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

\frac{V X}{4 l^{2}}

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\frac{V l^{2}}{2 X}

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\frac{V X}{2 l^{2}}

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\frac{V X^{2}}{2 l}

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Solution

The correct option is B $\frac{V l^{2}}{2 X}$
The frequency of an organ pipe with length $l$
$f_{1} = \frac{v}{2 l}$
The frequency of an organ pipe with length $(l + x)$
$f_{2} = \frac{v}{2 (l + x)}$
where $v$ is the speed of sound in air
Then, for the beat frequency heard,
$f_{1} - f_{2} = \frac{v}{2 l} - \frac{v}{2 (l + x)}$
$\Rightarrow f_{1} - f_{2} = \frac{v}{2 l} [1 - (1 + \frac{x}{l})^{- 1}]$
$\Rightarrow f_{1} - f_{2} = \frac{v}{2 l} [1 - 1 + \frac{x}{l}] = \frac{v x}{2 l^{2}}$

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

The correct option is B Vl22XThe frequency of an organ pipe with length lf1=v2lThe frequency of an organ pipe with length (l+x)f2=v2(l+x)where v is the speed of sound in airThen, for the beat frequency heard,f1−f2=v2l−v2(l+x)⇒f1−f2=v2l[1−(1+xl)−1]⇒f1−f2=v2l[1−1+xl]=vx2l2

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