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Question: A point source of power \[5\,{\text{W}}\] is emitting sound waves in a medium. If the medium does no...

A point source of power 5W5\,{\text{W}} is emitting sound waves in a medium. If the medium does not absorb any energy, then the intensity of the sound wave at a distance of 5m5\,{\text{m}} from the source is
A. 5Wm25\,{\text{W}} \cdot {{\text{m}}^{ - 2}}
B. 15Wm2\dfrac{1}{5}\,{\text{W}} \cdot {{\text{m}}^{ - 2}}
C. 125Wm2\dfrac{1}{{25}}\,{\text{W}} \cdot {{\text{m}}^{ - 2}}
D. 120πWm2\dfrac{1}{{20\pi }}\,{\text{W}} \cdot {{\text{m}}^{ - 2}}

Explanation

Solution

Use the formula for the intensity of the sound wave at a distance r from the source of the sound. This equation gives the relation between the power output of the sound source and the distance of the wave from the sound source.

Formula used:
The intensity of the sound wave at a distance from the source of sound is given by
I=P4πr2I = \dfrac{P}{{4\pi {r^2}}} …… (1)
Here, II is the intensity of the source at a distance rr from the source of sound and PP is the power output of the sound source.

Complete step by step answer:
The power output PP of the source of sound is 5W5\,{\text{W}}.
P=5WP = 5\,{\text{W}}
Determine the intensity of the sound wave at a distance of 5m5\,{\text{m}} from the source of sound.
Rewrite equation (1) for the intensity of sound.
I=P4πr2I = \dfrac{P}{{4\pi {r^2}}}

Substitute 5W5\,{\text{W}} for PP and 5m5\,{\text{m}} for rr in the above equation.
I=5W4π(5m)2I = \dfrac{{5\,{\text{W}}}}{{4\pi {{\left( {5\,{\text{m}}} \right)}^2}}}
I=54π(25)\Rightarrow I = \dfrac{5}{{4\pi \left( {25} \right)}}
I=5100π\Rightarrow I = \dfrac{5}{{100\pi }}
I=120πWm - 2\therefore I = \dfrac{1}{{20\pi }}\,{\text{W}} \cdot {{\text{m}}^{{\text{ - 2}}}}

Hence, the correct option is D.

Additional information:
The intensity of the sound wave is the ratio of the power of the sound wave to the area in which the sound wave travels.The larger the intensity of sound, larger is the amplitude of the sound and the listener will hear a larger sound. Any normal person with a normal hearing ability can hear the sound with minimum intensity 1012Wm - 2{10^{ - 12}}\,{\text{W}} \cdot {{\text{m}}^{{\text{ - 2}}}}. The equation (1) for the intensity of sound waves at a distance r from the source is also used for the intensity of the light wave at a distance r from the source of light.

Note: The students may substitute the value of π\pi while calculating the intensity of the sound wave using equation (1). But if the value of π\pi is substituted in the respective equation, the final intensity of the sound after solving the equation will be different and the students may get confused that the correct answer is not given in the options. So avoid directly substituting the value ofπ\pi in the equation.