While attempting to tune the note C at 2093 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?Lower possible frequency=2087 Hz/ Higher possible frequency=2099 Hz

(b) When she tightens the string slightly, she hears 7.00 beats/s. What is the frequency of the string now? 2100 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.)

Expert Solution

a )

we have equation is using here is

number of beats = frequency of oscillator frequency of piano wire

6 = 2093 f_p

f_p = 2093 6

= 2099 Hz or 2087 Hz

Lower possible frequency=2087 Hz

and

Higher possible frequency=2099 Hz

b )

the frequency of the string now is

f_p ' = 7 + f_o

= 7 + 2093

= 2100 Hz

the frequency of the string now is = 2100 Hz

c )

f = k T

2093 = k T_o

T_o = 2093² / k²

T₁ = 2100² / k²

T % = ( ( T₀ - T₁ ) / T₁ ) x 100

= ( ( 2093² - 2100 ) / 2100 ) x 100

T % = - 0.6655 ( decrease in the tension )

genius_generous answered 2 years ago

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