Questions
A loudspeaker that produces signals from 50 to 500 Hz is placed at the open end of a closed tube of length 1.1 m. The lowest and the highest frequency that excites resonance in the tube are fl and fh, respectively. The velocity of sound is 330 m/s. Then
detailed solution
Correct option is C
For a closed tubefn=nv4L, where n = 1,3,5,7,...(odd numbers)L=1.1m,v=330m/sfn=n×3304×1.1=500Hz⇒n=6.66 S maximum value of n is 5 and minimum value is 1. Hence fl = fmin = 1×3304×1.1 Hz = 75 Hzand fh =fmax= 5×3304×1.1 m = 375 HzTalk to our academic expert!
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An organ pipe P1 closed at one end vibrating in its first overtone and another pipe P2 open at the both ends vibrating in its third overtone are in resonance with a given tuning fork. The ratio of the length of P1 to that of P2 is then n = ?
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