A radio set at a sound level of 20 dB is suddenly increased to 80 dB. how many more time greater is the pressure amplitude at 80 dB? I do not know what to do please help
20Log(Vo/Vi) = 20db,
Log(Vo/Vi) = 1,
Vo/Vi = 10^1 = 10. = Voltage ratio.
20Log(Vo/Vi) = 80db,
Log(Vo/Vi) = 4,
Vo/Vi = 10^4 = 10,000.
Ratio = 10^4 / 10^1 = 10^3 = 1,000.
Therefore, the signal level at 80db is
1,000 times the signal level at 20db.
To determine how many times greater the pressure amplitude is at 80 dB compared to 20 dB, we need to understand the relationship between decibels (dB) and the pressure amplitude.
The decibel scale is logarithmic, which means that each 10 dB increase represents a tenfold increase in the sound intensity or pressure amplitude. This relationship is expressed by the formula:
dB = 10 * log10(P / Pref),
where P is the sound pressure and Pref is the reference pressure (typically set as 20 micropascals). In this case, we can assume that the reference pressure remains constant.
Given that the sound level increases from 20 dB to 80 dB, we can calculate the change in pressure amplitude.
1. Convert the initial and final dB values into pressure amplitudes using the formula:
P1 = Pref * 10^(dB1 / 10)
P2 = Pref * 10^(dB2 / 10)
where P1 and P2 are the pressure amplitudes at 20 dB and 80 dB, respectively.
2. Calculate the ratio of the pressure amplitudes at 80 dB to 20 dB:
Ratio = P2 / P1
This ratio will represent how many times greater the pressure amplitude is at 80 dB compared to 20 dB.
Let's calculate it step by step:
1. Calculate the initial and final pressure amplitudes:
P1 = 20 micropascals * 10^(20 / 10) = 20 micropascals
P2 = 20 micropascals * 10^(80 / 10) = 20 micropascals * 10^8
2. Calculate the ratio:
Ratio = P2 / P1 = (20 micropascals * 10^8) / 20 micropascals = 10^8
Therefore, the pressure amplitude at 80 dB is 10^8 times greater than at 20 dB.
In summary, the pressure amplitude at 80 dB is 10^8 (100,000,000) times greater than at 20 dB.