Question

Sol. (A) As obtained in earlier problem (Q. 14), Total energy of dx element:

= (2μA²ω² sin² Kx sin²ωt + 2μA²ω² cos² Kx cos²ωt)dx

So, Total energy in fundamental frequency (1 loop), at t = 0

E₁ = $\int_{0}^{\lambda/2}{2\mu A^{2}\omega^{2}(\sin^{2}Kx\sin^{2}\omega t + \cos^{2}Kx\cos^{2}\omega t)dx}$

= $2\mu A^{2}\omega^{2}\int_{0}^{\lambda/2}{(0 + \cos^{2}Kx)}dx$ (For t = 0)

= $\mu A^{2}\omega^{2}\int_{0}^{\lambda/2}{(\cos 2Kx + 1)dx}$

= $\mu A^{2}\omega^{2}\left\lbrack \frac{\sin 2Kx}{2K} + x \right\rbrack_{0}^{\lambda/2}$

= $\mu A^{2}(2\pi f_{0})^{2}\left( 0 + \frac{\lambda}{2} \right)$

= μA² 4π²f₀² L $\left( L = \frac{\lambda}{2} \right)$

Now, Total energy in n^th harmonic (n loops, L = $\frac{n\lambda}{2}$), at t = 0

E_n = $n\int_{0}^{\lambda/2}{2\mu A^{2}\omega^{2}}(\sin^{2}Kx\sin^{2}\omega t + \cos^{2}Kx\cos^{2}\omega t)dx$

= $n\mu A^{2}\omega^{2}\int_{0}^{\lambda/2}{(0 + \cos^{2}{{Kx})}}dx$ (t = 0)

= $n\mu A^{2}(2\pi f)^{2}\frac{\lambda}{2}$

= nμA² 4π² (nf₀)²$\left( \frac{L}{n} \right)$ $\left( L = \frac{n\lambda}{2} \right)$

⇒ E_n = n²μA² 4π² f₀² L = E_Total

= n²E₁

(C) As obtained earlier

KE of dx element = 2μA²ω²sin²Kx sin²ωt dx

So KE_Total = $n\int_{0}^{\lambda/2}{2\mu A^{2}\omega^{2}\sin^{2}Kx\sin^{2}\omega tdx}$

= $n\mu A^{2}\omega^{2}\sin^{2}\omega t\int_{0}^{\lambda/2}{(1 - \cos 2Kx)}dx$

= $n\mu A^{2}\omega^{2}\sin^{2}\omega t\left( \frac{\lambda}{2} \right)$

= $n\mu A^{2}4\pi^{2}n^{2}f_{0}^{2}\left( \frac{L}{n} \right)\sin^{2}\omega t$

= n²μA² 4π² f₀²L sin²ωt

= E_Total sin²ωt

So Average KE = E_Total< sin²ωt >

= $\frac{1}{2}E_{Total}$