14) While attempting to tune a string to 529 Hz, a piano tuner hears 2beats/s be

Question

14) While attempting to tune a string to 529 Hz, a piano tuner hears 2beats/s between a reference oscillator and the string. (a) What are the possible frequencies of thestring?
Hz (lowest possible frequency)
Hz (highest possiblefrequency)
(b) When she tightens the string slightly, she hears 3 beats/s. What is the frequency of the stringnow?
Hz
(c) By what percentage should the piano tuner now change thetension in the string to bring it into tune? (Give a positiveanswer for an increase in the tension, or a negative answer for adecrease in the tension.)
% (a) What are the possible frequencies of thestring?
Hz (lowest possible frequency)
Hz (highest possiblefrequency)
(b) When she tightens the string slightly, she hears 3 beats/s. What is the frequency of the stringnow?
Hz
(c) By what percentage should the piano tuner now change thetension in the string to bring it into tune? (Give a positiveanswer for an increase in the tension, or a negative answer for adecrease in the tension.)
% Hz (lowest possible frequency)
Hz (highest possiblefrequency)
(b) When she tightens the string slightly, she hears 3 beats/s. What is the frequency of the stringnow?
Hz
(c) By what percentage should the piano tuner now change thetension in the string to bring it into tune? (Give a positiveanswer for an increase in the tension, or a negative answer for adecrease in the tension.)
%

Explanation / Answer

beat freq is the difference between the frequencies of the twosounds heard, so the two possible other frequencies are .       527     and    531 . (b)   the freq must have gone higher, and since thebeat freq increased by one the new freq mustbe    532 . (c) tension is prop to freq squared so we get .         new tension / oldtension = (529/532)2 =  0.988754 . So the tension must be reduced by    0.988754 -1 =   -0.01125   =     -1.125%

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