = \frac{195v}{40}$ and $f_2 = \frac{v}{\lambda_2} = \frac{v}{40/193} = \frac{193v}{40}$.Since each note produces $9$ beats per second with the third note, we have:$|f_1 - f_0| = 9$ and $|f_2 - f_0| = 9$.This implies $f_1 - f_0 = 9$ and $f_0 - f_2 = 9$ (assuming $f_1 > f_0 > f_2$).Adding these two equations:$(f_1 - f_0) + (f_0 - f_2) = 9 + 9$$f_1 - f_2 = 18$$\frac{195v}{40} - \frac{193v}{40} = 18$$\frac{2v}{40} = 18$$\frac{v}{20} = 18$$v = 360 \,m/s$.

The wavelengths of two sound notes in air are $\frac{40}{195} \,m$ and $\frac{40}{193} \,m$. Each note produces $9$ beats per second separately with a third note of fixed frequency. The velocity of sound in air in $m/s$ is

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