Human beings can typically detect a difference in sound level of 2.0 dB. What is the ratio of the amplitudes of two sounds whose levels differ by this amount. Is this even solvable?
20 log (A1/A2) = 2
log(A1/A2) = 0.1
A1/A2 = 10^(0.1) = 1.2589
Yes, this question is solvable using the formula for the decibel (dB) scale. The formula for the ratio of amplitudes (A1/A2) corresponding to a given difference in sound level (ΔdB) is:
A1/A2 = 10^(ΔdB / 20)
In this case, the difference in sound level is 2.0 dB. Plugging this value into the formula:
A1/A2 = 10^(2.0 / 20)
= 10^0.1
≈ 1.122
Therefore, the ratio of the amplitudes of the two sounds whose levels differ by 2.0 dB is approximately 1.122.
Yes, this question is solvable. The ratio of the amplitudes of two sounds can be determined by using the formula for sound level in decibels (dB):
Sound Level (dB) = 10 * log10(A^2 / A0^2)
Where:
- A is the amplitude of the sound being measured
- A0 is the reference amplitude (the minimum sound level that humans can detect, usually defined as 0 dB)
In this case, we are given that the difference in sound level is 2.0 dB. Let's call the amplitude of the two sounds A1 and A2, with A1 being the higher amplitude. The difference in sound level can be calculated as follows:
2.0 dB = 10 * log10(A1^2 / A0^2) - 10 * log10(A2^2 / A0^2)
Simplifying the equation:
0.2 = log10(A1^2 / A0^2) - log10(A2^2 / A0^2)
0.2 = log10((A1 / A0)^2 / (A2 / A0)^2)
0.2 = log10((A1 / A0)^2) - log10((A2 / A0)^2)
0.2 = 2 * log10(A1 / A0) - 2 * log10(A2 / A0)
0.1 = log10(A1 / A0) - log10(A2 / A0)
0.1 = log10((A1 / A0) / (A2 / A0))
Taking the antilogarithm of both sides:
10^0.1 = (A1 / A0) / (A2 / A0)
10^0.1 = A1 / A2
Now we can solve for the ratio A1 / A2 by evaluating 10^0.1:
A1 / A2 = 10^0.1
Using a calculator or any computing device capable of calculating exponents, we find:
A1 / A2 ≈ 1.25893
Therefore, the ratio of the amplitudes of two sounds whose levels differ by 2.0 dB is approximately 1.25893.