A steel wire in a piano has a length of 0.600 m and a mass of 5.200 ✕ 10-3 kg. To what tension must this wire be stretched so that the fundamental vibration corresponds to middle C (fC = 261.6 Hz on the chromatic musical scale) ?

854.39 N

Explanation:

The formula for the fundamental frequency of a stretched string is given as,

f = 1/2L√ (T/m) ... Equation 1

Where f = fundamental frequency, L = Length of the wire, T = Tension, m = mass per unit length.

Given: f = 261.6 Hz, L = 0.6 m, m = (5.2*10⁻³/0.6) = 8.67*10⁻³ kg/m.

Substitute into equation 1

261.6 = 1/0.6√ (T/8.67*10⁻³)

Making T the subject of the equation,

T = (261.6*0.6*2) ² (8.67*10⁻³)

T = 854.39 N

Hence the tension of the wire is 854.39 N.