A steel wire in a piano has a length of 0.6000 m and a mass of 5.100*10^-3 kg. T
ID: 2014053 • Letter: A
Question
A steel wire in a piano has a length of 0.6000 m and a mass of 5.100*10^-3 kg. To what tension must this wire be stretched in order that the fundamental vibration correspond to middle C (fC = 261.6 Hz on the chromatic musical scale)?
F=?, L=0.6000 m, m=5.100*10-3 kg, n=0.50031N where g=9.81m/s2, fc=261.6 Hz
I honestly can not figure out where to begin. I believe this is a standing waves problem so I'm almost positive that it should use the following equation, but I can not figure out a proper derivative or manipulation of the equation to find a seemingly correct answer.
fn= (n/2L)*(F/) where n=the number of harmonic (1,2,3...)
Explanation / Answer
Length of the wire L = 0.6 m Mass of the wire m = 5.1*10^-3 Kg Linear density of wire = m/L = (5.1*10^-3 Kg)/(0.6 m) = 8.5*10^-3 Kg/m Fundamental frequency f = 261.6 Hz Let F be the tension required along the string. Fundamental frequency f = (1/2L) (F/) 261.6 Hz = (1/2* 0.6 m) (F/ 8.5*10^-3 Kg/m) (F/ 8.5*10^-3 Kg/m) = 313.92 F = 837.64 NRelated Questions
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