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(A) For an open organ pipe of length $L$,the fundamental frequency is given by $f = v / (2L)$,where $v$ is the speed of sound in air.When the pipe is

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$f$

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(A) For an open organ pipe of length $L$,the fundamental frequency is given by $f = v / (2L)$,where $v$ is the speed of sound in air.When the pipe is immersed in water up to half its length $(L/2)$,the water acts as a closed end at that point.Thus,the effective length of the air column becomes $L' = L/2$.Since the pipe is now closed at one end (by water) and open at the other,it behaves as a closed organ pipe of length $L' = L/2$.The fundamental frequency of a closed organ pipe is given by $f' = v / (4L')$.Substituting $L' = L/2$ into the formula,we get $f' = v / (4 * (L/2)) = v / (2L)$.Comparing this with the original frequency $f = v / (2L)$,we find that $f' = f$.

The length of an open organ pipe is $L$ and its fundamental frequency is $f$. If it is now immersed into water up to half of its length,what will be the new fundamental frequency of the organ pipe?

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