Asteel 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

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation