PROBLEM (a) Find the frequencies of the fundamental, second, and third harmonics

Question

PROBLEM (a) Find the frequencies of the fundamental, second, and third harmonics of a steel wire 1 00 m long with a mass per unit length of 2.00 x 1013 kg/m and under a tension of 80.0 N. (b) Find the wavelengths of the sound waves created by the vibrating wire for all three modes Assume the speed of sound in air is 345 m/s. (c) Suppose the wire is carbon steel with a density of 7.80 X 103 kg/m a cross-sectional area A 2.56 x 10-7 m2, and an elastic limit of 2.80 X 1 0 Pa Find the fundamental frequency if the wire is tightened to the elastic limit. Neglect any stretching of the wire which would slightly reduce the mass per unit length). STRATEGY (a) It's easiest to find the speed of waves on the wire then substitute Equation 14.15 to find the first harmonic. The next two are multiples of the first, given by Equation 14.17. (b) The frequencies of the sound waves are the same as the frequencies of the vibrating wire, but the wavelengths are different. Use vs 3 fa, where vs s the speed of sound in air, to find the wavelength in air. (c) Find the force corresponding to the elastic limit and substitute it into Equation 14.16 SOLUTION (A) Find the first three harmonics at the given tension Calculate the speed of the wave on 80 N 1/2 1/2 2.00 x 104 m/s the wire: 2.00 x 10 kg/m Find the wire's fundamental v 2.00 x 102 m/s 1.000 X 1 Hz frequency (1.00 m) 2L 2fi 2.000 X 1 Hz, fa 3f. 3.00 X 1 Hz Find the next two harmonics by multiplication

Explanation / Answer

PRACTISE IT

L= 1.26 m

mass per unit length= 2.50x10-3

T= 90

so , v= sqrt( F/mu) = 189.73 m/s

f1= v/2L = 75.29 Hz

f2=2*f1 = 150.58 Hz

f3=3*f3= 225.87 Hz

lambda 1 = v / f1= 339/75.29 Hz= 4.50m

lambda 2 = v / f21= 339/150.58 Hz=2.25m

lambda 3 = v / f3= 339/225.87 Hz= 1.5 m

c) F=2.66x10-7 * 2.7 x108 = 71.82N

f= 1/2L * sqrt( F/mu) = 67.25 Hz

EXercise

a) T= 120N

L= 1.26 m

mass per unit length= 2.50x10-3

so , v= sqrt( F/mu) = 219.08m/s

f1= v/2L =86.94 Hz

f2= 173.88Hz

b) speed of sound = 339m

lambda larger= v / f1= 339/86.94 Hz= 3.899 m

lambda smaller = v / f2= 339/173.88=1.949 m

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