Let's solve the problem step by step. --- ## **Step 1: Wave Equations and General Solutions** ### **Region I (\( -L \leq x \leq 0 \), density \(\rho\)):** The wave equation is: \[ \frac{\partial^2 y_1}{\partial t^2} = \frac{T}{\rho} \frac{\partial^2 y_1}{\partial x^2} \] General solution (for standing waves, since boundary at \(x = -L\) is driven): \[ y_1(x,t) = [A \sin(k(x+L))] \sin(\omega t) \] where \(k = \omega/v_1\), \(v_1 = \sqrt{T/\rho}\). ### **Region II (\( x > 0 \), density \(\rho'\)):** The wave equation is: \[ \frac{\partial^2 y_2}{\partial t^2} = \frac{T}{\rho'} \frac{\partial^2 y_2}{\partial x^2} \] Only outgoing waves (right-going): \[ y_2(x,t) = B \sin(k'(x)) \sin(\omega t) \] where \(k' = \omega/v_2\), \(v_2 = \sqrt{T/\rho'}\). But since only outgoing wave, we use: \[ y_2(x,t) = C \sin(\omega t - k' x) \] --- ## **Step 2: Boundary and Continuity Conditions** - **At \(x = -L\):** The string is driven: \(y_1(-L, t) = \chi \sin(\omega t)\) - **At \(x = 0\):** Displacement and slope (force) must be continuous. Let's write the general solution in region I as: \[ y_1(x, t) = A \sin[k(x+L)] \sin(\omega t) \] At \(x = -L\): \[ y_1(-L, t) = A \sin(0) \sin(\omega t) = 0 \] But this cannot match the driven displacement unless we add a cosine term: \[ y_1(x, t) = [a \sin(k(x+L)) + b \cos(k(x+L))] \sin(\omega t) \] Apply boundary: \[ y_1(-L, t) = b \cos(0) \sin(\omega t) = b \sin(\omega t) \implies b = \chi \] So, \[ y_1(x, t) = [a \sin(k(x+L)) + \chi \cos(k(x+L))] \sin(\omega t) \] --- ## **Step 3: Matching at \(x = 0\)** - **Displacement continuity:** \[ y_1(0, t) = y_2(0, t) \] - **Slope (force) continuity:** \[ T \left. \frac{\partial y_1}{\partial x} \right|_{x=0} = T \left. \frac{\partial y_2}{\partial x} \right|_{x=0} \] ### **Compute values at \(x = 0\):** - \(y_1(0, t) = [a \sin(kL) + \chi \cos(kL)] \sin(\omega t)\) - \(\frac{\partial y_1}{\partial x} \Big|_{x=0} = a k \cos(kL) - \chi k \sin(kL)\) Let the outgoing wave in region II be: \[ y_2(x, t) = C \sin(\omega t - k' x) \] - At \(x = 0\): \(y_2(0, t) = C \sin(\omega t)\) - \(\frac{\partial y_2}{\partial x} \big|_{x=0} = -C k' \cos(\omega t)\) --- ## **Step 4: Apply Matching Conditions at \(x = 0\):** (Assume all terms have the same time dependence, so we can equate coefficients.) 1. **Displacement:** \[ a \sin(kL) + \chi \cos(kL) = C \] 2. **Slope:** \[ a k \cos(kL) - \chi k \sin(kL) = -C k' \] --- ## **Step 5: Solve for \(a\) and \(C\)** From (1): \[ C = a \sin(kL) + \chi \cos(kL) \] Plug into (2): \[ a k \cos(kL) - \chi k \sin(kL) = -[a \sin(kL) + \chi \cos(kL)] k' \] \[ a k \cos(kL) - \chi k \sin(kL) + a k' \sin(kL) + \chi k' \cos(kL) = 0 \] \[ a [k \cos(kL) + k' \sin(kL)] + \chi [-k \sin(kL) + k' \cos(kL)] = 0 \] \[ a = \chi \frac{k \sin(kL) - k' \cos(kL)}{k \cos(kL) + k' \sin(kL)} \] --- ## **Step 6: Plug in \(a\) to Find Displacement at \(x=0\)** Recall: \[ y_1(0, t) = [a \sin(kL) + \chi \cos(kL)] \sin(\omega t) \] So, \[ y_1(0, t) = \left[ \chi \frac{k \sin(kL) - k' \cos(kL)}{k \cos(kL) + k' \sin(kL)} \sin(kL) + \chi \cos(kL) \right] \sin(\omega t) \] \[ = \chi \left[ \frac{ (k \sin(kL) - k' \cos(kL)) \sin(kL) + (k \cos(kL) + k' \sin(kL)) \cos(kL) } {k \cos(kL) + k' \sin(kL)} \right] \sin(\omega t) \] Expand numerator: \[ k \sin^2(kL) - k' \sin(kL) \cos(kL) + k \cos^2(kL) + k' \sin(kL) \cos(kL) \] The \(k' \sin(kL) \cos(kL)\) terms cancel: \[ k (\sin^2(kL) + \cos^2(kL)) = k \] So: \[ y_1(0, t) = \chi \frac{k}{k \cos(kL) + k' \sin(kL)} \sin(\omega t) \] --- ## **Step 7: Substitute \( k = \omega/v_1 = \omega \sqrt{\rho/T} \), \( k' = \omega \sqrt{\rho'/T} \), and \( \omega = \frac{\pi}{2L} \sqrt{T/\rho} \) (given)** - \(k = \frac{\pi}{2L}\) - \(k' = \omega \sqrt{\rho'/T} = \frac{\pi}{2L} \sqrt{\frac{\rho'}{\rho}}\) Also, \(kL = \frac{\pi}{2}\), so \(\cos(kL) = 0\), \(\sin(kL) = 1\). Plug in: \[ y_1(0, t) = \chi \frac{\frac{\pi}{2L}}{k \cdot 0 + k' \cdot 1} \sin(\omega t) = \chi \frac{\frac{\pi}{2L}}{ \frac{\pi}{2L} \sqrt{\frac{\rho'}{\rho}} } \sin(\omega t) = \chi \frac{1}{ \sqrt{\frac{\rho'}{\rho}} } \sin(\omega t) = \chi \sqrt{\frac{\rho}{\rho'}} \sin(\omega t) \] --- ## **Final Answer** \[ \boxed{ y(0, t) = \chi \sqrt{\frac{\rho}{\rho'}} \sin(\omega t) } \] with \(\omega = \frac{\pi}{2L} \sqrt{\frac{T}{\rho}}\). --- **Summary of Approach:** - Write general solutions in both regions. - Match boundary conditions at \(x = -L\) and \(x = 0\). - Solve for the amplitude at \(x = 0\). - Substitute the provided value of \(\omega\) to get the final result.