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asambeis [7]
3 years ago
13

A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats

per second. The tuner increases the tension on the as-yet untuned string, and now when they are played together the beat frequency is
1.0s−1.
(a) What was the original frequency of the untuned string?
(b) By what percentage did the tuner increase the tension on that string?
Physics
1 answer:
ololo11 [35]3 years ago
4 0

Answer:

Part a)

f_B = 290 Hz

Part B)

percentage increase is

percentage = 1.38%

Explanation:

Part a)

As we know that the beat frequency is

f_A - f_B = 3

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase

So we have

293 - f_B = 3

f_B = 290 Hz

Part B)

percentage increase in the tension of the string will be given as

f_A - f_B' = 1

f_B' = 292 Hz

now we have

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

so we have

T_1 = C (290)^2

T_2 = C(292)^2

so we have

\frac{\Delta T}{T} = \frac{292^2 - 290^2}{290^2}

percentage increase is

percentage = 1.38

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