An amplified guitar has a sound intensity level that is 11 dB greater than the same
unamplified sound. What is the ratio of the amplified intensity to the unamplified intensity?
The sound intensity level is measured in decibels (dB) and is given by the formula:
L = 10 * log10(I / I0)
where L is the sound intensity level in decibels, I is the sound intensity, and I0 is the reference intensity.
In this case, we are given that the amplified guitar has a sound intensity level that is 11 dB greater than the unamplified sound. Let's denote the sound intensity of the unamplified guitar as I_unamp and the sound intensity of the amplified guitar as I_amp.
Therefore, we can write the following equation:
L_amp = L_unamp + 11
Using the formula for sound intensity level, we can rewrite this equation as:
10 * log10(I_amp / I0) = 10 * log10(I_unamp / I0) + 11
Canceling out the common terms, we have:
log10(I_amp / I0) = log10(I_unamp / I0) + 1.1
By using the properties of logarithms, we can simplify the equation further:
I_amp / I0 = 10 * (I_unamp / I0)
Simplifying the equation, we get:
I_amp = 10 * I_unamp
So, the ratio of the amplified intensity (I_amp) to the unamplified intensity (I_unamp) is 10:1.
In other words, the amplified intensity is 10 times greater than the unamplified intensity.
To find the ratio of the amplified intensity to the unamplified intensity, we need to first understand the concept of decibels (dB) and how they relate to intensity levels.
The decibel scale is a logarithmic scale that compares the intensity of a sound to a reference level. In this case, the reference level is the unamplified sound.
The sound intensity level (SIL) in decibels is given by the formula:
SIL = 10 * log10(I / I₀)
where I represents the actual sound intensity and I₀ is the reference intensity.
Given that the amplified guitar has a sound intensity level that is 11 dB greater than the unamplified sound, we can write the equation as:
SIL amplified = SIL unamplified + 11 dB
Now, let's solve for the ratio of the amplified intensity (I amplified) to the unamplified intensity (I unamplified).
Using the formula for SIL, we have:
SIL amplified = 10 * log10(I amplified / I₀)
SIL unamplified = 10 * log10(I unamplified / I₀)
Substituting the equations for the sound intensity levels into our equation, we get:
10 * log10(I amplified / I₀) = 10 * log10(I unamplified / I₀) + 11 dB
Cancel out the common factor of 10:
log10(I amplified / I₀) = log10(I unamplified / I₀) + 1.1
Apply the logarithmic property:
I amplified / I₀ = 10^(log10(I unamplified / I₀) + 1.1)
Simplify the right side expression:
I amplified / I₀ = 10^log10(I unamplified / I₀) * 10^1.1
I amplified / I₀ = (I unamplified / I₀) * 10^1.1
Finally, cancel out the I₀ terms:
I amplified = I unamplified * 10^1.1
Therefore, the ratio of the amplified intensity to the unamplified intensity is 10^1.1.