1. A sound has an intensity of 5.92 „e 1025W/m2.What is the loudness of the sound in decibels? Use .
2. Suppose you decrease the intensity of a sound by 45%. By how many decibels would the loudness be decreased?
To calculate the loudness of the sound in decibels (dB), you need to use the formula:
dB = 10 * log10(I/Io)
Where I is the intensity of the sound and Io is the reference intensity, which is typically set at 1 * 10^(-12) W/m^2.
Let's solve the first question:
1. The intensity of the sound is given as 5.92 * 10^25 W/m^2.
2. Convert the given intensity to the reference intensity by dividing it:
Io = 1 * 10^(-12) W/m^2
Intensity ratio = I/Io = (5.92 * 10^25 W/m^2) / (1 * 10^(-12) W/m^2) = 5.92 * 10^37
3. Calculate the loudness in decibels using the formula:
dB = 10 * log10(I/Io) = 10 * log10(5.92 * 10^37) ≈ 369.86 dB
Therefore, the loudness of the sound is approximately 369.86 decibels.
Now let's move on to the second question:
To calculate the change in loudness in decibels, you need to use the formula:
ΔdB = 10 * log10(I2/I1)
Where I1 is the initial intensity of the sound and I2 is the final intensity after the decrease.
1. The intensity of the sound is decreased by 45%, which means it is reduced to 55% of its original value.
2. Convert the percentage to a decimal by dividing it by 100: 45% = 0.45.
3. Calculate the final intensity after the decrease:
I2 = (1 - 0.45) * I1
4. Calculate the change in loudness in decibels using the formula:
ΔdB = 10 * log10(I2/I1)
Plug in the values:
ΔdB = 10 * log10((1 - 0.45) * I1 / I1) = 10 * log10(0.55) ≈ -1.8 dB
Therefore, the loudness would decrease by approximately 1.8 decibels.