Answer:
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.
We have that for the Question "To what tension must this wire be stretched so that the fundamental vibration corresponds to middle C "
Answer:
From the question we are told
a piano has a length of 0.6m and a mass of [tex]5.2 * 10^{−3} kg[/tex]
Generally the equation for frequency is mathematically given as
[tex]F = \frac{V}{2L}[/tex]
where,
[tex]V = \sqrt\frac{T}{U}[/tex]
Therefore,
[tex]261.6 = \frac{V}{2*0.76}\\\\V = 261.6*2*0.6\\\\V = 313.92m/s[/tex]
so,
[tex]v^2 = \frac{T}{U}\\\\T = V^2 * U\\\\T = 512.43N[/tex]
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