An open pipe resonates to a frequency 'f1' and a closed pipe resonates to a frequency 'f2'. If they are joined together to form a longer tube, then it will resonate to a frequency of (Neglect end corrections)

l = l1+l2f=v4l f1=v2l1 and f2=v4l2 v4f=v2f1+v4f2⇒1f=2f1+1f2;f=f1f22f2+f1

An open pipe resonates to a frequency 'f1' and a closed pipe resonates to a frequency 'f2'. If they are joined together to form a longer tube, then it will resonate to a frequency of (Neglect end corrections)

l = l1+l2f=v4l f1=v2l1 and f2=v4l2 v4f=v2f1+v4f2⇒1f=2f1+1f2;f=f1f22f2+f1

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An open pipe resonates to a frequency 'f₁' and a closed pipe resonates to a frequency 'f₂'. If they are joined together to form a longer tube, then it will resonate to a frequency of (Neglect end corrections)

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$\frac{f_{1} f_{2}}{f_{2} + 2 f_{1}}$

$\frac{2 f_{1} f_{2}}{f_{1} + f_{2}}$

$\frac{2 f_{1} f_{2}}{2 f_{2} + f_{1}}$

$\frac{f_{1} f_{2}}{f_{1} + 2 f_{2}}$

answer is D.

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Detailed Solution

l = l₁+l₂

$f = \frac{v}{4 l} f_{1} = \frac{v}{2 l_{1}} and f_{2} = \frac{v}{4 l_{2}} \frac{v}{4 f} = \frac{v}{2 f_{1}} + \frac{v}{4 f_{2}}$

$\Rightarrow \frac{1}{f} = \frac{2}{f_{1}} + \frac{1}{f_{2}}; f = \frac{f_{1} f_{2}}{2 f_{2} + f_{1}}$

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An open pipe resonates to a frequency 'f1' and a closed pipe resonates to a frequency 'f2'. If they are joined together to form a longer tube, then it will resonate to a frequency of (Neglect end corrections)

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An open pipe resonates to a frequency 'f₁' and a closed pipe resonates to a frequency 'f₂'. If they are joined together to form a longer tube, then it will resonate to a frequency of (Neglect end corrections)