Answer :
To find the difference in wavelength between the wave produced in ice and the wave produced in water, we need to follow a systematic approach. Here are the steps:
1. Understand the given data:
- Frequency of the sound wave: \( f = 550 \) Hz
- Speed of sound in ice: \( v_{\text{ice}} = 3200 \) m/s
- Speed of sound in water: \( v_{\text{water}} = 1500 \) m/s
2. Calculate the wavelength in each medium:
The formula to calculate the wavelength (\( \lambda \)) is:
[tex]\[ \lambda = \frac{v}{f} \][/tex]
where \( v \) is the speed of sound in the medium, and \( f \) is the frequency of the sound wave.
3. Calculate the wavelength in ice:
[tex]\[ \lambda_{\text{ice}} = \frac{v_{\text{ice}}}{f} = \frac{3200 \, \text{m/s}}{550 \, \text{Hz}} \approx 5.818 \, \text{m} \][/tex]
4. Calculate the wavelength in water:
[tex]\[ \lambda_{\text{water}} = \frac{v_{\text{water}}}{f} = \frac{1500 \, \text{m/s}}{550 \, \text{Hz}} \approx 2.727 \, \text{m} \][/tex]
5. Find the difference in wavelengths:
[tex]\[ \Delta \lambda = \lambda_{\text{ice}} - \lambda_{\text{water}} = 5.818 \, \text{m} - 2.727 \, \text{m} \approx 3.091 \, \text{m} \][/tex]
Therefore, the difference in wavelength between the wave produced in ice and the wave produced in water is approximately \(3.1 \, \text{m}\).
Among the given options:
- \(2.1 \, \text{m}\)
- \(3.1 \, \text{m}\)
- \(5.2 \, \text{m}\)
- \(8.5 \, \text{m}\)
The correct answer is [tex]\(3.1 \, \text{m}\)[/tex].
1. Understand the given data:
- Frequency of the sound wave: \( f = 550 \) Hz
- Speed of sound in ice: \( v_{\text{ice}} = 3200 \) m/s
- Speed of sound in water: \( v_{\text{water}} = 1500 \) m/s
2. Calculate the wavelength in each medium:
The formula to calculate the wavelength (\( \lambda \)) is:
[tex]\[ \lambda = \frac{v}{f} \][/tex]
where \( v \) is the speed of sound in the medium, and \( f \) is the frequency of the sound wave.
3. Calculate the wavelength in ice:
[tex]\[ \lambda_{\text{ice}} = \frac{v_{\text{ice}}}{f} = \frac{3200 \, \text{m/s}}{550 \, \text{Hz}} \approx 5.818 \, \text{m} \][/tex]
4. Calculate the wavelength in water:
[tex]\[ \lambda_{\text{water}} = \frac{v_{\text{water}}}{f} = \frac{1500 \, \text{m/s}}{550 \, \text{Hz}} \approx 2.727 \, \text{m} \][/tex]
5. Find the difference in wavelengths:
[tex]\[ \Delta \lambda = \lambda_{\text{ice}} - \lambda_{\text{water}} = 5.818 \, \text{m} - 2.727 \, \text{m} \approx 3.091 \, \text{m} \][/tex]
Therefore, the difference in wavelength between the wave produced in ice and the wave produced in water is approximately \(3.1 \, \text{m}\).
Among the given options:
- \(2.1 \, \text{m}\)
- \(3.1 \, \text{m}\)
- \(5.2 \, \text{m}\)
- \(8.5 \, \text{m}\)
The correct answer is [tex]\(3.1 \, \text{m}\)[/tex].
