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(C) The fundamental frequency of an open organ pipe of length $l$ is given by $f_1 = \frac{v}{2l}$.The fundamental frequency of an open organ pipe of

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$\frac{vx}{2l^2}$

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(C) The fundamental frequency of an open organ pipe of length $l$ is given by $f_1 = \frac{v}{2l}$.The fundamental frequency of an open organ pipe of length $(l+x)$ is given by $f_2 = \frac{v}{2(l+x)}$.The beat frequency is the difference between the two frequencies: $f_{beat} = |f_1 - f_2|$.$f_{beat} = \frac{v}{2l} - \frac{v}{2(l+x)} = \frac{v}{2} \left( \frac{1}{l} - \frac{1}{l+x} \right)$.$f_{beat} = \frac{v}{2} \left( \frac{l+x-l}{l(l+x)} \right) = \frac{v}{2} \left( \frac{x}{l(l+x)} \right)$.Since $x Therefore,$f_{beat} \approx \frac{v}{2} \left( \frac{x}{l^2} \right) = \frac{vx}{2l^2}$.

An open organ pipe of length $l$ is sounded together with another organ pipe of length $l + x$ in their fundamental tones $(x << l)$. The beat frequency heard will be (speed of sound is $v$):

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