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A 1.0-g string that is 0.64 m long is fixed at both ends and is under tension. This string produces a 100-Hz tone when it vibrates in the third harmonic. The speed of sound in air is 344 m/s. The tension in the string is closest to

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onyebuchinnaji

The tension in the given string of 1.0 g and length of 0.64 m is closest to 2.84 N

Tension in the string

The tension in the string at nth harmonic is calculated as follows;

[tex]f_n = \frac{n}{2l} \sqrt{\frac{T}{\mu} }[/tex]

where;

  • n th harmonic = 3
  • L is length of the string = 0.64 m
  • μ is mass per unit length = (1 x 10⁻³ kg)/(0.64 m) = 0.00156 kg/m
  • f is frequency = 100 Hz
  • T is tension in the string = ?

[tex]f_n = \frac{n}{2l} \sqrt{\frac{T}{\mu} }\\\\\frac{f_n(2l)}{n} = \sqrt{\frac{T}{\mu} }\\\\\frac{f_n^2(4l^2)}{n^2} = \frac{T}{\mu} \\\\T = \mu (\frac{f_n^2(4l^2)}{n^2})\\\\T = 0.00156(\frac{100^2 \times 4\times 0.64^2}{3^2} )\\\\T = 2.84 \ N[/tex]

Learn more about tension in string here: https://brainly.com/question/24994188

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