# Step-by-Step Solution break down the and questions on "W on a String into two main, with explanations at each step. --- ## Step1: Understand the Table and Harmonics **Explanation The table asks for information aboutics, frequency, number of wavelengths and wavelength. For a string fixed at both ends: - **n-th harmonic:** The string forms `n` half-wavelengths. - **Number of wavelengths:** For the n-th harmonic, the length of the string \( L \) contains \( n/2 \) wavelengths, so: \[ \text{Number of wavelengths} = \frac{n}{2} \] \[ \text{Wavelength} (\lambda) = \frac{2L}{n} \] - **Frequency:** The frequency of the n-th harmonic is: \[ f_n = n f_1 \] where \( f_1 \) is the fundamental frequency. --- ## Step 2: Fill in the Table (General Form) **Explanation:** Let’s fill the table with formulas and an example, assuming the string length \( L \) and fundamental frequency \( f_1 \) are known. | Harmonics | Frequency (\(f_n\)) | # of Wavelengths (\(n/2\)) | Wavelength (\(2L/n\)) | |-----------|---------------------|----------------------------|-----------------------| | 1 | \(f_1\) | 1/2 | \(2L/1\) | | 2 | \(2f_1\) | 2/2 = 1 | \(2L/2 = L\) | | 3 | \(3f_1\) | 3/2 | \(2L/3\) | | 4 | \(4f_1\) | 2 | \(2L/4 = L/2\) | | 5 | \(5f_1\) | 5/2 | \(2L/5\) | | 6 | \(6f_1\) | 3 | \(2L/6 = L/3\) | **Pattern noticed:** - The frequency increases linearly with the harmonic number. - The wavelength decreases as the harmonic number increases. --- ## Final Answer - **Step 1:** The n-th harmonic on a string fixed at both ends contains \( n/2 \) wavelengths, with wavelength \( \lambda = 2L/n \). - **Step 2:** The frequencies are integer multiples of the fundamental (\( f_n = n f_1 \)), and the pattern shows frequency increases with harmonic number while wavelength decreases. **To fully complete the table, you’ll need the actual measured values for frequency and string length \( L \). Use the above formulas to fill in your data.**