A wire is made out of a special steel alloy with Y = 2.00 ×10 Pa , density = 780

Question

A wire is made out of a special steel alloy with Y = 2.00 ×10 Pa , density = 7800 kg/m ,

radius 1.00 mm, and coefficient of linear thermal expansion = 2.50 ×105 K1 . The initial temperature of the wire is 20°C. As a constant tension of 3.53 x 104 N is applied to the wire, the wire is measured to be exactly 60.0 cm long.

(a) What is the fundamental frequency of the wire? Assume the speed of sound is 344 m/s.

(b) Where should you pluck the wire to achieve the third harmonic (second overtone) without also exciting the fundamental mode, and what is its frequency?
(c) If the wire is heated to the melting point of steel (1363oC), by how much does the fundamental frequency change?

Explanation / Answer

(a)

A = r² = *(1*10^-3)^2=3.1416*10^-6 m²

linear mass density µ = *A = 7800kg/m³ * 3.1416*10^-6 6m² = 0.0245044227 kg/m
wave velocity v = (T / µ) = (3.53 x 10^4 N / 0.0245044227 /m) = 1200.23 m/s

fundamental wavelength = 2L = 2*0.6=1.2 m, so
fundamental frequency f = v / = 1200.23 m/s / 1.2m = 1000.19 Hz

assuming speed of sound as 344m/s   

a fundamental wavelength = 2L = 2*0.6=1.2 m, so
fundamental frequency f = v / = 344 m/s / 1.2m = 286.67Hz

b) To excite the third, you have to pluck between two nodes, but not at the center of the wire. Therefore you should pluck 1/6 of the way from the bottom or the top to excite the third harmonic, which is
f = 344m/s / (2*0.6m / 3) = 860 Hz

c) change in temp=1363-20=1343

the elongation is hence= *L* change in temp= 1343*2.50 ×105*0.6=0.020145

new length= 0.6+0.020145=0.620145

a fundamental wavelength = 2L = 2*0.620145=1.24029 m, so

fundamental frequency f = v / = 344 m/s / 1.24029 m = 277.35Hz

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