Question

a person , with his ear to the ground , see a huge stone strike the concrete payment. a moment later two sounds are heard from the impact. one travels in the air and the other in the concrete, and they are 1.1 is apart. how far away did the impact occur?

Answer 1

Answer on Question#49417 - Physics - Acoustics

A person, with his ear to the ground, see a huge stone strike the concrete payment. A moment later two sounds are heard from the impact. One travels in the air and the other in the concrete, and they are $\Delta t = 1.1\mathrm{s}$ apart. How far away did the impact occur?

Solution:

The difference in time $\Delta t$ is caused by the different speed of sound in the air ( $c_{a}$ ) and in the concrete ( $c_{c}$ ). The time needed for the sound to overcome the distance $L$ :

$$t = \frac {L}{c}$$

So, the difference in time $\Delta t$ can be defined as

$$\Delta t = \frac {L}{c _ {a}} - \frac {L}{c _ {c}} = \frac {(c _ {c} - c _ {a})}{c _ {c} c _ {a}} L$$

And for the distance $L$ between the man and the impact we obtain the following formula

$$L = \frac {c _ {c} c _ {a}}{c _ {c} - c _ {a}} \Delta t$$

Since the speed of sound in the concrete is $c_{c} = 3400\frac{\mathrm{m}}{\mathrm{s}}$ and in the air $c_{a} = 343\frac{\mathrm{m}}{\mathrm{s}}$ , we can now calculate this distance

$$L = \frac {c _ {c} c _ {a}}{c _ {c} - c _ {a}} \Delta t = \frac {3 4 0 0 \frac {\mathrm {m}}{\mathrm {s}} \cdot 3 4 3 \frac {\mathrm {m}}{\mathrm {s}}}{3 4 0 0 \frac {\mathrm {m}}{\mathrm {s}} - 3 4 3 \frac {\mathrm {m}}{\mathrm {s}}} 1. 1 \mathrm {s} = 3 8 1. 5 \mathrm {m}$$

Answer: $L = \frac{c_{c}c_{a}}{c_{c} - c_{a}}\Delta t = 381.5\mathrm{m}$ .

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