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Sagot :
To find the frequency at which destructive interference occurs, we can follow these steps:
1. Determine the path difference:
The path difference is the difference in the distances from the person to the two speakers.
[tex]\[ \text{Path difference} = d_2 - d_1 = 13.20 \, \text{m} - 7.70 \, \text{m} = 5.50 \, \text{m} \][/tex]
2. Apply the condition for destructive interference:
For destructive interference, the path difference should satisfy the condition:
[tex]\[ \text{Path difference} = \left( n + \frac{1}{2} \right) \lambda \][/tex]
Given [tex]\( n = 1 \)[/tex]:
[tex]\[ \text{Path difference} = \left( 1 + \frac{1}{2} \right) \lambda = \frac{3}{2} \lambda \][/tex]
Rearrange to solve for the wavelength ([tex]\(\lambda\)[/tex]):
[tex]\[ \lambda = \frac{\text{Path difference}}{\frac{3}{2}} \][/tex]
Substituting the path difference:
[tex]\[ \lambda = \frac{5.50 \, \text{m}}{\frac{3}{2}} = \frac{5.50 \, \text{m}}{1.5} = 3.67 \, \text{m} \][/tex]
3. Calculate the frequency:
To find the frequency, we use the relationship between speed (v), wavelength ([tex]\(\lambda\)[/tex]), and frequency (f):
[tex]\[ v = \lambda f \][/tex]
Rearranging for frequency:
[tex]\[ f = \frac{v}{\lambda} \][/tex]
Given the speed of sound [tex]\( v = 343 \, \text{m/s} \)[/tex] and the wavelength [tex]\( \lambda = 3.67 \, \text{m} \)[/tex]:
[tex]\[ f = \frac{343 \, \text{m/s}}{3.67 \, \text{m}} = 93.55 \, \text{Hz} \][/tex]
So, the frequency at which destructive interference occurs is approximately:
[tex]\[ f \approx 93.55 \, \text{Hz} \][/tex]
1. Determine the path difference:
The path difference is the difference in the distances from the person to the two speakers.
[tex]\[ \text{Path difference} = d_2 - d_1 = 13.20 \, \text{m} - 7.70 \, \text{m} = 5.50 \, \text{m} \][/tex]
2. Apply the condition for destructive interference:
For destructive interference, the path difference should satisfy the condition:
[tex]\[ \text{Path difference} = \left( n + \frac{1}{2} \right) \lambda \][/tex]
Given [tex]\( n = 1 \)[/tex]:
[tex]\[ \text{Path difference} = \left( 1 + \frac{1}{2} \right) \lambda = \frac{3}{2} \lambda \][/tex]
Rearrange to solve for the wavelength ([tex]\(\lambda\)[/tex]):
[tex]\[ \lambda = \frac{\text{Path difference}}{\frac{3}{2}} \][/tex]
Substituting the path difference:
[tex]\[ \lambda = \frac{5.50 \, \text{m}}{\frac{3}{2}} = \frac{5.50 \, \text{m}}{1.5} = 3.67 \, \text{m} \][/tex]
3. Calculate the frequency:
To find the frequency, we use the relationship between speed (v), wavelength ([tex]\(\lambda\)[/tex]), and frequency (f):
[tex]\[ v = \lambda f \][/tex]
Rearranging for frequency:
[tex]\[ f = \frac{v}{\lambda} \][/tex]
Given the speed of sound [tex]\( v = 343 \, \text{m/s} \)[/tex] and the wavelength [tex]\( \lambda = 3.67 \, \text{m} \)[/tex]:
[tex]\[ f = \frac{343 \, \text{m/s}}{3.67 \, \text{m}} = 93.55 \, \text{Hz} \][/tex]
So, the frequency at which destructive interference occurs is approximately:
[tex]\[ f \approx 93.55 \, \text{Hz} \][/tex]
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