00:01
In this problem, we're calculating the sound intensity in decibels for a variety of sounds, and we're using this formula here, which relates to the level of sound density with the intensity of the sound and the intensity of the quietest audible sound.
00:20
And we have to go back into section 3 .3 to find that formula.
00:25
Okay, so for part a, we have a hearing threshold, which has a sound intensity level of i equals 10 to the next number.
00:32
Negative 12th.
00:34
So to find the level of sound in decibels, which we're calling beta, we're going to substitute the value of i into the equation, and we're going to substitute 10 to the negative 12th in for i -0.
00:50
And now we can simplify because 10 to the negative 12th over 10 to the negative 12th is just one.
00:57
And keep simplifying.
00:59
The log of 1 is 0, so we have 10 times 0, so it's just 0.
01:04
So it's just zero decibels.
01:06
Zero decibels is the hearing threshold.
01:12
That's kind of like our starting place.
01:16
From there we get gradually louder sounds.
01:19
The next one is rustling leaves and the intensity is 10 to the negative 11th so we can substitute that in to our formula and we're still going to substitute in every time 10 to the negative 12th for the barely audible sound as i not.
01:36
Simplify and what we want to do here is subtract the power on 10, and that gives us 10 to the first.
01:45
10 to the first is just 10.
01:47
And the log of 10 is 1, so we have 10 times 1, which is 10.
01:52
So the sound intensity is 10 decibels for rustling leaves.
02:03
The next one is conversation, where the sound intensity is 10 to the negative 6th.
02:13
So we substitute that in over 10 to the negative 1th...