Question

Find vilocity of sound in a gas when the 2 waves of wavelenth 0.8 and 0.81 metre produce 4 beat per second

Answer 1

The beat frequency (4 beat per second = 4 Hz ) corresponds to the difference between the frequencies of the two waves:
$4 Hz = f_1 - f_2$ (1)

The frequency of the first wave can be written as
$f_1= \frac{v}{\lambda_1}$
where v is the speed of sound in the gas and $\lambda_1 = 0.8 m$ is the wavelength of the first wave, while the frequency of the second wave is
$f_2 = \frac{v}{\lambda_2}$
where $\lambda_2=0.81 m$ is the wavelength of the second wave. If we substitute into the first equation (1), we find
$4 Hz = \frac{v}{\lambda_1}- \frac{v}{\lambda_2} = v (\frac{1}{\lambda_2}- \frac{1}{\lambda_2} )$

and if we re-arrange it, we can find the velocity of the sound in the gas:
$v= \frac{4 Hz}{ \frac{1}{\lambda_1}- \frac{1}{\lambda_2} }= \frac{4 Hz}{ \frac{1}{0.8 m} - \frac{1}{0.81 m} }=259.2 m/s$