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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 ? |
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Answer» Solution :As the tension in one of the strings is changed , its fundamental frequency changes. THEREFORE, when both strings are played , they will have different frequencies and beats will be heard. We must combine our understanding of the waves under boundary conditions model for strings with our new knowledge of beats . Set up a ratio of the fundamental frequencies of the two strings using equation `(f_(2))/( f_(1)) = (( v_(2)// 2L))/(( v_(1) //2 L)) = ( v_(2))/( v_(1))` Use equation ` v = sqrt ( T//mu)` to substitute for the wave SPEEDS on the strings . `( f_(2))/( f_(1)) = ( sqrt (T_(2)//mu))/( sqrt (T_(1) //mu)) = sqrt((T_(2))/(T_(1)))` Incorporate that the tension in one string is `1.0 %` larger than the other , that is , `T_(2) = 1.010 T_(1)`. `(f_(2))/( f_(1)) = sqrt((1.010 T_(1))/( T_(1))) = 1.005` Solve for the frequency of the tightened string : `f_(2) = 1.005 f_(1) = 1.005 ( 440 Hz) = 442 Hz` Find the beat frequency using Eq. (iii) `f_(beat) = 442 Hz - 440 Hz = 2 Hz` Notice that a `1.0%` mistuning in tension leads to an easily audible beat frequency of `2 Hz`. A PIANO 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. |
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