A person rings a metallic bell near a strong concrete wall. He hears the echo after 0.3 s. If the sound moves with a speed of 340 m/s, how far is the wall from him?
51 m
Understanding the Echo Phenomenon
An echo is a sound heard after the sound wave is reflected from a surface back to the listener. When a person rings a metallic bell near a wall, the sound travels from the person to the wall and then reflects back to the person's ears. The time taken for the echo to be heard is the total time the sound wave travels from the source to the reflecting surface and back to the source.
Problem Analysis: Calculating Wall Distance
We are given the following information in the problem:
- Time taken to hear the echo (\(t\)): 0.3 seconds
- Speed of sound (\(v\)): 340 meters per second
We need to find the distance between the person and the wall. Let's call this distance \(d\).
Relating Distance, Speed, and Echo Time
The sound wave travels from the person to the wall (distance \(d\)) and then from the wall back to the person (distance \(d\)). So, the total distance covered by the sound wave is \(d + d = 2d\). The relationship between distance, speed, and time is given by the formula:
\(\text{Distance} = \text{Speed} \times \text{Time}\)
In the case of an echo, the total distance traveled is \(2d\). Therefore, the formula becomes:
\(2d = v \times t\)
Step-by-Step Calculation
Using the formula \(2d = v \times t\), we can substitute the given values:
\(2d = 340 \text{ m/s} \times 0.3 \text{ s}\)
First, calculate the total distance covered by the sound:
\(2d = 102 \text{ m}\)
Now, to find the distance to the wall (\(d\)), we divide the total distance by 2:
\(d = \frac{102 \text{ m}}{2}\)
\(d = 51 \text{ m}\)
So, the distance of the wall from the person is 51 meters.
Comparing with Given Options
Let's look at the provided options:
- 102 m
- 11 m
- 51 m
- 30 m
Our calculated distance is 51 m, which matches one of the options.
Summary of Calculation
The distance to the wall is half of the total distance the sound travels for the echo.
Total distance = Speed of sound \(\times\) Time for echo
Total distance = 340 m/s \(\times\) 0.3 s = 102 m
Distance to wall = Total distance / 2
Distance to wall = 102 m / 2 = 51 m
| Quantity | Symbol | Value |
|---|---|---|
| Time for echo | \(t\) | 0.3 s |
| Speed of sound | \(v\) | 340 m/s |
| Distance to wall | \(d\) | ? |
| Step | Calculation | Result |
|---|---|---|
| 1 | Calculate total distance traveled by sound (\(2d = v \times t\)) | \(2d = 340 \times 0.3 = 102 \text{ m}\) |
| 2 | Calculate distance to the wall (\(d = 2d / 2\)) | \(d = 102 / 2 = 51 \text{ m}\) |
Revision Table: Sound and Echo Concepts
| Concept | Description | Formula (related) |
|---|---|---|
| Sound Wave | A vibration that travels through a medium (like air) carrying energy. | Speed = Frequency \(\times\) Wavelength |
| Echo | A reflected sound wave heard after the original sound. Requires a reflecting surface and sufficient distance. | \(2d = v \times t\) (for distance calculation) |
| Speed of Sound | How fast sound travels through a medium. Varies with temperature and medium properties. | \(v\) (typically given in m/s) |
| Time of Flight | The total time taken for the sound to travel to the surface and back for an echo. | \(t\) (in seconds) |
Additional Information: Factors Affecting Echoes
Several factors influence whether an echo is heard clearly:
- Distance to the Reflecting Surface: The surface must be far enough away so that the reflected sound arrives at least 0.1 seconds after the original sound. This is approximately the persistence of hearing, allowing the brain to distinguish the two sounds. At a sound speed of 340 m/s, this requires a minimum distance of about \((340 \text{ m/s} \times 0.1 \text{ s}) / 2 = 17 \text{ m}\). In this problem, the distance is 51 m, which is sufficient.
- Size and Nature of the Reflecting Surface: Large, hard, smooth surfaces (like a concrete wall, as mentioned in the question) are good reflectors of sound.
- Absorption by the Medium: Sound energy is absorbed by the medium it travels through, especially over long distances or in certain materials (like fog).
- Intensity of the Original Sound: The original sound must be loud enough for the reflected sound to be detectable after traveling twice the distance.
Echoes are used in various applications, such as SONAR (Sound Navigation And Ranging) for underwater depth finding and medical ultrasound imaging.
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