While attempting to tune the note C at 4186 Hz, a piano tuner hears 6.00 beats/s

Question

While attempting to tune the note C at 4186 Hz, a piano tuner hears 6.00 beats/s between a reference oscillator and the string. (a) What are the possible frequencies of the string? (Give the frequencies to the nearest integer.) lower possible frequency Hz higher possible frequency Hz (b) When she tightens the string slightly, she hears 7.00 beats/s. What is the frequency of the string now? (Give the frequency to the nearest integer.) Hz (c) By what percentage should the piano tuner now change the tension in the string to bring it into tune? (Give a positive answer for an increase in the tension, or a negative answer for a decrease in the tension.) %

Explanation / Answer

possible frequencies are 4186+6 and 4186-6

then lower psossible frequency is f1 = 4180 Hz

and higher possible frequency is 4192 Hz

B) by tightening th estring the frequency increases

So new frequency is 4186+7 = 4193 Hz

C) since frequency f is directly proprtional to the square root of the tension

(f1/f2)^2 = (T1/T2)

(4186/4193)^2 = T1/T2

T2/T1 = (4193/4186)^2

(T2/T1)-1 = (4193/4186)^2-1

(T2-T1)/T1 = 0.003347

0.33 % of the tension should be changed

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