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The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe L metre long. The length of the open pipe will be
A
$\frac{L}{2}$
B
4 L
C
L
D
2 L
Explanation
Equate $3v/2l_0$ with $3v/4L$.
Detailed Solution
Second overtone (3rd harmonic) of the open pipe: $\frac{3\lambda}{2} = l_0 \Rightarrow \lambda = \frac{2l_0}{3}$
First overtone (3rd harmonic) of the closed pipe: $\frac{3\lambda}{4} = l_C \Rightarrow \lambda = \frac{4l_C}{3} = \frac{4L}{3}$
So $\frac{2l_0}{3} = \frac{4L}{3} \Rightarrow l_0 = 2L$
First overtone (3rd harmonic) of the closed pipe: $\frac{3\lambda}{4} = l_C \Rightarrow \lambda = \frac{4l_C}{3} = \frac{4L}{3}$
So $\frac{2l_0}{3} = \frac{4L}{3} \Rightarrow l_0 = 2L$