00:01
In this question, we have a sound source that operates at a frequency of 2 kilohertz and a power rating of 20 watts.
00:12
And it emits a sound uniformly in all directions.
00:16
The speed of sound is given as 340 meters per second and the density of air 1 .2 kilograms per meter cubed.
00:23
And we're asked to find a variety of different quantities.
00:27
So in part a, we're asked to find what is the intensity at a, distance of six meters from the source.
00:33
So there's lots of givens in this question and lots of things to find.
00:38
So i think in particular, we really want to focus on writing down our givens and making sure we know what those are.
00:47
So we've got 2 ,000 hertz or 2 kilohertz for the frequency.
00:53
We know that the power is the 20 watts.
00:58
And we also know the speed.
01:00
So that's going to be 340 meters per second and then finally we know the density as well.
01:09
And in the first part, we're told that it's at a distance of six meters and we would like to know what the intensity is.
01:19
So since we know the power that the source is emitting, we can easily calculate the intensity because that energy is going to spread out over a circle of radius r.
01:32
So the intensity is equal to the power or, yes, the power divided by 4 pi are squared.
01:38
So we're just going to plug everything in.
01:40
So 20 watts divided by 4 pi, 6 meters squared.
01:51
So that's going to give us an intensity of, whoops, 0 .04 watts per meter squared.
02:04
So that's our intensity.
02:06
So that's the final answer for part a.
02:11
Okay, and then we'll go on to part b where we're asked to find the pressure amplitude at the point six meters away from the sound source.
02:24
So now we're trying to find the pressure, which is a small case p.
02:28
I'm going to put p not just so we can be very clear about the different p's going on here.
02:35
And for this, i'm going to just merely use the formula that relates the intensity to the pressure amplitude to the pressure, amplitude.
02:47
So intensity is p not squared divided by two row v.
02:54
So we can go ahead and rearrange that.
02:57
That's going to be p not is the square root of 2 i row v and i'll just go ahead and plug everything in here.
03:08
So we've got the intensity from the previous part.
03:15
The density was given in the question and then also given in the question was the speed of sound.
03:23
So we can go ahead and calculate that.
03:30
Okay, so we get a pressure amplitude of six.
03:42
And if you work out the units, that's going to be in our standard units for pressure.
03:47
So six newtons per meter squared force per area.
03:53
And so that's our final answer for part b as well.
03:58
Okay, and then we go on to part c, where we are asked to find the displacement amplitude...