Questions
If the length of a closed organ pipe is 1m and velocity of sound is 330 m/s, then the frequency for the second note is
detailed solution
Correct option is B
For closed pipe n1=v4l=3304HzSecond note = 3n1=3×3004HzTalk to our academic expert!
Similar Questions
A tube of certain diameter and of length 48 cm is open at both ends. Its fundamental frequency of resonance is found to be 320 Hz. The velocity of sound in air is 320 m/s. One end of the tube is now closed, considering the effect of end correction, calculate the lowest frequency of resonance for the tube (in Hz).
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