Questions
Stationary waves are set up in air column. Velocity of sound in air is 330 m/s and frequency is 165 Hz. Then distance between the nodes is
detailed solution
Correct option is B
v=330 m/s; n=165 Hz. Distance between two successive nodes = λ2=v2n=3302×165=1mTalk 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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