Question

Two identical speakers emit sound waves of frequency 660 Hz uniformly in all directions. The audio output of each speaker is 1 m W and the speed of sound in air 330 m/s. A point P is a distance 2m from one speaker and 3m from the other. Find
a) Intensities ${I_1}$ and ${I_2}$ from each speaker from P separately.
b) If they are in coherence, the resultant I at P.
c) The resultant I at P if the two waves differ by a phase of $\pi$ at the time of emission.

Answer 1

Hint
Use the formula ${I_1} = \dfrac{P}{{4\pi {R^2}}}$ to take out individual intensity due to speakers. Then use the formulas $I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \Delta \;\phi \;\;$ and $\,\;\phi \;\; = \dfrac{{2\pi }}{{\;\lambda }}\Delta x$ to solve the problem. For part b) take the phase difference as 0.

Complete step by step answer
Given, Frequency = 660 Hz
Speed of sound in air = 330 m/s
Power of each speaker = 1 m W
a) Intensities ${I_1}$ and ${I_2}$ from each speaker from P separately.
$\Rightarrow {I_1} = \dfrac{P}{{4\pi {R^2}}}$
$\Rightarrow {I_1} = \dfrac{{1 \times {{10}^{ - 3}}}}{{4\pi \times {{(2)}^2}}}$
$\Rightarrow {I_1} = 19.9 \times {10^{ - 6}}{\text{W/}}{{\text{m}}^{\text{2}}}$
(putting the values from the question)
For second speaker:
$\Rightarrow {I_2} = \dfrac{P}{{4\pi {R^2}}}$
$\Rightarrow {I_2} = \dfrac{{1 \times {{10}^{ - 3}}}}{{4\pi \times {{(3)}^2}}}$
$\Rightarrow {I_2} = 8.85 \times {10^{ - 6}}{\text{W/}}{{\text{m}}^{\text{2}}}$
b) If they are in coherence, the resultant I at P.
$\Rightarrow I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \Delta \;\phi \;\;$
$\Rightarrow {\text{and}}\,\;\phi \;\; = \dfrac{{2\pi }}{{\;\lambda }}\Delta x$
We know
$\Rightarrow v = f\;\lambda$
$\Rightarrow \;\lambda = \dfrac{v}{f} = \dfrac{{330}}{{660}} = \dfrac{1}{2}$
$\Rightarrow \Delta x = 3 - 2 = 1$ m
Hence,
$\Rightarrow \Delta \;\phi \;\; = \dfrac{{2\pi }}{{\dfrac{1}{2}}} = 4\pi$
Hence the net intensity becomes,
$\Rightarrow I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \Delta \;\phi \;\;$
$\Rightarrow I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos (4\pi )$
$\Rightarrow I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}}$
Putting the values we have,
$\Rightarrow I = 19.9 + 8.85 + 2\sqrt {19.9 \times } 8.85$
$\Rightarrow I = 55.3 \times {10^{ - 6}}W/{m^2}$
c) The resultant I at P if the two waves differ by a phase of $\pi$ at the time of emission.
Here, $\;\phi \;\; = \pi$
Hence resultant intensity becomes,
$\Rightarrow I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \Delta \;\phi \;\;$
$\Rightarrow I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos (\pi )$
$\Rightarrow I = {I_1} + {I_2} - 2\sqrt {{I_1}{I_2}}$
Putting values we get,
$\Rightarrow I = 19.9 + 8.85 - 2\sqrt {19.9 \times } 8.85$
$\Rightarrow I = 28.7 \times {10^{ - 6}}W/{m^2}$.

Note
That the formula ${I_1} = \dfrac{P}{{4\pi {R^2}}}$ is only applicable for point sources. If the source is a line source or a sheet, then we can’t use this formula. We will have to use different formulas for taking out intensity.