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A tube 1m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.3 m long and has a mass of 0.01 kg. It is held fixed an both ends and vibrates in its fundamental mode. It sets the air column in the tube into vibration all its fundamental frequency by resonance Find the tension in the wire. Speed of sound in air: 330 m/s |
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Answer» Solution :FUNDAMENTAL frequency of closed pipe `= v/(4l)` `=330/(4xx1) =82.5 Hz` B) At resonance, given : fundamental frequency of stretched wire (fixed at both ends) =fundamental frequency of air column `:. v/(2L) = 82.5 Hz` `:. Sqrt(T//MU)/(2l) =82.5 (or)` `T = mu (2 xx0.3 xx 82.5)^2` `=81.675 N` |
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