A steel wire in a piano has a length of 0.6000 m and a mass of

3.600

10-3 kg. To what tension must this wire be stretched in

order that thefundamental vibration correspond to middle C

(fC = 261.6 Hz on the chromatic musical

scale) ?

The tension of the wire = 3.5*10³ N

Explanation:

The frequency of a string is given as

f = √ (T/m) / 2πL ... Equation 1

Where f = fundamental frequency, L = length of the steel wire, T = Tension of the steel wire, m = mass of the steel wire

Making T the subject of the equation above.

T = f²4π²L²m ... Equation 2

Given: f = 261.6 Hz, L = 0.6 m, m = 3.6*10⁻³ kg and π = 3.143.

Substituting these values into equation 2,

T = 261.6²*4 * (3.143) ² * (0.6) ²*3.6*10⁻³

T = 68434.56*4*9.878*0.36*3.6*10⁻³

T = 3504366.3*10⁻³

T = 3504.37 N

T ≈ 3.5*10³ N

Thus the tension of the wire = 3.5*10³ N