Update 8_many_atoms_solutions.md

src/8_many_atoms_solutions.md

@@ -17,7 +17,7 @@ pi = np.pi
 
 ### Subquestion 1
 
-Accoustic branch corresponds with (-) in the equation given in the lecture notes. Use the small angle approximation $\sin(x) \approx x$ to ease calculations. For the Taylor polynomial take $\omega^2 = f(x) \approx f(0) + f'(0)k + f''(0)k^2$ (some terms vanish, computation is indeed quite tedious). You should find: $$|v_g| = \sqrt{\frac{\kappa a^2}{2(m_a+m_2)}}$$
+Accoustic branch corresponds with (-) in the equation given in the lecture notes. Use the small angle approximation $\sin(x) \approx x$ to ease calculations. For the Taylor polynomial take $\omega^2 = f(x) \approx f(0) + f'(0)k + \frac{1}{2} f''(0)k^2$ (some terms vanish, computation is indeed quite tedious). You should find: $$|v_g| = \sqrt{\frac{\kappa a^2}{2(m_a+m_2)}}$$
 
 ### Subquestion 2
 
@@ -142,7 +142,7 @@ Follow the same procedure as before but now using Ansatz $$ \begin{pmatrix} u_{1
 
 ### Subquestion 4
 
-The eigenvalues are given by:
+Using the eigenvectors from $X$ we find the eigenvalues:
 
 $$ \omega^2 = \begin{pmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \end{pmatrix} = \frac{1}{m} \begin{pmatrix} q \\ k_3 + \frac{3q}{2} - \frac{\sqrt{4k_3^2 - 4k_3q + 9q^2}}{2} \\ k_3 + \frac{3q}{2} + \frac{\sqrt{4k_3^2 - 4k_3q + 9q^2}}{2} \end{pmatrix} $$