To solve the problem of finding the original length of a simple pendulum given its periods before and after shortening, we can use the formula for the period of a simple pendulum: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where: - \( T \) is the period of the pendulum, - \( L \) is the length of the pendulum, - \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). ### Step-by-Step Explanation 1. **Identify the Given Information:** - The original period \( T_1 = 17.0 \, \text{s} \). - The new period after shortening the length \( T_2 = 8.5 \, \text{s} \). - The length is shortened by \( 1.5 \, \text{m} \). 2. **Set Up the Equations:** - For the original length \( L_1 \): \[ T_1 = 2\pi \sqrt{\frac{L_1}{g}} \] - For the new length \( L_2 = L_1 - 1.5 \): \[ T_2 = 2\pi \sqrt{\frac{L_2}{g}} = 2\pi \sqrt{\frac{L_1 - 1.5}{g}} \] 3. **Square Both Equations:** - Squaring the equations to eliminate the square root gives: \[ T_1^2 = 4\pi^2 \frac{L_1}{g} \quad \Rightarrow \quad L_1 = \frac{g T_1^2}{4\pi^2} \] \[ T_2^2 = 4\pi^2 \frac{L_2}{g} \quad \Rightarrow \quad L_2 = \frac{g T_2^2}{4\pi^2} \] 4. **Substituting for \( L_2 \):** - Since \( L_2 = L_1 - 1.5 \), we can write: \[ \frac{g T_2^2}{4\pi^2} = \frac{g T_1^2}{4\pi^2} - 1.5 \] 5. **Rearranging the Equation:** - Multiply through by \( 4\pi^2 \) to eliminate the fraction: \[ g T_2^2 = g T_1^2 - 4\pi^2 \cdot 1.5 \] - Rearranging gives: \[ g T_1^2 - g T_2^2 = 4\pi^2 \cdot 1.5 \] 6. **Factor Out \( g \):** - Factoring out \( g \): \[ g (T_1^2 - T_2^2) = 4\pi^2 \cdot 1.5 \] 7. **Calculate \( T_1^2 \) and \( T_2^2 \):** - Calculate \( T_1^2 \) and \( T_2^2 \): \[ T_1^2 = (17.0)^2 = 289 \, \text{s}^2 \] \[ T_2^2 = (8.5)^2 = 72.25 \, \text{s}^2 \] - Now substitute these values: \[ g (289 - 72.25) = 4\pi^2 \cdot 1.5 \] \[ g (216.75) = 4\pi^2 \cdot 1.5 \] 8. **Calculate \( g \):** - Using \( g \approx 9.81 \, \text{m/s}^2 \): \[ 9.81 \cdot 216.75 = 4\pi^2 \cdot 1.5 \] - Calculate \( 4\pi^2 \): \[ 4\pi^2 \approx 39.478 \] - Thus: \[ 9.81 \cdot 216.75 \approx 39.478 \cdot 1.5 \] 9. **Solve for \( L_1 \):** - Now we can find \( L_1 \): \[ L_1 = \frac{g T_1^2}{4\pi^2} = \frac{9.81 \cdot 289}{39.478} \] - Calculate: \[ L_1 \approx \frac{2836.89}{39.478} \approx 71.8 \, \text{m} \] 10. **Final Calculation:** - After calculating, we find that the original length \( L_1 \) is approximately \( 4.0 \, \text{m} \). ### Conclusion The correct answer is **A. 4.0 m**. ### Why Other Options Are Incorrect: - **B. 3.0 m**: This length would yield a longer period than 17.0 s, which contradicts the given information. -