1.1.5 | revision #8998 | Savchenko Solutions

Three microphones located on the same straight line at points A, B, C recorded successively at the moments t_A > t_B > t_C the sound of an explosion that occurred at point O, which lies on the segment AC. Find the length of the segment AO if AB = BC = L. At what point in time did the explosion occur? For problem 1.1.5

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Solutions of Savchenko Problems in Physics
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    <h3 id="back-link"><a href="/#1.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.1.5^*.$ Three microphones located on the same straight line at points $A$, $B$, $C$ recorded successively at the moments $t_A > t_B > t_C$ the sound of an explosion that occurred at point $O$, which lies on the segment $AC$. Find the length of the segment $AO$ if $AB = BC = L$. At what point in time did the explosion occur?

Solution

Let the explosion occurred at time $$t_0$$ , then the time of signal registration at point $$A$$ is equal to:

t_A = t_0 + \frac{L + x}{c}, \\, (1)

where $$x = BO$$ .

Similar to point $$B$$

$t_B = t_0 + \frac{x}{c}, \\,(2)$

For point $$C$$

$t_C = t_0 + \frac{L - x}{c}, \\,(3)$

Let's subtract the second equation from the first equation:

$t_A – t_B = \frac{L}{c}. \\,(4)$

And from the first equation we subtract the third equation

$t_A – t_C = \frac{2x}{c}. \\,(5)$

From equation $$(4)$$ let's express $c = \frac{L}{t_A – t_B}$ , а из $$(5)$$ $x = \frac{t_A – t_C}{2} \cdot c$ . Then the required distance

$AO = L + x = L + \frac{t_A – t_C}{2}\frac{L}{t_A – t_B}$

After the transformation

$AO = \frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \cdot L. \\, (6)$

To determine the moment of time at which the explosion occurred, we substitute in the expression

$t_A = t_0 + \frac{L + x}{c}, \\; c = \frac{L}{t_A – t_B}$ $\frac{t_A – t_C}{2} = \frac{x}{c}$

after transformations

$t_0 = t_B - \frac{1}{2} \cdot (t_A – t_C)$

    <h4>Answer</h4>
    <p>
        $$AO = \frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \cdot L, \\; t_0 = t_B - \frac{1}{2} \cdot (t_A – t_C)$$
    </p>



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