PRACTICE IT Use the worked example above to help you solve this problem. (a) Find the frequencies of the fundamental, second, and third harmonics of a steel wire 1.18 m long with a mass per unit length of 2.40 x 10-3 kg/m and under a tension of 83.9 N. f₁ = Hz √₂ = Hz f3= Hz (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 342 m/s. A₁ = m d₂ = m A3 = m (c) Suppose the wire is carbon steel with a density of 7.89 x 10³ kg/m³, a cross-sectional area A = 2.68 x 10-7 m², and an elastic limit of 2.50 x 108 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). Hz

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Chapter1: Units, Trigonometry. And Vectors
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SOLUTION
(A) Find the first three harmonics at the given tension.
Calculate the speed of the wave on the
wire:
Find the wire's fundamental frequency:
Find the next two harmonics by
multiplication:
=
V =
f₁
Substitute the values of F, μ, and L.
F
1/2
80 N
(=) ¹/² = (2.00 x 10-³ kg/m
=
(B) Find the wavelength of the sound waves produced.
Solve vs
fλ for the wavelength and
substitute the frequencies.
F
==
A
V
f₁
2L
(C) Find the fundamental frequency corresponding to the elastic limit.
Calculate the tension in the wire from
the elastic limit:
f₁
f₂ = 2f₁ = 2.00 × 10² Hz, ƒ3 = 3ƒ₁ = 3.00 × 10² Hz
A₁ = v₁/f₁ = (345 m/s)/(1.00 x 10² Hz) = 3.45 m
1₂ = vs/f₂ = (345 m/s)/(2.00 x 10² Hz) = 1.73 m
^₂ = vs/f3 = (345 m/s)/(3.00 x 10² Hz) = 1.15 m
=
2.00 × 10² m/s
2(1.00 m)
elastic limit →→ F = (elastic limit) A
F = (2.80 x 108 Pa) (2.56 × 10-7m²) = 71.7 N
1
V
2L
F
1/2
-= 1.00 x 10² Hz
μ
= 2.00 x 10² m/s
1
2(1.00 m)
71.7 N
2.00 x 10-³ kg/m
= 94.7 Hz
Transcribed Image Text:SOLUTION (A) Find the first three harmonics at the given tension. Calculate the speed of the wave on the wire: Find the wire's fundamental frequency: Find the next two harmonics by multiplication: = V = f₁ Substitute the values of F, μ, and L. F 1/2 80 N (=) ¹/² = (2.00 x 10-³ kg/m = (B) Find the wavelength of the sound waves produced. Solve vs fλ for the wavelength and substitute the frequencies. F == A V f₁ 2L (C) Find the fundamental frequency corresponding to the elastic limit. Calculate the tension in the wire from the elastic limit: f₁ f₂ = 2f₁ = 2.00 × 10² Hz, ƒ3 = 3ƒ₁ = 3.00 × 10² Hz A₁ = v₁/f₁ = (345 m/s)/(1.00 x 10² Hz) = 3.45 m 1₂ = vs/f₂ = (345 m/s)/(2.00 x 10² Hz) = 1.73 m ^₂ = vs/f3 = (345 m/s)/(3.00 x 10² Hz) = 1.15 m = 2.00 × 10² m/s 2(1.00 m) elastic limit →→ F = (elastic limit) A F = (2.80 x 108 Pa) (2.56 × 10-7m²) = 71.7 N 1 V 2L F 1/2 -= 1.00 x 10² Hz μ = 2.00 x 10² m/s 1 2(1.00 m) 71.7 N 2.00 x 10-³ kg/m = 94.7 Hz
PRACTICE IT
Use the worked example above to help you solve this problem.
(a) Find the frequencies of the fundamental, second, and third harmonics of a steel wire 1.18 m
long with a mass per unit length of 2.40 x 10-³ kg/m and under a tension of 83.9 N.
f₁ =
Hz
f₂ =
Hz
f3=
Hz
(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 342 m/s.
2₁:
^₂ =
13 =
m
m
m
(c) Suppose the wire is carbon steel with a density of 7.89 × 10³ kg/m³, a cross-sectional area A
= 2.68 x 10-7 m², and an elastic limit of 2.50 × 108 Pa. Find the fundamental frequency if the wire
is tightened to the elastic limit. Neglect any stretching the wire (which would slightly reduce the
mass per unit length).
Hz
Transcribed Image Text:PRACTICE IT Use the worked example above to help you solve this problem. (a) Find the frequencies of the fundamental, second, and third harmonics of a steel wire 1.18 m long with a mass per unit length of 2.40 x 10-³ kg/m and under a tension of 83.9 N. f₁ = Hz f₂ = Hz f3= Hz (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 342 m/s. 2₁: ^₂ = 13 = m m m (c) Suppose the wire is carbon steel with a density of 7.89 × 10³ kg/m³, a cross-sectional area A = 2.68 x 10-7 m², and an elastic limit of 2.50 × 108 Pa. Find the fundamental frequency if the wire is tightened to the elastic limit. Neglect any stretching the wire (which would slightly reduce the mass per unit length). Hz
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