Show that, if one sound is 10 decibels louder than a second sound, than the first sound is 10 times as intense as the second sound.
That is how decibels are defined. A ten decibel difference is a factor of ten in intesity (power/area)
dB2 - dB1 = 10 log(10) I2/I1
To show that, we need to understand the relationship between sound intensity and decibels. Sound intensity is a measure of the energy carried by sound waves per unit of area. It is typically measured in watts per square meter (W/m^2).
Decibels (dB) are a logarithmic unit used to express the ratio between two sound intensities. The formula to convert sound intensity in watts per square meter (I) to decibels (dB) is as follows:
dB = 10 * log10(I/I0)
Where I0 is the reference intensity, which is typically set at the threshold of human hearing, I0 = 10^(-12) W/m^2.
Now, let's consider the two sounds, where the first sound is 10 decibels louder than the second sound. Let's denote the sound intensities as I1 and I2 respectively.
According to the formula, we can express the decibel difference between the two sounds as follows:
dB = 10 * log10(I1/I0) - 10 * log10(I2/I0)
Since the first sound is 10 decibels louder than the second sound, we can write:
10 = 10 * log10(I1/I0) - 10 * log10(I2/I0)
Now, let's simplify this equation:
1 = log10(I1/I0) - log10(I2/I0)
According to the logarithmic properties, we can rewrite this equation as:
1 = log10[(I1/I0)/(I2/I0)]
Applying the inverse logarithm (common logarithm base 10) to both sides to eliminate the log10, we have:
10^1 = (I1/I0)/(I2/I0)
10 = I1/I2
Therefore, I1 = 10 * I2
This shows that the first sound is 10 times as intense as the second sound.