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  • The fundamental frequency of a closed organ pipe of length 10 cm is equal to the second overtone of an organ pipe open at both ends. The length of the organ pipe open at both ends is:<div class

The fundamental frequency of a closed organ pipe of length 10 cm is equal to the second overtone of an organ pipe open at both ends. The length of the organ pipe open at both ends is:

Option: 1

80 cm


Option: 2

100 cm


Option: 3

120 cm


Option: 4

140 cm


Answers (1)

best_answer

Frequency:

        \frac{\nu}{4l} = \frac{\nu}{4 * 20} = \frac{\nu}{80}

If L is the length of the organ pipe open at both ends, the frequency of the fundamental tone = \frac{\nu}{2l}; then the frequency of the second overtone, the third harmonic = 3 \cdot \frac{\nu}{2l}

According to the question,

\frac{\nu}{80} = 3 \cdot \frac{\nu}{2l} \implies l = \frac{3}{2} * 80 = 120 cm

 

Posted by

shivangi.bhatnagar

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