Respuesta :
a) 1 W/m^2
The sound intensity level ([tex]L_I[/tex]) is defined as
[tex]L_I = 10 Log_{10} (\frac{I}{I_0}) dB[/tex]
where
I is the intensity of the sound
[tex]I_0 = 1 pW/m^2 = 1\cdot 10^{-12}W/m^2[/tex] is the reference sound intensity
dB is the decibel
For the sound wave in the problem we have
[tex]L_I = 120 dB[/tex]
So we can solve the formula to find I, the sound intensity:
[tex]I=I_0 10^{L_I/10}=(1\cdot 10^{-12}W/m^2) 10^{120/10}=1 W/m^2[/tex]
b) 314 W
The intensity is related to the power by
[tex]I=\frac{P}{4 \pi r^2}[/tex]
where
P is the power
r is the radius of the area considered (so, the distance from the source)
In this case, we have
r = 5 m
So the power of the sound is
[tex]P=4\pi r^2 I=4 \pi (5 m)^2 (1 W/m^2)=314 W[/tex]
c) 0.02 W/m^2
The sound intensity follows an inverse square law:
[tex]\frac{I_1}{I_2} = \frac{r_2^2}{r_1^}[/tex]
where we have
[tex]I_1 = 1 W/m^2[/tex] is the sound intensity at r = 5 m from the source
[tex]I_2 = ?[/tex] is the sound intensity at r = 35 m from the source
Solving the equation for I2, we find
[tex]I_2 = I_1 \frac{r_1^2}{r_2^2}=(1 W/m^2)\frac{(5 m)^2}{(35 m)^2}=\frac{1}{49} W/m^2=0.02 W/m^2[/tex]
d) 100.3 dB
The sound intensity level is given by
[tex]L_I = 10 Log_{10} (\frac{I}{I_0}) dB[/tex]
where in this case we have
[tex]I=0.02 W/m^2[/tex]
Substituting this value into the equation, we find
[tex]L_I = 10 Log_{10} (\frac{(0.02 W/m^2)}{(1\cdot 10^{-12} W/m^2)}) dB=100.3 dB[/tex]
The value of the sound intensity for the rock band playing an outdoor concert at different place as,
- a.) The intensity of sound (I) 5 m from the loudspeaker is 1 W/m².
- b.) The power produced by the loud speaker is 314.14 W.
- c.) The intensity (I) 35 m from the speaker is 0.02 W/m².
- d.) The sound intensity level 35 m from the speaker is 100.3 dB
What is the intensity of the sound?
Sound energy is the form of energy, produces when the object is vibrates. Intensity of the sound is the intensity flowing per unit area which is perpendicular to the direction of sound wave travelling in.
The intensity of the sound wave is inversely proportional to the distance of the object and source of sound wave.
- a.) The intensity of sound (I) 5 m from the loudspeaker
The sound intensity can be given as,
[tex]\beta=10\log \dfrac{I}{I_o}[/tex]
As the rock band playing an outdoor concert produces sound at 120 dB and the distance is 5 meters. Therefore, put the value in the above formula as,
[tex]\beta=10\log \dfrac{I}{10^{-12}}\\\beta=1\rm W/m^2[/tex]
Thus, the intensity of sound (I) 5 m from the loudspeaker is 1 W/m².
- b.) The power produced by the loud speaker-
Power producing by the loud speaker can be find out as,
[tex]P=I\times4\pi r^2\\P=I\times4\pi (5)^2\\P=314.15\rm W[/tex]
Thus, the power produced by the loud speaker is 314.14 W.
- c.) The intensity (I) 35 m from the speaker-
The ratio of intensity at 5 meter and intensity at 35 meter is equal to the ratio of square of distance of both. Therefore,
[tex]\dfrac{I_{35}}{I_{5}}=\dfrac{5^2}{35^2}\\\dfrac{I_{35}}{1}=\dfrac{5^2}{35^2}\\I_{35}=0.02\rm W/m^2[/tex]
Thus, the intensity (I) 35 m from the speaker is 0.02 W/m².
- d.) The sound intensity level 35 m from the speaker
The sound intensity can be given as,
[tex]\beta=10\log \dfrac{I}{I_o}[/tex]
To find the sound intensity level 35 m from the speaker, put the value of the intensity (I) 35 m from the speaker in the above formula as,
[tex]\beta=10\log \dfrac{0.02}{10^{-12}}\\\beta=1\rm W/m^2\\\beta=100.3\rm dB[/tex]
Thus, the sound intensity level 35 m from the speaker is 100.3 dB.
Hence, the value of the sound intensity for the rock band playing an outdoor concert at different place as,
- a.) The intensity of sound (I) 5 m from the loudspeaker is 1 W/m².
- b.) The power produced by the loud speaker is 314.14 W.
- c.) The intensity (I) 35 m from the speaker is 0.02 W/m².
- d.) The sound intensity level 35 m from the speaker is 100.3 dB.
Learn more about the sound intensity here;
https://brainly.com/question/14261338
