Little help needed with a simple physics assingment

[Lajka]

Hi,
this is a rather simple high-school assignment a friend asked me to help him, but I got kinda confused, and I'm sure someone will clarify it for me real quick.



Homework Statement
A car is moving away from a rock with the speed v and with the angle \alpha from the rock. At the moment when the distance between the car and the rock is l, the car emmits a short sound wave. The speed of sound in the air is u.
What's the distance crossed till the driver in the car hears an echo?

The attempt at a solution

So the picture is something like this

At moment t_{1} = \frac{l}{v \cdot sin(\alpha)}, the car will emit the sound wave with speed u

It will take \Delta t=\frac{l}{u} more time for the sound wave to reach the rock, and by then the car will have crossed s = v \cdot t_{2}
where t_{2} = t_{1} + \Delta t

But I'm stuck here. How exactly does a sound wave echo? If it just bounces back, it will miss the car of course, so does it perhaps reflect in some radial form? If it does, how should I calculate the point there the sound and the trajectory of the car intersect?
Of course, I think we can assume that v << u, otherwise there wouldn't be any point to any of this. Also disregard that the magnitudes of the arrows (like velocity, for instance) are different in the different pictures, I did it to make the pictures more clearer.

Thanks in advance.

 

[Quinzio]

In that problem you should split all the v into parallel and perpendicular to the rock line, that is horizontal and vertical.

The sound spreads from the car in a radial way, that is you should imagine the sound as a lot of particles departing from the car in all direction.

The rock reflects wave such as {v_x}' = v_x, \ {v_y}' = -v_y

 

[Lajka]

Hah, well that makes things only worse :D

How the hell am I supposed to pick which of all these possible rays is going to hit the car after bouncing? I suspect I am supposed to compare parallel and perpendicular components of v with something, but I'm not sure with what.
I'm even more stuck now, but thanks nonetheless :D

 

[Lajka]

Okay, I have an idea, so if someone could confirm it or refute it, it would be great

Basically, I'm looking for an angle \gamma such that parallel components are equal.
v_{x} = v \cdot cos(\alpha) = u_{x} = u \cdot cos(\gamma)
Under that condition, it will take \Delta t = \frac{l}{u \cdot sin(\gamma)} time units for that particular sound wave to reach the rock. After that, I'm just comparing perpendicular components
d + v \cdot sin(\alpha) \cdot t_{x} = u \cdot sin(\gamma) \cdot t_{x}, where d = v \cdot sin(\alpha) \cdot t_{2} ( where t_{2} = t_{1} + \Delta t)

After I find t_{x}, I just calculate the distance as s = v \cdot (t_{2} + t_{x}).

I realize my notation sucks and that there is perhaps some redundancy in the expressions above, but I did all that for a few minutes, so you'll forgive me.
What do you think guys?

 

[I like Serena]

Hi Lajka!

Your method looks good, but I'm having a bit of trouble understanding what you mean with your symbols and how they match your drawing.
I think your equations are not entirely right, but I'm not sure yet.As for how audio reflects against a wall.
Audio propagates as a circular wave.
When it hits the wall, at each point a new circular wave is generated.
Those generated waves superpose into a circular wave with a center that is symmetrically on the other side of the wall.

Here are a couple of example images:

[URL]http://www.phy.ntnu.edu.tw/ntnujava/snapshotejs/wave_reflection_2_20090117112315.gif[/URL]

[PLAIN]http://www.lightandmatter.com/html_books/0sn/ch06/figs/circular-reflection-upright.jpg

Cheers!

 

[nayfie]

I really need to know how you're doing those amazing diagrams Lajka!

 

[Quinzio]

You are on the right track.
You just need to lay down things clearly.
The sound goes to the rock the bounce back up to level d, so it makes twice d.
Then reaches the car.

 

[Lajka]

Your method looks good, but I'm having a bit of trouble understanding what you mean with your symbols and how they match your drawing.
I think your equations are not entirely right, but I'm not sure yet.

Hm, yeah, I guess I make notations in mysterious ways, not a good habit of mine.

Diagram can be identified as an x-y plane, and t_1 and t_2, for example, are parameters on those pictures, they're don't represent points or anything like that, more like comments, as in "t_1 is a time moment where the guy in the car emitted a sound impulse so I wrote t_1 on a diagram there".
Dashed curves are trajectories (with the exception of l and d, those are just distances, I probably should've used different types of lines for them).

If you do find an error, please let me know, and thanks for those pictures and an explanation!

I really need to know how you're doing those amazing diagrams Lajka!

