For two sounds whose sound levels differ by 69 dB, find the ratios (greater value / smaller value) of the following values.
(a) the intensities
Intensity Final/Intensity Initial =
log^-1 (69 / 10) = 7.9x10^6
(b) the pressure amplitudes
(c) the particle displacement amplitudes
I am not sure how to find the ratios for b and c. I reviewed the equations relating to pressure and displacement amplitudes but I cannot come up with anything.
Both the pressure-amplitude and displacement ratios are the square root of the intensity (power) ratio.
Ah! thanks so much :)
You're welcome! I'm glad I could help.
To find the ratios for the pressure amplitudes and particle displacement amplitudes, you can use the relationship between intensity, pressure, and particle displacement. The intensity of a sound wave is directly proportional to the square of both the pressure amplitude and the particle displacement amplitude.
So, if the ratio of intensities is 7.9x10^6 (as you calculated in part (a)), then the ratio of pressure amplitudes and particle displacement amplitudes can be found by taking the square root of the intensity ratio.
Let's calculate:
(b) Ratio of Pressure Amplitudes:
To find the ratio of pressure amplitudes, take the square root of the intensity ratio:
Ratio of pressure amplitude = √(Intensity Final / Intensity Initial)
= √(7.9x10^6)
≈ 2804.7
So, the ratio of pressure amplitudes is approximately 2804.7.
(c) Ratio of Particle Displacement Amplitudes:
Similarly, to find the ratio of particle displacement amplitudes, take the square root of the intensity ratio:
Ratio of particle displacement amplitude = √(Intensity Final / Intensity Initial)
= √(7.9x10^6)
≈ 2804.7
So, the ratio of particle displacement amplitudes is also approximately 2804.7.
Therefore, the ratios for part (b) and (c) are both approximately 2804.7.
If you have any more questions, feel free to ask!