Let's break down the problem step-by-step: ### **Given:** - **String 1 linear density (\(\mu_1\))**: \(2.0 \, \text{g/m} = 0.002 \, \text{kg/m}\) - **String 2 linear density (\(\mu_2\))**: \(4.0 \, \text{g/m} = 0.004 \, \text{kg/m}\) - **Total length**: \(L_1 + L_2 = 4.0 \, \text{m}\) - **Assume tension (\(T\)) is the same throughout the string** (since the strings are connected and must be under the same tension). - **Goal**: Find \(L_1\) and \(L_2\) so that a pulse sent from the knot reaches both ends at the same time. --- ### **Step 1: Find Wave Speeds** The speed of a wave on a string is given by: \[ v = \sqrt{\frac{T}{\mu}} \] Let \(v_1\) and \(v_2\) be the speeds on string 1 and string 2, respectively: \[ v_1 = \sqrt{\frac{T}{\mu_1}} \] \[ v_2 = \sqrt{\frac{T}{\mu_2}} \] --- ### **Step 2: Time for Pulse to Travel Each String** The time to travel each section is: \[ t_1 = \frac{L_1}{v_1} \] \[ t_2 = \frac{L_2}{v_2} \] Set \(t_1 = t_2\) for simultaneous arrival: \[ \frac{L_1}{v_1} = \frac{L_2}{v_2} \] \[ \frac{L_1}{\sqrt{\frac{T}{\mu_1}}} = \frac{L_2}{\sqrt{\frac{T}{\mu_2}}} \] Since \(T\) cancels out, \[ \frac{L_1}{\sqrt{1/\mu_1}} = \frac{L_2}{\sqrt{1/\mu_2}} \] \[ L_1 \sqrt{\mu_1} = L_2 \sqrt{\mu_2} \] --- ### **Step 3: Substitute Values** \[ L_1 \sqrt{0.002} = L_2 \sqrt{0.004} \] \[ L_1 \cdot 0.0447 = L_2 \cdot 0.0632 \] \[ \frac{L_1}{L_2} = \frac{0.0632}{0.0447} = 1.414 \] --- ### **Step 4: Use Total Length Constraint** \[ L_1 + L_2 = 4.0 \] Let \(L_1 = 1.414 L_2\): \[ 1.414 L_2 + L_2 = 4.0 \] \[ 2.414 L_2 = 4.0 \] \[ L_2 = \frac{4.0}{2.414} = 1.66 \, \text{m} \] \[ L_1 = 4.0 - 1.66 = 2.34 \, \text{m} \] --- ### **Final Answers:** \[ \boxed{L_1 = 2.34 \, \text{m}}, \quad \boxed{L_2 = 1.66 \, \text{m}} \] #### **Explanation:** - The pulse must travel farther on the lighter (faster) string, so \(L_1 > L_2\). - We used the relationship between string lengths and wave speeds, set the travel times equal, and used the total length constraint to solve for \(L_1\) and \(L_2\).