(a) We can find the tension in the string from the formula:
"f_n=\\dfrac{n}{2L}\\sqrt{\\dfrac{T}{\\mu}},""f_1=\\dfrac{1}{2L}\\sqrt{\\dfrac{T}{\\mu}},""T=4L^2f_1^2\\mu,""T=4\\cdot(0.6\\ m)^2\\cdot(220\\ Hz)^2\\cdot6.5\\cdot10^{-3}\\ \\dfrac{kg}{m}=453\\ N."(b) Let's first find the frequency of the 2nd, 3rd and 4th harmonics:
"f_2=\\dfrac{2}{2\\cdot0.6\\ m}\\sqrt{\\dfrac{453\\ N}{6.5\\cdot10^{-3}\\ \\dfrac{kg}{m}}}=440\\ Hz,""f_3=\\dfrac{3}{2\\cdot0.6\\ m}\\sqrt{\\dfrac{453\\ N}{6.5\\cdot10^{-3}\\ \\dfrac{kg}{m}}}=660\\ Hz,""f_4=\\dfrac{4}{2\\cdot0.6\\ m}\\sqrt{\\dfrac{453\\ N}{6.5\\cdot10^{-3}\\ \\dfrac{kg}{m}}}=880\\ Hz."
Let's find the velocity of the wave in the string:
"v=\\sqrt{\\dfrac{T}{\\mu}}=\\sqrt{\\dfrac{453\\ N}{6.5\\cdot10^{-3}\\ \\dfrac{kg}{m}}}=264\\ \\dfrac{m}{s}."Finally, we can find the wavelength of the 2nd, 3rd and 4th harmonics:
"\\lambda_2=\\dfrac{v}{f_2}=\\dfrac{264\\ \\dfrac{m}{s}}{440\\ Hz}=0.6\\ m,""\\lambda_3=\\dfrac{v}{f_3}=\\dfrac{264\\ \\dfrac{m}{s}}{660\\ Hz}=0.4\\ m,""\\lambda_4=\\dfrac{v}{f_4}=\\dfrac{264\\ \\dfrac{m}{s}}{880\\ Hz}=0.3\\ m."
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