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Sagot :
Answer:
The first four resonance frequency in the string are;
1) 50·√50 Hz
2) 100·√50 Hz
3)150·√50 Hz
4) 200·√50 Hz
Explanation:
The given parameters of the string are;
The density of the string, ρ = 0.01 kg/m
The tension force on the string, T = 5 N
The length of the string, l = 0.1 m
Therefore the mass of the string, m = Length of string × Density of the string
∴ m = 0.01 kg/m × 0.1 m = 0.001 kg
The formula for the fundamental frequency, f₁, is given as follows;
[tex]f_1 = \dfrac{\sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{\sqrt{\dfrac{T}{\rho} } }{2 \cdot L}[/tex]
Where;
f₁ = The fundamental frequency in the string
T = The tension in the string = 5 N
m = The mass of the string = 0.001 kg
L = The length of the string = 0.1 m
ρ = The density of the string = 0.01 kg/m
By plugging in the values of the variables, we have;
[tex]f_1 = \dfrac{\sqrt{\dfrac{5}{0.01} } }{2 \times 0.1} = 50 \cdot \sqrt{5}[/tex]
The first four harmonics are;
f₁, 2·f₁, 3·f₁, 4·f₁
Therefore, we have the first four resonance frequency of the string are as follows;
1 × 50·√50 Hz = 50·√50 Hz
2 × 50·√50 Hz = 100·√50 Hz
3 × 50·√50 Hz = 150·√50 Hz
4 × 50·√50 Hz = 200·√50 Hz
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