The fundamental frequency of an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is
(A) 12.5cm
(B) 13.2cm
(C) 16cm
(D) 8cm
Answer
Verified
118.8k+ views
Hint We should know that by the term fundamental frequency we mean the lowest frequency at which a medium will oscillate freely. Based on this concept we have to solve this question.
Complete step by step answer
We know that the fundamental frequency for an open organ pipe is $\dfrac{{\text{V}}}{{{\text{2l}}}}$
Also, we can therefore derive the frequency of third harmonic for closed organ pipe as $\dfrac{{{\text{3V}}}}{{{\text{4}}{{\text{l}}_1}}}$
Since both are given as equal in the question,
We can derive that,
$\dfrac{{\text{V}}}{{{\text{2l}}}} = \dfrac{{{\text{3V}}}}{{{\text{4}}{{\text{l}}_1}}}$
$\Rightarrow {\text{l = }}\dfrac{{{\text{2}}{{\text{l}}_1}}}{3}$
$\Rightarrow {\text{l = }}\dfrac{{2 \times 20}}{3}$
$\therefore {\text{l = 13}}{\text{.2cm}}$
Therefore, the correct answer is Option B.
Note The fundamental frequency is important because it will give us an idea about the strongest audible pitch reference which is predominant in any kind of complex waveform.
Complete step by step answer
We know that the fundamental frequency for an open organ pipe is $\dfrac{{\text{V}}}{{{\text{2l}}}}$
Also, we can therefore derive the frequency of third harmonic for closed organ pipe as $\dfrac{{{\text{3V}}}}{{{\text{4}}{{\text{l}}_1}}}$
Since both are given as equal in the question,
We can derive that,
$\dfrac{{\text{V}}}{{{\text{2l}}}} = \dfrac{{{\text{3V}}}}{{{\text{4}}{{\text{l}}_1}}}$
$\Rightarrow {\text{l = }}\dfrac{{{\text{2}}{{\text{l}}_1}}}{3}$
$\Rightarrow {\text{l = }}\dfrac{{2 \times 20}}{3}$
$\therefore {\text{l = 13}}{\text{.2cm}}$
Therefore, the correct answer is Option B.
Note The fundamental frequency is important because it will give us an idea about the strongest audible pitch reference which is predominant in any kind of complex waveform.
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