Homework Help: Finding the speed to receive sound constantly

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1. Jul 5, 2016

Soren4

1. The problem statement, all variables and given/known data
A source emits a very brief sound signal. A receiver $A$ moves along the $x$ axis with a varying velocity such that it receives constantly the echo from the reflecting wall. Find $v_A$ as a function of the position $x$, knowing the distance $D$ and the speed of sound $v_s$.

Answer $\bigg[v_A=v_s \frac{2 \sqrt{D^2+(\frac{x}{2})^2}}{x} \bigg]$
2. Relevant equations
$v=dx/dt$

3. The attempt at a solution
I tried to think that the velocity must be such that, in equal times $A$ finds itself in the position $x$ such that, at that istant, the sound has traveled to that position, after being reflected by the wall.

But this would mean $$\frac{v_A}{v_s}=\frac{x}{2 \sqrt{D^2+(\frac{x}{2})^2}}$$ i.e. the ratio of spaces traveled is equal to the ratio of velocites (the time taken to travel is the same).

But this is the exact opposite of the answer which makes it wrong. Can anyone suggest me what is the right way to approach this problem?

2. Jul 5, 2016

The Bill

Ask yourself: what is the velocity vector of the sound at point A?

3. Jul 5, 2016

Soren4

It's a vector $\vec{v_s}$ with magnitude $v_s$ and directed along the line connecting $A$ with the wall..

4. Jul 5, 2016

The Bill

And what is the projection of $\vec{v_s}$ along the x axis?

5. Jul 5, 2016

Soren4

Thanks a lot for the suggestions!

The projection of $\vec{v_s}$ is $$v_{s,x}=v_s \frac{x}{2 \sqrt{D^2+(\frac{x}{2})^2}}$$

Unfortunately I can't see the point still, if I set $v_{s,x}$ equal to $v_A$ I get what I tried to answer in the first post which does not agree with the answer given by textbook..

6. Jul 12, 2016

Soren4

@The Bill

I thought again about the problem, but still seems that the right expression is the opposite of the one in the solution.. If I may ask, would you be so kind as to give me some further suggestions on why what I tried is wrong?

Thanks a lot again for your help

7. Jul 12, 2016

haruspex

The Bill misled you by specifically asking for the projection of vs along the x axis. Why should it not be the projection of v along the vs direction that has to match vs in magnitude? Can you think of a basis for picking one or the other?

8. Jul 12, 2016

The Bill

I wasn't trying to mislead, I was just trying to lead the OP to the conclusion that the printed solution was wrong.

9. Jul 12, 2016

haruspex

but it isn't.

10. Jul 13, 2016

Soren4

Thanks so much for the reply!

Taking the projection of $v_A$ on $v_S$ gives the correct result! The two ways looks quite equivalent, the only thing I could think of is this
1. If I set $v_{A,projection}=v_{S}$ this means that $v_{A}>v_{S}$ (supersonic speed)
2. If I set $v_{A}=v_{S,projection}$ this means that $v_{A}<v_{S}$ (subsonic speed)
Case 1. is the one that gives the solution of the textbook so apparently $A$ is moving with supersonic speed.

But I could not think of a criteria to choose between the two if I do not know which of the two is greater. Should I look for something different?

11. Jul 13, 2016

haruspex

It has to be some logical argument, not merely to choose between those two options, but even to say that it must be one of those two.

Conceptually, it might be easier to think in terms of pulling a rope over a pulley. Let the pulley be at the mirror image of O in the wall, and consider the rope running from there to the moving receiver. We wish to draw the rope over the pulley at speed vS.
A component of movement of the receiver normal to the rope achieves nothing, since it does not change the distance from the pulley. Only the receiver's motion along the rope serves to draw the rope over the pulley. Therefore the receiver's velocity component in the direction of the rope has to be vS.