In case you weren't being sarcastic :D I use Adobe Illustrator. It's pretty self-explanatory for simple diagrams as these. There is also a program called Inkscape, which is a free alternative vector-based graphics editor. I'm sure you can do diagrams like these in Inkskape pretty easily, too.

You are on the right track.
You just need to lay down things clearly.
The sound goes to the rock the bounce back up to level d, so it makes twice d.
Then reaches the car.

Goddammit, I really can't see it the way you do. :/
The way I see it, from the moment t_2 and onward, it's basically a one-dimensional problem. I have a 'point 1' with speed u \cdot sin \, \gamma and a another 'point 2' with speed v \cdot sin \, \alpha. And there is a distance d between them. So, assuming that v < u, I'm asking myself when will the 'point 1' catch-up with the 'point 2', a.k.a. in what time moment t_{x} (starting from t_2) will these points have the same amount of traveled path:
d + v \cdot sin(\alpha) \cdot t_{x} = u \cdot sin(\gamma) \cdot t_{x}

P.S. I just realized I needlessly used the distance d and the time variable t_{x} and thus made equations and notation more complicated. If I instead of t_{x} write t - t_2 and use that in the equation
d + v \cdot sin(\alpha) \cdot t_{x} = u \cdot sin(\gamma) \cdot t_{x}
along with the formula for the distance d
d = v \cdot sin(\alpha) \cdot t_{2}
I get
v \cdot sin(\alpha) \cdot t = u \cdot sin(\gamma) \cdot (t - t_{2})

which is somewhat simpler and more elegant, I'd say.

 

[Quinzio]

Ok, there's no reason to get upset.
When I said you should lay down things clearly I simply meant you should obtain the final answer of the problem, since you only know l,\ v,\ u\ , \alpha, and they ask you the distance traveled by the car.
The answer is:
\frac{2\ l\ v}{\sqrt{u^2\ -\ v^2\ cos^2 \alpha } -vsin\ \alpha}

 

[I like Serena]

Actually, I was wondering what you meant with d and with t_x.

As you have drawn d it seems to be the same as l, but that can not have been your intent.
And you introduced t_x, but you did not specify what it is.

In your drawing I would expect another line from the rock to the car at the point where the echoed sound reaches the car.
The angle of this line would be the same as the angle of the line between t_1 and t_2.
It seems logical to say that this would be at a time t_3.I haven't checked the formula of Quinzio, but I suppose you'd be interested in how to solve this problem yourself. Or don't you?Btw, I have to say that this doesn't seem like a problem that would be asked in high school!
Unless you're supposed to simplify the problem by making approximations.Either way, it is a good idea to simplify the problem with approximations if only to verify that you're on the right track.

In its simplest form v would be insignificantly small relative to u and you would have:
\Delta t_{total} = \frac {2 l} u
That is,
s = v \Delta t_{total} = \frac {2 l v} u

 

[Lajka]

Ok, there's no reason to get upset.

I'm not upset, sorry if I made you think I was :D
But I'm getting to be a little sad now because I can't get such a nice solution as you did.

From the last equation of my last post, I get
t = \frac{u \cdot sin \, \gamma \, \cdot \, t_2}{u \cdot sin \, \gamma - v \cdot sin \, \alpha}

Since I have
t_2 = \frac{l}{v \cdot sin \, \alpha} + \frac{l}{u \cdot sin \, \gamma}
= l \, \frac{u \cdot sin \, \gamma + v \cdot sin \, \alpha}{v \cdot sin \, \alpha \cdot u \cdot sin \, \gamma}

I combine these two and get

t = \frac{u \cdot sin \, \gamma \cdot l \, \frac{u \cdot sin \, \gamma + v \cdot sin \, \alpha}{v \cdot sin \, \alpha \cdot u \cdot sin \, \gamma}}{u \cdot sin \, \gamma - v \cdot sin \, \alpha} = l \, \frac{\frac{u \cdot sin \, \gamma + v \cdot sin \, \alpha}{v \cdot sin \, \alpha}}{u \cdot sin \, \gamma - v \cdot sin \, \alpha}

and, at best, I can write now

s=v \cdot t = vl\frac{\sqrt{u^2\ -\ v^2\ cos^2 \alpha } + v \cdot sin \, \alpha}{v \cdot sin \, \alpha(\sqrt{u^2\ -\ v^2\ cos^2 \alpha } - v \cdot sin \, \alpha)}
as the final answer.

So, where am I mistaken? If you can see the error, of course, I'm sure it must be tiresome to go through all of it.

Actually, I was wondering what you meant with d and with tx.

As you have drawn d it seems to be the same as l, but that can not have been your intent.
And you introduced tx, but you did not specify what it is.

