The speed of sound in air (in m / s ) depends on temperature according to the approximate expres

The speed of sound in air (in m / s ) depends on temperature...

Key Concepts

Speed of Sound and Temperature Dependency

This concept explains that the speed of sound in a medium like air increases with temperature. Higher temperatures result in faster movement of air molecules, which in turn facilitates quicker propagation of sound waves. This dependency is often expressed with formulas that linearly approximate the variation in speed with temperature within specific ranges.

Atmospheric Temperature Gradient

An atmospheric temperature gradient refers to the change in temperature with altitude. In many atmospheric layers, temperature decreases (or sometimes increases) predictably with height. This gradient affects physical properties including the speed of sound, meaning that as the temperature changes with altitude, so does the speed at which sound travels through the air.

Integration in a Variable Medium

When the speed at which a wave propagates varies continuously over space due to changes in parameters like temperature, determining travel time requires integration. By dividing the path into infinitesimal segments where the speed is approximately constant, and then integrating the reciprocal of the speed over the total distance, one can calculate the overall time taken for the sound wave to travel through the medium.

Comparison of Constant and Variable Medium Propagation

Analyzing differences in travel times between a medium that is uniform (constant temperature) and one that has a gradient (variable temperature) is crucial. This comparison helps in understanding how variations in physical conditions, such as temperature, can affect the propagation of waves, leading to differences in arrival times despite the same spatial separation.

Transcript

00:01 Hello everyone.

00:02 So for this problem, we are given a function that tells us how the speed of sound relates to temperature.

00:15 So we have the speed of sound.

00:19 So the velocity of the speed of sound is equal to 331 .5 meters per second plus 0 .607 times temperature in celsius.

00:35 And this temperature here is changing with height.

00:41 So in our diagram, we have a plane 900 meters high, and every 500 meters, every rise in 150 meters gives us a increase of negative 1 degrees celsius, or a decrease in celsius.

01:03 So we're losing temperature as we go up in space or in height towards the plane.

01:13 So we are asked to find the speed of sound at which the sound of the plane will reach the ground.

01:23 Now, since this is changing with time, we have to write an expression for temperature.

01:29 So since temperature is changing with height, the temperature on the ground is 30 degrees.

01:37 So let's start with that.

01:39 So t is equal.

01:42 Let's write t c just to make sure that we label that we are on celsius, that the temperature is on celsius.

01:49 So we have 30 degrees in celsius.

01:54 That is our initial condition.

01:58 And we're going to put minus because we are subtracting 1 degree celsius with height.

02:09 So we're looking in the rest of the ground looking up towards the plane.

02:16 So in a sense, we're looking at the problem as if the person on the ground is making the noise and that's going towards the plane.

02:26 It will work either way.

02:27 But if you were to look at from the plane going down, it would be a plus sign.

02:38 But that'll be for our situation, we'll be looking.

02:42 Looking at how temperature is decreasing with height.

02:46 So let's write a negative sign or a negative subtraction.

02:55 So 30 degrees minus y over 150 meters, because for each degree in celsius, so i'm going to write each one degrees celsius there.

03:13 So i left out the one because this is a variable.

03:17 We don't know how high are we going to go here.

03:25 And in this case, we are going up to 900 meters.

03:30 So that will be 6 for this variable.

03:35 Why? but we want to make a equation, a general equation that will work for all heights for temperature.

03:44 And we don't actually want to just plug in 6 here.

03:47 So you get 6 from dividing 900.

03:50 And 150.

03:53 But what we're going to do instead is actually get this.

03:58 Since we have a function of t, we can actually plug that into the expression we were given in the beginning.

04:06 So what are we going to, so we're going to write v is now equal to 331 .5 plus 0 .607.

04:19 And then we're going to insert all this stuff here that we got for tc so parentheses 30 degrees in cc minus y i'm going to leave out the one because y multiplied by one is just y so we're put y over 150 meters and close the parentheses so now we have an expression for v that includes why now one issue that we have is that x is also changing so x is going from 150 to 900 oh this is actually 9 000 for this problem so in this problem we're given 9 ,000 meters here so we're going from 0 to 950 meters um so before when i said that it was 6 uh it won't we'll take 60 steps.

05:34 So 60 times 150 will give you 9 ,000.

05:39 But that doesn't matter because we made a variable here, why.

05:44 So that will work if we put, if we were going to 900 meters, we would just put, we would just change the variable and which we're going to integrate.

05:53 So we're going to, we are going to integrate this y.

05:57 But first, let's simplify this a bit by doing distributive properties.

06:01 So let's distribute this value here.

06:08 And once we do that, we'll get v is equal to 331 .5 plus.

06:19 So this, multiplying this and this will give you 18 .21, 18 .21.

06:28 And that'll be minus 0 .004 .05 meters.

06:35 So we can add these two values and get 349.

06:46 349 .71.

06:51 Let me make that 4 looks look nicer.

06:55 So 349 .71 minus this value here.

07:03 And this value contains a y as well.

07:09 Also be y.

07:13 And so this will have units of m minus 1, but we'll leave out the units because we know we'll end up getting, by multiplying these values, we'll get meters per second.

07:27 So this is meters per second, meters per second.

07:30 And once we include our y value, we'll get meters per second as well.

07:37 So 0 .004 .05.

07:41 Y.

07:42 And so now we have this expression here for our velocity of the speed of sound.

07:51 Now we want to, so this is a function for the speed of sound.

07:56 The distance traveled is 9 ,000 meters.

08:00 Let's use the distance formula to find the time traveled.

08:05 So to find the time traveled, so we're going to look at x equals vt.

08:10 But in our situation, you know, we would, since we have a changing distance and we have a changing velocity because our velocity depends on x.

08:28 Because since the velocity incorporated the temperature, which is changing with height, so we have a velocity that changes with x and x that changes with time.

08:39 So we have every value changing here.

08:42 So we have a change in x is equal to a change in v, which is multiplied by a change in time.

08:52 So, and we know that this is also equal to, you know, dx is equal to dv, dt.

09:02 So we can rearrange our equation here and get dt by itself.

09:09 Dt is equal to dx over dv.

09:18 With this, let's take the integral of this so that we can look at the, so we want to take the integral to look at this area here between the ground and the plane, and that will be the distance covered.

09:36 And also, and that will give us the correct time it would take for the sound to reach the plane or for the sound of the plane to reach the ground or vice versa.

09:49 So integrating this, we have the integral of dt is equal to the integral of dx over dv.

10:00 And we can plug in this...

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