Question

While attempting to tune the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string. (a) What are the possible frequencies of the string? (b) When she tightens the string slightly, she hears 3.00 beats/s. What is the frequency of the string now? (c) By what percentage should the piano tuner now change the tension in the string to bring it into tune?

Answer 1

Answer

given,

frequency = 523 Hz

piano tuner hears = 2.00 beats/s

a) possible frequency = 523 ± 2

            525 Hz and 521 Hz

b) On tightening piano wire, tension will be increased which will cause an increase of frequency of piano wire. Since no. of beats heard is also increased, the frequency of piano wire must be greater than the frequency of oscillator.

     f p'   =   5+ f₀

     f p'   =  3 + 523 Hz   =   526 Hz.

c)  f   =   k x √T

  Let the tension at 523 Hz be T₀ and at 526 Hz it be T₁

  523  =   k x √T₀  

  T₀    =   523²/ k²

  T₁  =     526²/ k²

  % change in tension

        = $\dfrac{T_0- T_1}{T_1}$

        = $\dfrac{523^2- 526^2}{526^2}$

        = 1.13 %