Question 14.8: HARMONICS OF A STRETCHED WIRE GOAL Calculate string harmonic...
HARMONICS OF A STRETCHED WIRE
GOAL Calculate string harmonics, relate them to sound, and combine them with tensile stress.
PROBLEM (a) Find the frequencies of the fundamental, second, and third harmonics of a steel wire 1.00 \mathrm{~m} long with a mass per unit length of 2.00 \times 10^{-3} \mathrm{~kg} / \mathrm{m} and under a ten sion of 80.0 \mathrm{~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 \mathrm{~m} / \mathrm{s}. (c) Suppose the wire is carbon steel with a density of 7.80 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}, a cross-sectional area A=2.56 \times 10^{-7} \mathrm{~m}^{2}, and an elastic limit of 2.80 \times 10^{8} \mathrm{~Pa}. Find the fundamental frequency if the wire is tight ened 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 into Equation 14.15
f_1 = \frac{v}{\lambda_1} = \frac{v}{2L} [14.15]
to find the first harmonic. The next two are multiples of the first, given by Equation 14.17.
f_n = nf_1 = \frac{n}{2L} \sqrt{\frac{F}{\mu}} \qquad n = 1,2,3, . . . [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 v_{s}=f \lambda, where v_{s} is the speed of sound in air, to find the wavelengths in air. (c) Find the force corresponding to the elastic limit and substitute it into Equation 14.16.
f_1 = \frac{1}{2 L} \sqrt{\frac{F}{\mu}} [14.16]
(a) Find the first three harmonics at the given tension.
Use Equation 13.18
v=\sqrt{\frac{F}{\mu}} [13.18]
to calculate the speed of the wave on the wire:
v=\sqrt{\frac{F}{\mu}}=\sqrt{\frac{80.0 \mathrm{N}}{2.00 \times 10^{-3} \mathrm{~kg} / \mathrm{m}}}=2.00 \times 10^{2} \mathrm{~m} / \mathrm{s}
Find the wire’s fundamental frequency from Equation 14.15:
\begin{aligned}&f_{1}=\frac{v}{2 L}=\frac{2.00 \times 10^{2} \mathrm{~m} / \mathrm{s}}{2(1.00 \mathrm{~m})}=1.00 \times 10^{2} \mathrm{~Hz}\end{aligned}
Find the next two harmonics by multiplication:
\begin{aligned}&f_{2}=2 f_{1}=2.00 \times 10^{2} \mathrm{~Hz}, f_{3}=3 f_{1}=3.00 \times 10^{2} \mathrm{~Hz}\end{aligned}
(b) Find the wavelength of the sound waves produced.
Solve v_{s}=f \lambda for the wavelength and substitute the frequencies:
\begin{aligned}&\lambda_{1}=v_{s} / f_{1}=(345 \mathrm{~m} / \mathrm{s}) /\left(1.00 \times 10^{2} \mathrm{~Hz}\right)=3.45 \mathrm{~m} \\&\lambda_{2}=v_{s} / f_{2}=(345 \mathrm{~m} / \mathrm{s}) /\left(2.00 \times 10^{2} \mathrm{~Hz}\right)=1.73 \mathrm{~m} \\&\lambda_{3}=v_{s} / f_{3}=(345 \mathrm{~m} / \mathrm{s}) /\left(3.00 \times 10^{2} \mathrm{~Hz}\right)=1.15 \mathrm{~m}\end{aligned}
(c) Find the fundamental frequency corresponding to the elastic limit.
Calculate the tension in the wire from the elastic limit:
\begin{aligned}&\frac{F}{A}=\text { elastic limit } \quad \rightarrow \quad F=(\text { elastic limit }) A \\&F=\left(2.80 \times 10^{8} \mathrm{~Pa}\right)\left(2.56 \times 10^{-7} \mathrm{~m}^{2}\right)=71.7 \mathrm{~N}\end{aligned}
Substitute the values of F, \mu, and L into Equation 14.16:
\begin{aligned}&f_{1}=\frac{1}{2 L} \sqrt{\frac{F}{\mu}} \\&f_{1}=\frac{1}{2(1.00 \mathrm{~m})} \sqrt{\frac{71.7 \mathrm{~N}}{2.00 \times 10^{-3} \mathrm{~kg} / \mathrm{m}}}=94.7 \mathrm{~Hz}\end{aligned}
REMARKS From the answer to part (c), it appears we need to choose a thicker wire or use a better grade of steel with a higher elastic limit. The frequency corresponding to the elastic limit is smaller than the fundamental!