To determine the wavelength of the second overtone of a string fastened tightly between two walls, we need to understand the relationship between the length of the string, the harmonics, and the wavelengths. ### Step-by-Step Explanation 1. **Understanding the Setup**: - The string is fixed at both ends, which means it can vibrate in specific modes called harmonics. The distance between the two walls is 24 cm, which is the length of the string (L = 24 cm). 2. **Identifying the Harmonics**: - The fundamental frequency (first harmonic) of a string fixed at both ends has a wavelength (λ) that is twice the length of the string: \[ \lambda_1 = 2L \] - The second harmonic (first overtone) has a wavelength that is equal to the length of the string: \[ \lambda_2 = L \] - The third harmonic (second overtone) has a wavelength that is two-thirds of the length of the string: \[ \lambda_3 = \frac{2}{3}L \] 3. **Calculating the Wavelengths**: - For the first harmonic: \[ \lambda_1 = 2L = 2 \times 24 \text{ cm} = 48 \text{ cm} \] - For the second harmonic: \[ \lambda_2 = L = 24 \text{ cm} \] - For the third harmonic (second overtone): \[ \lambda_3 = \frac{2}{3}L = \frac{2}{3} \times 24 \text{ cm} = 16 \text{ cm} \] 4. **Conclusion**: - The wavelength of the second overtone (third harmonic) is **16 cm**. ### Why the Other Options are Incorrect - **Option A: 24 cm**: This is the wavelength of the second harmonic (first overtone), not the second overtone. - **Option C: 12 cm**: This would correspond to a harmonic that is not applicable in this case. The second overtone is not equal to 12 cm. - **Option D: 8 cm**: This wavelength does not correspond to any harmonic of the string fixed at both ends. It is too short for the given length of the string. ### Summary of Key Points - The wavelength of the second overtone (third harmonic) for a string fixed at both ends is calculated using the formula \(\lambda_n = \frac{2}{n}L\), where \(n\) is the harmonic number. - For a string of length 24 cm, the second overtone corresponds to \(n = 3\), yielding a wavelength of 16 cm. - The first harmonic has a wavelength of 48 cm, and the second harmonic has a wavelength of 24 cm, which are both different from the second overtone. ### Revision Summary - The second overtone corresponds to the third harmonic of a string fixed at both ends. - The wavelength of the second overtone is given by \(\lambda_3 = \frac{2}{3}L\). - For a string length of 24 cm, the wavelength of the second overtone is 16 cm. - Understanding the relationship between harmonics and wavelengths is crucial for solving similar problems.