Questions
In a resonance tube the first resonance with a tuning fork occurs at 16 cm and second at 49 cm. If the velocity of sound is 330 m/s, the frequency of tuning fork is (in Hz)
detailed solution
Correct option is A
For closed pipe l1=v4n; l2=3v4n⇒v=2n (l2−l1)⇒n=v2(l2−l1)=3302×(0.49−0.16)=500 HzTalk 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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