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The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both ends is
A
80 cm
B
100 cm
C
120 cm
D
140 cm
Detailed Solution
Fundamental frequency of the closed pipe: $\frac{v}{4L_C}$
Second overtone of the open pipe: $\frac{3v}{2L_O}$
$\frac{v}{4L_C} = \frac{3v}{2L_O} \Rightarrow L_O = 6L_C$
$L_O = 6\times 20 = 120$ cm
Second overtone of the open pipe: $\frac{3v}{2L_O}$
$\frac{v}{4L_C} = \frac{3v}{2L_O} \Rightarrow L_O = 6L_C$
$L_O = 6\times 20 = 120$ cm