Questions
An open tube is in resonance with string (frequency of vibration of tube is n0). If tube is dipped in water so that 75% of length of tube is inside water, then the ratio of the frequency of tube to string now will be
detailed solution
Correct option is B
For open tube, n0=v2lFor closed tube length available for resonance is l'=l×25100=l4 ∴Fundamental frequency of water filled tube n=v4l'=v4×(l/4)=vl=2n0⇒nn0=2Talk 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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