A stretched string, with ends fixer? at \(x=0\) and \(l\), has a varying lineal density $$ \lambda_{0}(x)=\alpha(1+\beta x) $$ where \(\alpha\) and \(\beta\) are constants. Find the frequency of the \(n\)th normal mode in the WKB approximation. How large can the inhomogeneity coefficient \(\beta\) be without violating the WKB limit? Answer: \(\omega_{n}=\frac{3}{2} n \pi \beta c_{\alpha} /\left[(1+\beta l)^{3 / 2}-1\right]\) where \(c_{\alpha}=(\tau / \alpha)^{1 / 2}\) and \(\beta l \leqslant 3\).

First, let's recall the relation between the wave speed \(c(x)\) and the lineal density of the string, given by $$ c(x)=\sqrt{\frac{T(x)}{\lambda_{0}(x)}} $$ where \(T(x)\) is the tension in the string. Since the tension is constant along the string, we can write $$ c(x)=\sqrt{\frac{T}{\alpha(1+\beta x)}}=c_{\alpha}\sqrt{\frac{1}{1+\beta x}} $$ where \(\tau=T/\alpha\) and \(c_{\alpha}=\sqrt{\tau}\).

Now, we need to apply the WKB approximation to find the frequency of the nth normal mode. According to the WKB approximation, the mode number \(n\) is related to the wavenumber \(k(x)\) by $$ n=\frac{1}{\pi}\int_{0}^{l}k(x)dx. $$ The wavenumber \(k(x)\) is related to the wave speed \(c(x)\) and frequency \(\omega\) by $$ k(x)=\frac{\omega}{c(x)} $$ Substitute the expression of \(c(x)\) into the equation of \(k(x)\): $$ k(x)=\frac{\omega}{c_{\alpha}\sqrt{\frac{1}{1+\beta x}}}=\frac{\omega\sqrt{1+\beta x}}{c_{\alpha}} $$ Next, we substitute the expression of \(k(x)\) into the equation for the mode number \(n\): $$ n=\frac{1}{\pi}\int_{0}^{l}\frac{\omega\sqrt{1+\beta x}}{c_{\alpha}}dx=\frac{\omega}{c_{\alpha}\pi}\int_{0}^{l}\sqrt{1+\beta x}dx $$ We can solve the above integral by using the substitution method. Let \(u=1+\beta x\), then \(du=\beta dx\). And change the limits of the integral accordingly: $$ n=\frac{\omega}{c_{\alpha}\pi}\int_{1}^{1+\beta l}\frac{\sqrt{u}du}{\beta} $$ Now, we can integrate and solve for \(\omega\): $$ n=\frac{\omega}{3\beta c_{\alpha}\pi}\left[u^{3/2}\right]_{1}^{1+\beta l}\Rightarrow\omega_{n}=\frac{3n\pi\beta c_{\alpha}}{(1+\beta l)^{3/2}-1} $$

Lastly, we need to find the maximum value of inhomogeneity coefficient (\(\beta\)), without violating the WKB limit. The WKB approximation is valid if the following condition is satisfied: $$ \frac{d\ln\lambda_{0}(x)}{dx}\ll k(x) $$ Calculate the left side part of the inequality: $$ \frac{d\ln\lambda_{0}(x)}{dx}=\frac{d}{dx}\ln(\alpha+\alpha\beta x)=\frac{\beta}{1+\beta x} $$ Now, substitute the expression of \(k(x)\): $$ \frac{\beta}{1+\beta x}\ll\frac{\omega\sqrt{1+\beta x}}{c_{\alpha}} $$ Since \(\omega_{n}=\frac{3n\pi\beta c_{\alpha}}{(1+\beta l)^{3/2}-1}\), we can write: $$ \frac{\beta}{1+\beta x}\ll\frac{3n\pi\beta\sqrt{1+\beta x}}{(1+\beta l)^{3/2}-1} $$ By observing the inequality, we can notice that \(\beta l \leq 3\) is a valid condition for the WKB limit. In conclusion, the frequency of the nth normal mode in the WKB approximation is \(\omega_{n}=\frac{3}{2} n \pi \beta c_{\alpha} /\left[(1+\beta l)^{3 / 2}-1\right]\) where \(c_\alpha=(\tau / \alpha)^{1/2}\), and the maximum value of the inhomogeneity coefficient \(\beta\) that does not violate the WKB limit is \(\beta l \leq 3\).