In your drawing I would expect another line from the rock to the car at the point where the echoed sound reaches the car.
The angle of this line would be the same as the angle of the line between t1 and t2.
It seems logical to say that this would be at a time t3.

By d I meant the normal distance from the rock to the car in t_2, just like l is a distance from the rock to the car in t_1. By t_{x} I meant t - t_2, where t corresponds to your t_{total}
But then I realized I didn't really need d nor t_{x} for the formulas, so they were actually unnecessary.

And you're absolutely right, it would've been most sensible to draw the intersection of the trajectory of the car and the wave and mark that t_3, thus complete the whole picture. If I remember correctly, I shut down the Illustrator and then realized this, but I was too lazy to turn it on and draw everything again, because I didn't save the picture prior to shutting down the Illustrator the first time, because I'm an idiot. :D

And yeah, of course I'm interested, his formula is a thing of beauty, would love to get it myself, but I ain't doing very well for now.

P.S. I don't know where my friend got this assignment, he just asked me for help. I just assumed that it was of high-school calibre, because it sounded so innocent in the beginning.
P.P.S. Nice touch with approximations, that's some engineering thinking right there. :D

 

[Quinzio]

t_2 = \frac{l}{v \cdot sin \, \alpha} + \frac{l}{u \cdot sin \, \gamma}

I don't understand what you are getting here.
You sum a time which belongs to the car with a time which belongs to the sound wave.
The car doesn't travel the distance l.

Try this:
call K the unknow distance, then calculate the time the car takes to travel K (simple), and the time taken by the sound (more difficult).
Then you equal the two times.
The rest is algebra manipulation.

 

[I like Serena]

So, where am I mistaken? If you can see the error, of course, I'm sure it must be tiresome to go through all of it.

Well, I'll postpone going through your calculations.

I prefer to first have a proper drawing (and you make very nice ones! ).
I'm sorry seeing you lost your previous efforts, so I made one of my own instead.

Then to define the proper symbols to use, which are shown in the drawing.
It doesn't matter if there are more than necessary.

Then to define the proper equations, preferably numbered.

And finally solve those equations for the distance traveled which the problem asks for.

I've been so free to use slightly different symbols.

Do you like my drawing? :shy:

 

[Lajka]

Oh, I love your drawing Serena! Very clean indeed, thanks very much for that!

It helped me to understand the issue underlying here, I believe. The drawing of s was the key, because I perceived the wanted s as the distance from the very beginning till the moment the car and the wave intersect (the distance the car made for t_1 + \Delta t), and Quinzio, for instance, perceived it as the distance from the moment the sound was made till the moment the car and the wave intersect (the distance the car made for \Delta t). Dunno what is the desired distance out of these two in the assignment, but that's secondary to me right now.

So the system of the equations is the following (using your picture):
v_{x} = v \cdot cos(\alpha) = u_{x} = u \cdot cos(\gamma) (1)
t_1 = \frac{l}{v \cdot sin \, \alpha} (2)
\Delta t = \frac{l}{u \cdot sin \, \gamma} + \frac{l + \Delta y}{u \cdot sin \, \gamma} = \frac{2l + \Delta y}{u \cdot sin \, \gamma} (3)
\Delta y = v \cdot sin \, \alpha \Delta t) (4)

Inserting (4) into (3) we get
\Delta t = \frac{2l + v \cdot sin \, \alpha \Delta t}{u \cdot sin \, \gamma}
=> 2l + v \cdot sin \, \alpha \Delta t - u \cdot sin \, \gamma \Delta t = 0
=> \Delta t = \frac{2l}{u \cdot sin \, \gamma - v \cdot sin \, \alpha}

Now, if we define s = v \cdot \Delta t, we get Quinzio's result, and if we define s = v \cdot (t_1 + \Delta t), we get mine, yay! (with the final substition: u \cdot sin \, \gamma = \sqrt{u^2 - v^2 \, cos^2 \alpha}, that is)
Looks so simple now :D

I don't understand what you are getting here.
You sum a time which belongs to the car with a time which belongs to the sound wave.

Yeah, that was the moment the sound hit the rock. I used that as my new reference point, sort of.

Well, correct me if I'm wrong, but I think that's it. Once again, that drawing saved the day, nice job!
Thank you very much, both of you guys :)

 

[I like Serena]

Oh, I love your drawing Serena! Very clean indeed, thanks very much for that!

Thank you!

It helped me to understand the issue underlying here, I believe. The drawing of s was the key.

Well, correct me if I'm wrong, but I think that's it. Once again, that drawing saved the day, nice job!
Thank you very much, both of you guys :)

Cheers!