Formula given:
The decibel (dB) scale for measuring loudness, d, is given by the formula
d = 10 log base ten(I X 12 to the power of 12) where I is the intensity of sound in watts per square metre.
a) Find the number of decibels of sound produced by a jet engine at a distance of 50 metres if the intensity is 10 watts per square metre.
b) Find the intensity of sound if the sound level of a pneumatic drill 10 metres away is 90 decibels.
c) Find how the value of d changes if the intensity is doubled. Give your answer to the nearest decibel.
d) Find how the value of d changes if the intensity is 10 times as great.
e) By what factor does the intensity of sound have to be multiplied in order to add 20 decibels to the sound level?
To solve these problems, we will use the given formula for the decibel scale:
d = 10 * log10(I * 12^12)
where d represents the number of decibels and I represents the intensity of sound in watts per square meter.
a) To find the number of decibels of sound produced by a jet engine at a distance of 50 meters with an intensity of 10 watts per square meter, we substitute the given values into the formula:
d = 10 * log10(10 * 12^12)
Using a scientific calculator, calculate the value inside the logarithm:
10 * log10(10 * 12^12) ≈ 10 * log10(1.44 * 10^13) ≈ 10 * log10(1.44) + 10 * log10(10^13) ≈ 10 * 0.16 + 10 * 13 ≈ 1.6 + 130 ≈ 131.6 decibels.
Therefore, the number of decibels of sound produced by the jet engine at a distance of 50 meters with an intensity of 10 watts per square meter is approximately 131.6 decibels.
b) To find the intensity of sound if the sound level of a pneumatic drill 10 meters away is 90 decibels, we need to rearrange the formula and solve for I:
d = 10 * log10(I * 12^12)
Rearranging the formula gives us:
I = 10^(d/10) / (12^12)
Substituting the given values:
I = 10^(90/10) / (12^12)
Using a scientific calculator:
I ≈ 10^9 / (12^12) ≈ 47.4 * 10^(-9) watts per square meter.
Therefore, the intensity of sound if the sound level of a pneumatic drill 10 meters away is 90 decibels is approximately 47.4 * 10^(-9) watts per square meter.
c) To find how the value of d changes if the intensity is doubled, we calculate the difference between the values of d for the original intensity and the doubled intensity:
Let's say the original intensity is denoted as I1, and the doubled intensity is denoted as I2.
For I2 = 2 * I1, we can calculate:
d2 - d1 = 10 * log10(I2 * 12^12) - 10 * log10(I1 * 12^12)
= 10 * log10((2 * I1) * 12^12) - 10 * log10(I1 * 12^12)
= 10 * log10(2 * I1 / I1)
= 10 * log10(2)
≈ 10 * 0.301
≈ 3.01 decibels
Therefore, the value of d changes by approximately 3.01 decibels if the intensity is doubled.
d) To find how the value of d changes if the intensity is 10 times as great, we use a similar approach as in part c). Let's say the original intensity is denoted as I1 and the new intensity is denoted as I2.
For I2 = 10 * I1, we can calculate:
d2 - d1 = 10 * log10(I2 * 12^12) - 10 * log10(I1 * 12^12)
= 10 * log10((10 * I1) * 12^12) - 10 * log10(I1 * 12^12)
= 10 * log10(10 * I1 / I1)
= 10 * log10(10)
≈ 10 * 1
≈ 10 decibels
Therefore, the value of d changes by approximately 10 decibels if the intensity is 10 times as great.
e) To find by what factor the intensity of sound has to be multiplied to add 20 decibels to the sound level, we need to solve for the factor k in the equation:
d2 - d1 = 20
where d1 is the initial sound level.
Using the formula for decibels:
10 * log10(I2 * 12^12) - 10 * log10(I1 * 12^12) = 20
Simplifying:
log10((I2 * 12^12) / (I1 * 12^12)) = 2
Applying logarithmic properties:
(I2 * 12^12) / (I1 * 12^12) = 10^2
Simplifying:
I2 / I1 = 10^2
Therefore, the intensity of sound needs to be multiplied by a factor of 100 (10^2) to add 20 decibels to the sound level.