
Two sound sources \[{S_1}\] and \[{S_2}\] emit pure sinusoidal waves in phase. If the speed of sound is 330 m/s, then for what frequencies does constructive interference occur at P?

A. 500 Hz
B. 1000 Hz
C. 1200 Hz
D. 1500 Hz
Answer
162k+ views
Hint:To solve this question we have to use the relationship between the frequencies for constructive interference and the path difference of two given sound sources. Path difference can be obtained by using Pythagoras theorem.
Formula used:
Path difference for constructive interference is given as,
\[\Delta x = n\lambda = n\left( {\dfrac{v}{f}} \right)\]
Where n is an integer, \[\lambda \] is the wavelength, v is the speed of sound and f is the frequency.
Complete step by step solution:
Given the speed of sound v = 330 m/s
If the line PO is perpendicular to the line joining the two sound sources \[{S_1}\]and\[{S_2}\].
Separation between the two sound sources \[{S_1}\] and \[{S_2}\], \[{S_1}{S_2}\]= 4.68M

Image: Separation between the two sound sources \[{S_1}\] and \[{S_2}\].
From the given data, we have
\[{S_1}P = \sqrt {{3^2} + {4^2}} = 5\]
And,
\[{S_2}P = \sqrt {{4^2} + {{1.68}^2}} = 4.338\]
As we know that Path difference,
\[\Delta x = \left| {{S_1}P - {S_2}P} \right| \\ \]
\[\Rightarrow \Delta x = 5 - 4.338 \approx 0.66 \\ \]
Also, we know that for constructive interference,
\[\Delta x = n\lambda = n\left( {\dfrac{v}{f}} \right)\]
where n = 1,2, 3…...
Now the frequency is given as,
\[f = \dfrac{{nv}}{{\Delta x}} \\ \]
\[\Rightarrow f = \dfrac{{n \times 330}}{{0.66}} \\ \]
\[\Rightarrow f = 500n{\rm{ Hz}}\]
For n = 0, frequency, f = 0
For n = 1, frequency, f = 500 Hz
For n = 2, frequency, f = 1000 Hz
For n = 3, frequency, f = 1500 Hz
and so on.
Therefore, for the frequencies constructive interference occurs at P is 500 Hz, 100 Hz, 1500 Hz and so on.
Hence option A, B and D is the correct answer.
Note: The difference in distance travelled by the two waves is defined as the one full wavelength or the path difference is \[1\lambda \]. When the path difference (\[\Delta x\]) is one full wavelength then a crest meets a crest and constructive interference occurs.
Formula used:
Path difference for constructive interference is given as,
\[\Delta x = n\lambda = n\left( {\dfrac{v}{f}} \right)\]
Where n is an integer, \[\lambda \] is the wavelength, v is the speed of sound and f is the frequency.
Complete step by step solution:
Given the speed of sound v = 330 m/s
If the line PO is perpendicular to the line joining the two sound sources \[{S_1}\]and\[{S_2}\].
Separation between the two sound sources \[{S_1}\] and \[{S_2}\], \[{S_1}{S_2}\]= 4.68M

Image: Separation between the two sound sources \[{S_1}\] and \[{S_2}\].
From the given data, we have
\[{S_1}P = \sqrt {{3^2} + {4^2}} = 5\]
And,
\[{S_2}P = \sqrt {{4^2} + {{1.68}^2}} = 4.338\]
As we know that Path difference,
\[\Delta x = \left| {{S_1}P - {S_2}P} \right| \\ \]
\[\Rightarrow \Delta x = 5 - 4.338 \approx 0.66 \\ \]
Also, we know that for constructive interference,
\[\Delta x = n\lambda = n\left( {\dfrac{v}{f}} \right)\]
where n = 1,2, 3…...
Now the frequency is given as,
\[f = \dfrac{{nv}}{{\Delta x}} \\ \]
\[\Rightarrow f = \dfrac{{n \times 330}}{{0.66}} \\ \]
\[\Rightarrow f = 500n{\rm{ Hz}}\]
For n = 0, frequency, f = 0
For n = 1, frequency, f = 500 Hz
For n = 2, frequency, f = 1000 Hz
For n = 3, frequency, f = 1500 Hz
and so on.
Therefore, for the frequencies constructive interference occurs at P is 500 Hz, 100 Hz, 1500 Hz and so on.
Hence option A, B and D is the correct answer.
Note: The difference in distance travelled by the two waves is defined as the one full wavelength or the path difference is \[1\lambda \]. When the path difference (\[\Delta x\]) is one full wavelength then a crest meets a crest and constructive interference occurs.
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