Sure! Let's work through each question in **2 steps** with explanations after each step, and then provide the final answer at the end. --- ### **Question (a)** **What is the frequency (in Hz) of the first allowed mode?** #### **Step 1: Understand the Problem** A string of length \( L = .55\, \text{m} \) is fixed at both ends. The first allowed mode (fundamental frequency) is when the wavelength \( \lambda_1 = 2L \). #### **Step 2: Find the Frequency** - Given wave speed \( v = 36\, \text{m/s} \). - Wavelength for the first mode: \( \lambda_1 = 2L = 2 \times .55 = 1.10\, \text{m} \). - Frequency \( f_1 = \frac{v}{\lambda_1} = \frac{36}{1.10} = 32.73\, \text{Hz} \). **Explanation:** The fundamental frequency is determined by the speed of the wave and the length of the string. **Final Answer:** \[ \boxed{32.7\, \text{Hz}} \] --- ### **Question (b)** **What is the frequency (in Hz) of the fifth allowed mode?** #### **Step 1: Understand the Fifth Mode** For the \( n \)-th mode: \( \lambda_n = \frac{2L}{n} \). For \( n = 5 \): \[ \lambda_5 = \frac{2 \times .55}{5} = .22\, \text{m} \] #### **Step 2: Calculate the Frequency** \[ f_5 = \frac{v}{\lambda_5} = \frac{36}{.22} = 163.64\, \text{Hz} \] **Explanation:** Higher modes have shorter wavelengths and higher frequencies. **Final Answer:** \[ \boxed{163.6\, \text{Hz}} \] --- ### **Question (c)** **What is the wavelength of the third allowed mode?** #### **Step 1: Find the Wavelength Formula** For \( n = 3 \): \[ \lambda_3 = \frac{2L}{3} \] #### **Step 2: Substitute Values** \[ \lambda_3 = \frac{2 \times .55}{3} = \frac{1.10}{3} = .3667\, \text{m} \] **Explanation:** The wavelength for each mode decreases as the mode number increases. **Final Answer:** \[ \boxed{.367\, \text{m}} \] --- If you need the answers to the other questions shown, please let me know!