Let's solve **Problem 6.1** step by step, as requested. --- ### **Step 1: Discretized Analog of the Continuous String** #### **Continuous Case Recap** From the provided text: - For a continuous string fixed at \( x = 0 \) and \( x = \ell \), the displacement can be written as: \[ \psi(x) = \sum_{n=1}^\infty c_n \sin\left(\frac{n\pi x}{\ell}\right) \] where \[ c_n = \frac{2}{\ell} \int_0^\ell \psi(x) \sin\left(\frac{n\pi x}{\ell}\right) dx \] --- #### **Discrete (Beaded String) Case** For a string with \( W \) beads (mass points) and \( W+1 \) points (including the endpoints), the allowed positions are: \[ x_k = k a, \quad k = 0, 1, \ldots, W+1 \] where \( a = \ell/(W+1) \). Since the ends are fixed, we only consider \( k = 1 \) to \( W \) (interior beads). Let the displacement at bead \( k \) be \( \psi_k \). The normal modes for the discrete case are: \[ \psi_k = \sum_{n=1}^W c_n \sin\left(\frac{n\pi k}{W+1}\right) \] This is the **finite analog of equation (6.8)**. --- #### **Finding the Coefficients \( c_n \) (Discrete Analog of 6.10)** To find \( c_n \), use orthogonality of sine functions: \[ \sum_{k=1}^{W} \sin\left(\frac{n\pi k}{W+1}\right) \sin\left(\frac{n'\pi k}{W+1}\right) = \begin{cases} b & \text{if } n = n' \\ 0 & \text{if } n \neq n' \end{cases} \] where \( b \) is a constant to be determined. Multiply both sides by \( \sin\left(\frac{n\pi k}{W+1}\right) \) and sum over \( k \): \[ \sum_{k=1}^{W} \psi_k \sin\left(\frac{n\pi k}{W+1}\right) = c_n \sum_{k=1}^{W} \sin^2\left(\frac{n\pi k}{W+1}\right) \] So, \[ c_n = \frac{1}{b} \sum_{k=1}^{W} \psi_k \sin\left(\frac{n\pi k}{W+1}\right) \] --- ### **Step 2: Find the Value of \( b \)** Recall the sum: \[ b = \sum_{k=1}^W \sin^2\left(\frac{n\pi k}{W+1}\right) \] This sum is independent of \( n \) (for \( n \neq 0 \)), and is a standard result (see discrete orthogonality of sines): \[ \sum_{k=1}^W \sin^2\left(\frac{n\pi k}{W+1}\right) = \frac{W+1}{2} \] You can verify this result using trigonometric identities or see it in standard references on Fourier series on discrete grids. --- ### **Step 3: Taking the Limit \( W \to \infty \)** As \( W \to \infty \), \( a = \ell/(W+1) \to 0 \). The sum \[ \sum_{k=1}^W f(x_k) a \to \int_0^\ell f(x) dx \] and the argument of the sine becomes continuous: \[ \frac{n\pi k}{W+1} = \frac{n\pi x_k}{\ell} \] So, \[ c_n = \frac{2}{\ell} \int_0^\ell \psi(x) \sin\left(\frac{n\pi x}{\ell}\right) dx \] matching equation (6.10). --- ### **Summary of Answers** #### **Discrete Analog (for finite \( W \)):** \[ \boxed{ \psi_k = \sum_{n=1}^W c_n \sin\left(\frac{n\pi k}{W+1}\right), \qquad c_n = \frac{2}{W+1} \sum_{k=1}^W \psi_k \sin\left(\frac{n\pi k}{W+1}\right) } \] #### **Value of \( b \):** \[ \boxed{b = \frac{W+1}{2}} \] #### **Continuous Limit (\( W \to \infty \)):** \[ \psi(x) = \sum_{n=1}^\infty c_n \sin\left(\frac{n\pi x}{\ell}\right), \qquad c_n = \frac{2}{\ell} \int_0^\ell \psi(x) \sin\left(\frac{n\pi x}{\ell}\right) dx \] --- **If you need the derivation for the sum for \( b \), let me know!**