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(C) Let $v$ be the speed of sound and $n$ be the frequency of the sources. Since $\lambda = v/n$,we have $n = v/\lambda$.For the $1^{st}$ source,the o

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$\frac{2u}{\lambda}$

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(C) Let $v$ be the speed of sound and $n$ be the frequency of the sources. Since $\lambda = v/n$,we have $n = v/\lambda$.For the $1^{st}$ source,the observer is moving away with velocity $u$. The apparent frequency is $n_1 = n \left( \frac{v - u}{v} \right) = n \left( 1 - \frac{u}{v} \right)$.For the $2^{nd}$ source,the observer is moving towards it with velocity $u$. The apparent frequency is $n_2 = n \left( \frac{v + u}{v} \right) = n \left( 1 + \frac{u}{v} \right)$.The beat frequency is the difference between the two apparent frequencies:$|n_2 - n_1| = n \left( 1 + \frac{u}{v} \right) - n \left( 1 - \frac{u}{v} \right) = n \left( \frac{2u}{v} \right)$.Substituting $n = v/\lambda$,we get:Beat frequency $= \left( \frac{v}{\lambda} \right) \left( \frac{2u}{v} \right) = \frac{2u}{\lambda}$.

Two stationary sources each emit waves of wavelength $\lambda$. An observer moves from one source to the other with velocity $u$. The number of beats heard by the observer is:

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