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I found this question and I have a problem:

Two friends A and B are standing a distance $x$ apart in an open field and wind is blowing from A to B. A beats a drum and B hears the sound $t1$ time after he sees the event. A and B interchange their position and the experiment is repeated. This time B hears the drum $t2$ time after he sees the event. Calculate the velocity of sound in still air v and the velocity of wind $u$. Neglect the time light takes in travelling between the friends.

I proceeded like this: If the direction from A to B is taken as positive then,

$(v+u)=x/t_1$ When they interchange Since wind still blows in positive direction therefore $(u-v)=x/t_2$

But in the book it is given that it should be $v-u$...Why is that?..Does the distance between them also changes sign?...I understand that in a real situation velocity of sound must be greater than that of wind and so my second equation is wrong because on solving it leads to opposite of that.. What is the problem?

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The method of interchange is a common method in physics experiments to cancel systematic errors (wind speed in this case) or their leading-order effects. The central idea is symmetry.

If there were no wind, the speed of sound would be measured by simply $v=x/t$. But if there is wind, we use the method of interchange to measure two times $t_1,t_2$. Then we have

$$x=(v+u)t_1=(v-u)t_2.$$

Why is it not $u-v$? Intuitively, the speed of sound $v\gg u$ is much faster than the wind speed. Our formulas should reduce to $x=vt_1=vt_2$ in case $u=0$ (no wind). Then the speed of sound is

$$v=\frac{1}{2}\left(\frac{x}{t_1}+\frac{x}{t_2}\right),$$

which is simply the average of the two speeds calculated using $t_1$ and $t_2$ separately.

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  • $\begingroup$ But after interchanging why doesn't velocity of sound change sign since now it would be travelling in opposite direction though still towards B $\endgroup$
    – LM2357
    Commented Nov 12, 2017 at 3:21
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    $\begingroup$ It's actually $x=(v+u)t_1$ and $-x=(u-v)t_2$. Both $x$ (displacement) and $v$ change directions while the wind speed $u$ does not. Anyway the measurement is down the wind once and up against the wind once. $\endgroup$
    – Zhuoran He
    Commented Nov 12, 2017 at 3:24
  • $\begingroup$ One last question....why does displacement change sign? The displacement of wind is still towards the same direction..I understand the intuitive idea but I'm confused with direction $\endgroup$
    – LM2357
    Commented Nov 12, 2017 at 3:30
  • $\begingroup$ Well, because the two people switched positions.. $\endgroup$
    – Zhuoran He
    Commented Nov 12, 2017 at 3:30
  • $\begingroup$ It may help doing this exact same problem, but with something more concrete, like a tennis ball, instead of something abstract like sound. Set up your x-axis with x = 0 where person A is initially. It has to stay the same throughout the problem, so when they switch B is standing at x = 0. Use $ d$ for the distance traveled so you're not confusing the variable with constants. $\endgroup$
    – David Reed
    Commented Nov 12, 2017 at 4:16
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The distance doesn't change sign but the kinematics formulas don't use distance. They use DISPLACEMENT, and the displacement DOES change sign.the displacement is now $-x$ instead of $x$. In the first scenario sound moves $x$ units to the right. In the second it moves $x$ units to the left. Say you put $x = 0$ at where A is initially standing. Then for the first experiment,

$x_i = 0 \space $ and $\space x_f = x \implies \Delta x = x_f - x_i = x- 0 = x$

For the second experiment,

$x_i = x \space $ and $ \space x_f = 0 \implies \Delta x = x_f - x_i = 0 - x = -x$

For the second scenario, the velocity of sound relative to B is $\space u - v$

So $$-x = \Delta x = (u-v)\Delta t = (u-v)t_2 \implies x = (v-u)t_2$$

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