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Two identical piano strings of length 0.750 m are each tunned exactly to 440 Hz. The tension in one of the strings is then increased by 1.0 %. If they are now struck , what is the beat frequency between the fundamental of the two strings ?

Answer» <html><body><p></p>Solution :As the tension in one of the strings is changed , its fundamental frequency changes. <a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>, when both strings are played , they will have different frequencies and beats will be heard. <br/> We must combine our understanding of the waves under boundary conditions model for strings with our new knowledge of beats . <br/> Set up a ratio of the fundamental frequencies of the two strings using equation <br/> `(f_(2))/( f_(1)) = (( v_(2)// <a href="https://interviewquestions.tuteehub.com/tag/2l-300409" style="font-weight:bold;" target="_blank" title="Click to know more about 2L">2L</a>))/(( v_(1) //2 L)) = ( v_(2))/( v_(1))` <br/> Use equation ` v = sqrt ( T//mu)` to substitute for the wave <a href="https://interviewquestions.tuteehub.com/tag/speeds-650188" style="font-weight:bold;" target="_blank" title="Click to know more about SPEEDS">SPEEDS</a> on the strings . <br/> `( f_(2))/( f_(1)) = ( sqrt (T_(2)//mu))/( sqrt (T_(1) //mu)) = sqrt((T_(2))/(T_(1)))` <br/> Incorporate that the tension in one string is `1.0 %` larger than the other , that is , `T_(2) = 1.010 T_(1)`. <br/> `(f_(2))/( f_(1)) = sqrt((1.010 T_(1))/( T_(1))) = 1.005` <br/> Solve for the frequency of the tightened string : <br/> `f_(2) = 1.005 f_(1) = 1.005 ( <a href="https://interviewquestions.tuteehub.com/tag/440-316801" style="font-weight:bold;" target="_blank" title="Click to know more about 440">440</a> Hz) = 442 Hz` <br/> Find the beat frequency using Eq. (iii) <br/> `f_(beat) = 442 Hz - 440 Hz = 2 Hz` <br/> Notice that a `1.0%` mistuning in tension leads to an easily audible beat frequency of `2 Hz`. A <a href="https://interviewquestions.tuteehub.com/tag/piano-1154352" style="font-weight:bold;" target="_blank" title="Click to know more about PIANO">PIANO</a> tuner can use beats to tune a stringed instrument by beating a note against a reference tone of known frequency . The tuner can then adjust equals the frequency . The tuner can adjust then adjust the string tension until the frequency of the sound it emits equals the frequency of the reference tone . The tuner does so by tightening or loosening the string until the beats produced by it and the reference source become too infrequent to notice.</body></html>


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