In certain ranges of a piano keyboard , more than one string is tuned to the same note to provide extra loudness . For example , the note at 110 Hz has two strings at this frequency . If one string slips from its normal tension of 600 N "to" 540 N , what beat frequency is heard when the hammer strikes the two strings simultaneously ?

In certain ranges of a piano keyboard , more than one string is tuned

1.	In certain ranges of a piano keyboard , more than one string is tuned to the same note to provide extra loudness . For example , the note at 110 Hz has two strings at this frequency . If one string slips from its normal tension of 600 N "to" 540 N , what beat frequency is heard when the hammer strikes the two strings simultaneously ?
Answer» Solution :Directly noticeable beat frequencies are usually only a few hertz , so we should not except a frequency much greater than this . Combining the VELOCITY and the tension equations `v = f lambda and v = sqrt (T//mu)` , we find that the frequency is ` f= sqrt((T)/( mu lambda^(2)))` Since ` mu and lambda ` are constant , we can apply that equation to both frequencies , and then DIVIDE the two equaitons to GET the proportion . ` (f_(1))/(f_(2)) = sqrt ((T_(1))/(T_(2)))` With ` f_(1) = 110 HZ , T_(1) = 600 N , and T_(2) = 540 N , we have f_(2) = ( 110 Hz) sqrt(( 540 N))/( 600 N) = 104.4 Hz` The beat frequency is `f_(b) = { f_(1) - f_(2)\| = 110 Hz - 104.4 Hz = 5.6 Hz` As excepted , the beat frequency is only a few cycles per second .