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

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fl=50Hz

fh=500Hz

fl=75Hz

fh=375Hz

answer is C.

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Detailed Solution

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 Hz

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