A person burned a firecracker in front of a cliff and heard its echo 5 s after it burst. The distance of the cliff from the person, if the speed of the sound is 340 m/s, is close to
Calculating Cliff Distance Using Echo Time and Sound Speed
This problem involves understanding how echoes work and applying the relationship between distance, speed, and time. An echo is produced when sound waves reflect off a surface, like a cliff. The time measured for the echo is the total time it takes for the sound to travel from the source (the person) to the reflecting surface (the cliff) and then back to the source (where the person hears the echo).
Understanding the Echo Process
- Sound travels from the person towards the cliff.
- The sound reflects off the cliff.
- The reflected sound (echo) travels back from the cliff to the person.
- The total time given (5 seconds) is for this entire round trip.
Given Information:
- Time taken to hear the echo (total round trip time), \(t = 5 \text{ s}\).
- Speed of sound, \(v = 340 \text{ m/s}\).
Calculating the Distance
Let the distance from the person to the cliff be \(d\). The sound travels this distance \(d\) towards the cliff and then travels the same distance \(d\) back as an echo. So, the total distance traveled by the sound for the echo is \(2d\).
The formula relating distance, speed, and time is:
\(\text{Distance} = \text{Speed} \times \text{Time}\)
In this case, the total distance is \(2d\) and the time is the round trip time \(t\). So, we have:
\(2d = v \times t\)
To find the distance to the cliff (\(d\)), we rearrange the formula:
\(d = \frac{v \times t}{2}\)
Step-by-Step Calculation
Substitute the given values into the formula:
\(d = \frac{340 \text{ m/s} \times 5 \text{ s}}{2}\)
First, calculate the product of speed and time:
\(v \times t = 340 \text{ m/s} \times 5 \text{ s} = 1700 \text{ m}\)
This 1700 m is the total distance the sound traveled (to the cliff and back). Now, divide by 2 to find the distance to the cliff:
\(d = \frac{1700 \text{ m}}{2}\)
\(d = 850 \text{ m}\)
Conclusion
The distance of the cliff from the person is 850 meters.
| Quantity | Symbol | Value |
|---|---|---|
| Speed of Sound | \(v\) | 340 m/s |
| Echo Time (Round Trip) | \(t\) | 5 s |
| Distance to Cliff | \(d\) | ? |
| Formula Used | Calculation | Result |
|---|---|---|
| \(d = \frac{v \times t}{2}\) | \(d = \frac{340 \text{ m/s} \times 5 \text{ s}}{2}\) | \(d = 850 \text{ m}\) |
Revision Table: Key Concepts for Echo Calculation
| Concept | Description |
|---|---|
| Echo | Reflection of sound waves off a surface. |
| Echo Time | The total time taken for sound to travel from the source to the reflector and back. |
| Total Distance for Echo | Twice the distance from the source to the reflector (\(2d\)). |
| Formula | \(2d = \text{Speed} \times \text{Time}\) or \(d = \frac{\text{Speed} \times \text{Time}}{2}\) |
Additional Information: Factors Affecting Speed of Sound and Echoes
- The speed of sound in air is mainly affected by temperature. Higher temperature increases the speed of sound.
- Humidity also affects the speed of sound slightly.
- Echoes are clearer and louder when the reflecting surface is hard and smooth.
- For a distinct echo to be heard, the reflecting surface must be at a sufficient distance from the sound source. This is because the sound and its echo must be separated by at least 0.1 seconds for the human ear to distinguish them as separate sounds.
- Echoes are used in various applications, such as SONAR (Sound Navigation and Ranging) used by ships and submarines to detect objects underwater, and in medical ultrasound imaging.
- Reverberation is the persistence of sound due to multiple reflections, unlike a distinct echo which is a single clear reflection.
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