The decibel level of a sound is given by dB=10log(EE0), where E0=10−12wattsm2 and E is the intensity of the sound.
(a) Find the intensity of a 80-decibel sound.
E= watts per square meter.
(b) How many times more intense is the sound from (a) than a 50-decibel sound?
A sound at 80 decibels is ____ times more intense than a sound at 50 decibels.
(c) The loudness (perceived volume by the human ear) of a sound of d decibels doubles each time the intensity of a sound increases by a factor of ten. To the human ear, how many times louder does a 80-decibel sound seem than a 50-decibel sound?
A sound at 80 decibels sounds______ times louder than a sound at 50 decibels.
To find the intensity of a sound measured in watts per square meter (W/m²) given its decibel level, we can use the formula dB = 10log(E/E₀), where dB is the decibel level, E is the intensity of the sound, and E₀ is a reference intensity of 10^(-12) W/m².
(a) To find the intensity of an 80-decibel sound, we need to rearrange the formula and solve for E:
dB = 10log(E/E₀)
80 = 10log(E/E₀)
Divide both sides of the equation by 10:
8 = log(E/E₀)
Next, we need to rewrite the logarithmic equation in exponential form. In exponential form, log base 10 is equivalent to 10 raised to the power of the logarithm. So, rewrite the equation as:
10^8 = (E/E₀)
Since E₀ is defined as 10^(-12) W/m², substitute this value in:
10^8 = (E/10^(-12))
Solve for E:
E = 10^8 * 10^(-12) = 10^(-4) W/m²
Therefore, the intensity of an 80-decibel sound is 10^(-4) W/m².
(b) To find how many times more intense the sound from part (a) is than a 50-decibel sound, we can use the formula:
Intensity ratio = 10^( (dB₂ - dB₁)/10 )
Let's substitute dB₁ = 50 and dB₂ = 80 into the formula:
Intensity ratio = 10^( (80-50)/10 )
= 10^3
Therefore, a sound at 80 decibels is 1000 times more intense than a sound at 50 decibels.
(c) To find how many times louder a sound at 80 decibels seems than a sound at 50 decibels, we can use the fact that the perceived loudness doubles each time the intensity increases by a factor of ten. Since we found that the intensity ratio is 10^3, the loudness ratio would be 2^3 = 8.
Therefore, a sound at 80 decibels sounds 8 times louder than a sound at 50 decibels.
To solve these problems, we will use the given equation: dB = 10 * log(E/E0), where dB is the decibel level, E is the intensity of the sound, and E0 is the reference intensity.
(a) To find the intensity of a 80-decibel sound, we can rearrange the equation:
80 = 10 * log(E/E0)
Divide both sides of the equation by 10:
8 = log(E/E0)
Exponentiate both sides of the equation with base 10:
10^8 = E/E0
Multiply both sides of the equation by E0:
E = 10^8 * E0
E = 10^8 * 10^(-12) watts per square meter
E = 10^(-4) watts per square meter
Therefore, the intensity of an 80-decibel sound is E = 10^(-4) watts per square meter.
(b) To find how many times more intense the 80-decibel sound is compared to a 50-decibel sound, we can use the equation:
dB1 - dB2 = 10 * log(E1/E2)
Substituting the values:
80 - 50 = 10 * log(E/10^(-4))
Simplify:
30 = 10 * log(E/10^(-4))
Divide by 10:
3 = log(E/10^(-4))
Exponentiate with base 10:
10^3 = E/10^(-4)
Multiply both sides by 10^(-4):
10^3 * 10^(-4) = E
Simplify:
10^(-1) = E
Therefore, the sound at 80 decibels is 10^(-1) = 0.1 times more intense than a sound at 50 decibels.
(c) To find how many times louder a sound at 80 decibels seems compared to a sound at 50 decibels, we compare their intensities using the equation:
loudness difference = log(E1/E2)
Substituting the given values:
loudness difference = log(E/10^(-4))
For doubling the perceived loudness, we need a factor of 10 increase in intensity:
10 = E/10^(-4)
Multiply both sides by 10^(-4):
10 * 10^(-4) = E
Simplify:
10^(-3) = E
Therefore, a sound at 80 decibels seems 10^(-3) = 0.001 times louder than a sound at 50 decibels.
Your E's are confusing.
And what about that E0 = 10-12wattsm2 ??