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Difficulty: MediumSound Waves, Echoes, Pitch, Loudness, and Quality

A surveyor standing 255 m255\text{ m} away from the base of a vertical canyon wall emits a short acoustic signal. If the speed of sound in air is 340 m/s340\text{ m/s}, after what time interval will the surveyor detect the reflected echo?

  1. A
    0.75 s0.75\text{ s}
  2. 1.50 s1.50\text{ s}Answer
  3. C
    1.33 s1.33\text{ s}
  4. D
    3.00 s3.00\text{ s}

Answer

1.50 s1.50\text{ s}
Sound must travel to the cliff face and reflect back to the surveyor, covering a total distance of 2×255 m=510 m2 \times 255\text{ m} = 510\text{ m}. Using the speed formula t=dvt = \frac{d}{v}, the elapsed time is 510 m340 m/s=1.50 s\frac{510\text{ m}}{340\text{ m/s}} = 1.50\text{ s}.

Step-by-Step Solution

1
Determine the total distance traveled by the sound wave
stotal=2×255 m=510 ms_{total} = 2 \times 255\text{ m} = 510\text{ m}
An echo requires sound to travel to the obstacle and reflect back to the source.
2
Calculate the time taken using the wave speed formula
t=stotalv=510 m340 m/s=1.50 st = \frac{s_{total}}{v} = \frac{510\text{ m}}{340\text{ m/s}} = 1.50\text{ s}
Time is equal to total distance divided by the speed of propagation.

Key Concept

Echo reflection and two-way sound propagation distance
Estimated Time:1m 0s
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Topics

Sound Waves, Echoes, Pitch, Loudness, and Quality

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