To determine the velocity of the transverse wave in a stretched string, we can use the relationship between wave speed, frequency, and wavelength. Let's break down the problem step-by-step. ### Step 1: Understand the relationship between wave speed, frequency, and wavelength The formula that relates wave speed (v), frequency (f), and wavelength (λ) is given by: \[ v = f \cdot \lambda \] Where: - \( v \) is the wave speed in meters per second (m/s), - \( f \) is the frequency in hertz (Hz), - \( \lambda \) is the wavelength in meters (m). ### Step 2: Determine the wavelength for the fundamental frequency For a string fixed at both ends, the fundamental frequency (first harmonic) has a wavelength that is twice the length of the string. This is because the fundamental mode of vibration forms one complete wave along the length of the string. Given: - Length of the string \( L = 0.6 \, \text{m} \) The wavelength \( \lambda \) for the fundamental frequency is: \[ \lambda = 2L = 2 \times 0.6 \, \text{m} = 1.2 \, \text{m} \] ### Step 3: Substitute the values into the wave speed formula Now that we have the frequency and the wavelength, we can substitute these values into the wave speed formula. Given: - Frequency \( f = 220 \, \text{Hz} \) - Wavelength \( \lambda = 1.2 \, \text{m} \) Now, substituting these values into the formula: \[ v = f \cdot \lambda = 220 \, \text{Hz} \cdot 1.2 \, \text{m} \] ### Step 4: Calculate the wave speed Now we perform the multiplication: \[ v = 220 \cdot 1.2 = 264 \, \text{m/s} \] ### Conclusion: The correct answer Thus, the velocity of the transverse wave in the string is: **B. 264 m/s** ### Explanation of Other Options - **Option A (528 m/s)**: This value is incorrect because it suggests a wave speed that would require a frequency or wavelength that is not consistent with the given parameters. It could arise from a misunderstanding of the relationship between frequency and wavelength. - **Option C (132 m/s)**: This value is also incorrect. It could be mistakenly derived from halving the correct wave speed, which does not apply in this context. - **Option D (66 m/s)**: This option is incorrect as well. It is significantly lower than the calculated wave speed and does not align with the fundamental frequency and string length provided. ### Common Pitfalls 1. **Misunderstanding Wavelength**: Remember that for the fundamental frequency, the wavelength is twice the length of the string. 2. **Incorrectly Applying the Formula**: Ensure that you are using the correct relationship between wave speed, frequency, and wavelength. 3. **Units**: Always check that your units are consistent (e.g., meters for length, seconds for time). ### Revision Summary - The wave speed in a stretched string can be calculated using \( v = f \cdot \lambda \). - For a string of length \( L \), the wavelength for the fundamental frequency is \( \lambda = 2L \). - In this case, with \( L = 0.6 \, \text{m} \) and \( f = 220 \, \text{Hz} \), the wave speed is \( 264 \, \text{m/s} \). - Always ensure to understand the physical setup and relationships in wave mechanics to avoid common mistakes.