Two narrow organ pipes, are open (length l $_{1}$ ) and the other closed (length l $_{2}$ ) are sounded in their respective fundamental modes. The beat frequency heard is 5 Hz. If now pipes are sounded in their first overtone then also the beat frequency heard is 5 Hz. Then

\frac{l_{1}}{l_{2}} = \frac{1}{2}

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\frac{l_{1}}{l_{2}} = \frac{1}{1}

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\frac{l_{1}}{l_{2}} = \frac{3}{2}

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\frac{l_{1}}{l_{2}} = \frac{2}{3}

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Solution

The correct option is B $\frac{l_{1}}{l_{2}} = \frac{1}{1}$
In a closed pipe the $1^{s t}$ normal mode and first overtone is $\frac{v}{4 l_{2}}$ and $\frac{3 v}{4 l_{2}}$ .
In a open pipe the $1^{s t}$ normal mode and first overtone is $\frac{v}{2 l_{1}}$ and $\frac{v}{l_{1}}$ .
Now, $\frac{3 v}{4 l_{2}} - \frac{v}{4 l_{2}} = 5$ and $\frac{v}{l_{1}} - \frac{v}{2 l_{1}} = 5$ .

Equating both of them we can see that $\frac{l_{1}}{l_{2}} = \frac{1}{1}$ .

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Similar questions

Two narrow organ pipes, one open (length $l_{1}$ ) and the other closed (length $l_{2}$ ) are sounded in the respective fundamental modes. The beat frequency heard is 5 Hz. If now the pipes are sounded in the first overtones, then also the beat frequency heard 5 Hz. Then