Question:
A bat emits an ultrasonic sound wave at $33.0$ kHz and receives an echo $0.1$ s later. What is the distance of the bat from the object-producing echo? (Speed of sound in air is $330$ m/s)
A bat emits an ultrasonic sound wave at $33.0$ kHz and receives an echo $0.1$ s later. What is the distance of the bat from the object-producing echo? (Speed of sound in air is $330$ m/s)
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Always remember to divide by 2 in echo problems! The wave has to travel to the target and back, so the time given is for a round trip.
Updated On: May 1, 2026
- $10.0$ m
- $20.0$ m
- $33.0$ m
- $66.0$ m
- $16.5$ m
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Solution and Explanation
Concept:
An echo is the reflection of sound that arrives at the listener with a delay after the direct sound. The total time taken ($t$) is the time for sound to travel to the object and back. Therefore, the distance to the object ($d$) is half the total distance travelled by the wave ($2d = v \times t$).
Step 1: {Identify the given parameters.}
The speed of sound in air ($v$) is $330$ m/s. The total time for the echo to return ($t$) is $0.1$ s. Note that the frequency ($33.0$ kHz) is extra information not needed for calculating distance
Step 2: {Apply the echo distance formula.}
Total distance ($2d$) is velocity multiplied by time: $$2d = v \times t$$ $$2d = 330 \text{ m/s} \times 0.1 \text{ s}$$
Step 3: {Solve for the distance $d$.}
$$2d = 33 \text{ m}$$ $$d = \frac{33}{2} = 16.5 \text{ m}$$ The bat is $16.5$ meters away from the object.
Step 1: {Identify the given parameters.}
The speed of sound in air ($v$) is $330$ m/s. The total time for the echo to return ($t$) is $0.1$ s. Note that the frequency ($33.0$ kHz) is extra information not needed for calculating distance
Step 2: {Apply the echo distance formula.}
Total distance ($2d$) is velocity multiplied by time: $$2d = v \times t$$ $$2d = 330 \text{ m/s} \times 0.1 \text{ s}$$
Step 3: {Solve for the distance $d$.}
$$2d = 33 \text{ m}$$ $$d = \frac{33}{2} = 16.5 \text{ m}$$ The bat is $16.5$ meters away from the object.
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