In an attempt to get your name in Guinness World Records, you

build a bass viol with strings of length 5.00 m between fixed points.

One string, with linear mass density 40.0 g>m, is tuned to a 20.0 Hz

fundamental frequency (the lowest frequency that the human ear can

hear). Calculate (a) the tension of this string, (b) the frequency and

wavelength on the string of the second harmonic, and (c) the frequency

and wavelength on the string of the second overtone.

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Step 1 Step 1:Given that,Leength of strings (\mathrm{L})=5 \mathrm{~m}Linear mass density =40(\mathrm{~g} / \mathrm{m})Fundamental frequency (\mathrm{f})=20(\mathrm{~Hz})isa: Here we need to find the tension of the string :isAs we know that,The velocity:v=f \lambda.eqn 1And,\lambda=\frac{n L}{2}Now,\begin{aligned}\lambda & =\frac{5}{2} \\\lambda & =2.5(\mathrm{~m})\end{aligned}Now,Putting the values in eqn 1 :We get,\begin{array}{l}v=20 \times 2.5 \\v=50(\mathrm{~m} / \mathrm{sec})\end{array}Step 2Step 2 :As we know that,v=\left(\frac{T}{P}\right)^{0.5}Putting the values in aboveeqn:We get,2500=\frac{T}{40 \times 10^{-3}}Hence,T=1600 N(i)b: Here we need to find the wavelength and second frequency of the spring:Hence,\lambda=5(m)And,f_{2}=40 \mathrm{~Hz}Step 3c: Here we need to find wavelength and third frequency:As we know that,L=\frac{3 \lambda}{2}Hence,\lambda=\frac{2 \times 5}{3}=3.33Hence,\text { frequency }\left(f_{3}\right)=60 \mathrm{~Hz} ...