Two sound waves of wavelengths 99 cm and 100 | vedclass

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(D) The speed of sound $v = 330 \ m \ s^{-1}$.The wavelengths are $\lambda_1 = 0.99 \ m$ and $\lambda_2 = 1.00 \ m$.The frequencies are $f_1 = \frac{v

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$3$

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(D) The speed of sound $v = 330 \ m \ s^{-1}$.The wavelengths are $\lambda_1 = 0.99 \ m$ and $\lambda_2 = 1.00 \ m$.The frequencies are $f_1 = \frac{v}{\lambda_1} = \frac{330}{0.99} = \frac{33000}{99} = \frac{1000}{3} \ Hz$ and $f_2 = \frac{v}{\lambda_2} = \frac{330}{1.00} = 330 \ Hz$.The beat frequency $f_b = |f_1 - f_2| = |\frac{1000}{3} - 330| = |\frac{1000 - 990}{3}| = \frac{10}{3} \ Hz$.Beat frequency is defined as the number of beats per second. Thus,$f_b = \frac{	ext{Number of beats}}{t}$.Given that $10$ beats are produced in $t$ seconds,we have $\frac{10}{3} = \frac{10}{t}$.Therefore,$t = 3 \ s$.

Two sound waves of wavelengths $99 \ cm$ and $100 \ cm$ produce $10$ beats in a time of $t$ seconds. If the speed of sound in air is $330 \ m \ s^{-1}$,then the value of $t$ in seconds is

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