While attempting to tune the note C at 277 Hz, a piano tuner hears 6.00 beats/s
ID: 1525101 • Letter: W
Question
While attempting to tune the note C at 277 Hz, a piano tuner hears 6.00 beats/s between a reference oscillator and the string. What are the possible frequencies of the string? (Give the frequencies to the nearest integer.) lower possible frequency higher possible frequency 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 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
given data:
frequency = 277 Hz
beats/sec=6
a) possible frequencies of the string
The beat frequency is 1/2 the difference of the two frequencies.
(1/2)(f1+/-f2) = 6
case 1 : f1+f2 = 12
277+f2 = 12
f2 = 265 Hz ----> lower possible frequency
case 2:f1-f2 = 12
277-f2 = 12
f2 =289 Hz----> higher possible frequency.
b) (1/2)(f1+/-f2) = 7
case 1 : f1+f2 = 14
277+f2 = 14
f2 = 263 Hz ----> lower possible frequency
case 2:f1-f2 = 12
277-f2 = 12
f2 =291 Hz----> higher possible frequency.
c) here is the equation for funfdamental frequency
f = (T/(m/L))^0.5/2L
f- frequency
T- tension
m- string mass
L- string length
from the equation
2Lf = (T/(m/L))^0.5
4mLf^2= T
the 4mL units are all constant so if we choose the mL product to equal 1/4 then the
equation becomes
(f)^2 = T
so at 211 hz the tension is 263^2 = 69169
and at 206 hz the tension is 291^2 = 75951
the percentage change is the difference divided by the original
(75951-69169) / 75951 = 8.9 % decrease in tension
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