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1 vote
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)?

User Zistoloen
by
4.9k points

1 Answer

3 votes

Answer:

The tension of the wire = 3.5×10³ N

Step-by-step 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

User Sayak Mukhopadhyay
by
5.6k points