EXERCISE HINTS: GETTING STARTED I'M STUCK! (a) Find the fundamental frequency and second harmonic if the tension in the wire is increased to 118 N. (Assume the wire doesn't stretch or break.) ffundamental Hz f2nd harmonic (b) Using a sound speed of 342 m/s, find the wavelengths of the sound waves produced. λ (larger) = m Hz λ (smaller) = m

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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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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:
F 1/2
80 N
= (-) ¹/² = ( 2.00 x 10-³ kg/m
V =
f₁
Substitute the values of F, μ, and L.
(B) Find the wavelength of the sound waves produced.
Solve = fλ for the wavelength and
substitute the frequencies.
f₂ = 2f₁ = 2.00 × 10² Hz, f3 = 3f₁:
=
V 2.00 x 10² m/s
2L 2(1.00 m)
(C) ind the fundamental frequency corresponding to the elastic limit.
Calculate the tension in the wire from
the elastic limit:
f₁
f₁
A₁ = v₁/f₁ = (345 m/s)/(1.00 x 10² Hz):
=
^₂ = vs/f₂ = (345 m/s)/(2.00 x 10² Hz) = 1.73 m
A₂ = v₁/f3 = (345 m/s)/(3.00 x 10² Hz) = 1.15 m
=
1
2L
F
== elastic limit → F = (elastic limit) A
A
F = (2.80 x 108 Pa) (2.56 x 10-7m²) = 71.7 N
1
2(1.00 m)
1/2
= 1.00 x 10² Hz
= 2.00 x 10² m/s
71.7 N
2.00 x 10-3 kg/m
·V.
3.00 x 10² Hz
3.45 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: F 1/2 80 N = (-) ¹/² = ( 2.00 x 10-³ kg/m V = f₁ Substitute the values of F, μ, and L. (B) Find the wavelength of the sound waves produced. Solve = fλ for the wavelength and substitute the frequencies. f₂ = 2f₁ = 2.00 × 10² Hz, f3 = 3f₁: = V 2.00 x 10² m/s 2L 2(1.00 m) (C) ind the fundamental frequency corresponding to the elastic limit. Calculate the tension in the wire from the elastic limit: f₁ f₁ A₁ = v₁/f₁ = (345 m/s)/(1.00 x 10² Hz): = ^₂ = vs/f₂ = (345 m/s)/(2.00 x 10² Hz) = 1.73 m A₂ = v₁/f3 = (345 m/s)/(3.00 x 10² Hz) = 1.15 m = 1 2L F == elastic limit → F = (elastic limit) A A F = (2.80 x 108 Pa) (2.56 x 10-7m²) = 71.7 N 1 2(1.00 m) 1/2 = 1.00 x 10² Hz = 2.00 x 10² m/s 71.7 N 2.00 x 10-3 kg/m ·V. 3.00 x 10² Hz 3.45 m 94.7 Hz
13
=
(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 × 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
EXERCISE
HINTS: GETTING STARTED I I'M STUCK!
(a) Find the fundamental frequency and second harmonic if the tension in the wire is increased to 118 N.
(Assume the wire doesn't stretch or break.)
ffundamental
Hz
f2nd harmonic
342 237.69
X
Your response differs from the correct answer by more than 10%. Double check your
calculations. m
=
λ (smaller)
Hz
(b) Using a sound speed of 342 m/s, find the wavelengths of the sound waves produced.
λ (larger) =
m
m
Transcribed Image Text:13 = (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 × 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 EXERCISE HINTS: GETTING STARTED I I'M STUCK! (a) Find the fundamental frequency and second harmonic if the tension in the wire is increased to 118 N. (Assume the wire doesn't stretch or break.) ffundamental Hz f2nd harmonic 342 237.69 X Your response differs from the correct answer by more than 10%. Double check your calculations. m = λ (smaller) Hz (b) Using a sound speed of 342 m/s, find the wavelengths of the sound waves produced. λ (larger) = m m
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