Question

A steel wire in a piano has a length of 0.683 m and a mass of 4.24 grams. To what tension must the wire be stretched so that the fundamental vibration corresponds to the musical note middle C? The frequency of middle C=262 Hz

Answer 1

The fundamental vibration must have a wavelength of exactly double the wire, or 1.366 m.

And for a wave, v = f?, speed = frequency times wavelength, so the speed of this wave must be
v = (262 Hz)(1.366 m) = 357.892 m/s.

Also, if T is the tension and ? the linear mass density, then v

Answer 2

f = v/(2*L)

v = 2*L*f

= 2*0.683*262

= 357.9 m/s

we know, v = sqrt(T/mue)

T = v^2*mue

= v^2*m/L

= 357.9^2*4.24*10^-3/0.683

= 795 N

Answer 3