Question

The A string on a violin has a fundamental frequency of 440 Hz. The length of the vibrating portion is 32 cm, and it has mass 0.35 g. Under what tension must the string be placed?

Answer 1

Answer:

86.75 N

Explanation:

f = 440 Hz

l = 32 cm = 0.32 m

m = 0.35 g = 0.35 x 10^-3 kg

μ = m / l = 0.35 x 10^-3 / 0.32 = 1.094 x 10^-3 kg/m

The formula for the fundamental frequency is given by

$f =\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

Where, T be the tension

$440 =\frac{1}{2\times 0.32}\sqrt{\frac{T}{1.094\times 10^{-3} }}$

$281.6 = \sqrt{\frac{T}{1.094\times 10^{-3} }}$

T = 86.75 N

Thus, the tension in the string is 86.75 N.

Answer 2

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.

$f_0 = \frac{1}{2l} \sqrt{\frac{T}{\mu} }$

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

$f_0 = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\2lf_0 = \sqrt{\frac{T}{\mu} } \\\\(2lf_0)^2 = \frac{T}{\mu}\\\\T = \mu(2lf_0)^2\\\\T = (1.094 \times 10^{-3})(2 \times 0.32 \times 440)^2\\\\T = 86.73 \ N$

Thus, the tension the string must be placed is 86.73 N.

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