Answer :
Final answer:
The tension required for the A string on a violin to vibrate at a fundamental frequency of 440 Hz is approximately 86.55 N.
Explanation:
The tension required for the A string of a violin to vibrate at its fundamental frequency can be determined using the formula for the wave speed (v) on a string, which is based on the tension (T) and the linear density (μ) of the string:
v = sqrt(T/μ)
From this, we can rearrange to find the tension:
T = μv^2
Firstly, we need to calculate the linear density of the string with the given mass and length:
μ = mass/length = 0.35 g / 32 cm = 1.09375 g/cm
Then convert to SI units: μ = 0.00109375 kg/m
The wave speed is twice the product of the fundamental frequency and the string length (because the wavelength of the fundamental frequency wave is twice the string length):
v = 2 * length * frequency = 2*0.32m*440Hz = 281.6 m/s
Finally, we can substitute v and μ into the tension formula, and get:
T = μv^2 = 0.00109375 kg/m * (281.6 m/s)^2 = 86.55 N
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The tension the string must be placed is 86.73 N.
Tension in the string
The tension in the string is determined by applying the following formula for fundamental frequency of wave in string.
[tex]f_0 = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]
where;
- L is the length of the string
- T is the tension in the string
- μ is mass per unit length = (0.35 x 10⁻³) / (0.32) = 1.094 x 10⁻³ kg/m
[tex]f_0 = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\2lf_0 = \sqrt{\frac{T}{\mu} } \\\\(2lf_0)^2 = \frac{T}{\mu}\\\\T = \mu(2lf_0)^2\\\\T = (1.094 \times 10^{-3})(2 \times 0.32 \times 440)^2\\\\T = 86.73 \ N[/tex]
Thus, the tension the string must be placed is 86.73 N.
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