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The length of an open organ pipe is twice the length of another closed organ pipe. The fundamental frequency of the open pipe $100 \mathrm{Hz}$. The frequency of the third harmon of the closed pipe is
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The correct answer is:
$300 \mathrm{Hz}$
Let the length of closed organ pipe 1 be. Then,
$$
v_{n}=(2 n-1) \frac{v}{4 l}
$$
Its third harmonics (put $n=2$ ) in above question.
$$
v_{3}=\frac{3}{4} \frac{v}{1}
$$
Now, the length of open organ pipe is 21 , then
$$
v_{n}=n \frac{v}{2(2 l)}=\frac{n v}{4 l}
$$
Its fundamental frequency,
$$
v_{0}=\frac{v}{2 l}=100
$$
$$
v_{3}=\frac{3 v}{4 l}=3 \times 100=300 \mathrm{Hz}
$$
$$
v_{n}=(2 n-1) \frac{v}{4 l}
$$
Its third harmonics (put $n=2$ ) in above question.
$$
v_{3}=\frac{3}{4} \frac{v}{1}
$$
Now, the length of open organ pipe is 21 , then
$$
v_{n}=n \frac{v}{2(2 l)}=\frac{n v}{4 l}
$$
Its fundamental frequency,
$$
v_{0}=\frac{v}{2 l}=100
$$
$$
v_{3}=\frac{3 v}{4 l}=3 \times 100=300 \mathrm{Hz}
$$
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