Question

12.2 A steel wire has density ? = 7800 kg/m3 , radius 0.80 mm, and length 90.0 cm. A constant tension of 2.50 x 103 N is applied to the wire, and both ends are fixed.

(a) What is the fundamental frequency of the wire?

(b) Where should you pluck the wire to achieve the 5th harmonic, and what is its frequency?

(c) What length of pipe closed at one end would have the same fundamental frequency as the wire in at (a)? Assume the speed of sound is 344 m/s.

Answer #1

µ = ?A = ??r²

µ = 7800kg/m² * ?(0.0008m)² = 0.01568 kg/m

(a) velocity v = ?(T / µ) = ?(2500N / 0.01568kg/m) = 399.297
m/s

fundamental ? = 2L = .9 m, and so

fundamental f = v / ? = 443.66 Hz

(b) The fifth has 5 antinodes, so you should pluck 1/10 of the
length away from the end at 7.50 cm.

The frequency will be 5*443Hz = 2215 Hz

(c) A pipe closed at one end has a fundamental wavelength that is 4
times its length (not twice, like a string or open pipe).

? = 4L = v / f

so L = v / 4f = 344m/s / 4*443.66Hz = 0.1938 m = 19.38 cm

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4.200 10-3 kg. To what tension must this wire be stretched so that
the fundamental vibration corresponds to middle C (fC = 261.6 Hz on
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