An echo is heard after 5 seconds of the production of sound which moves with a speed of 340 m/s. What is the distance of the mountain from the source of sound which produced the echo?
0.85 km
Calculating Mountain Distance Using Echo Sound
This problem involves the concept of an echo, which is a sound reflection. When a sound is produced near a mountain, it travels towards the mountain, reflects off it, and returns to the source as an echo. The time taken for the echo to be heard is the total time for the sound to travel from the source to the mountain and back to the source.
Understanding the Echo Phenomenon
An echo occurs because sound waves bounce off surfaces, much like how light reflects off a mirror. For a distinct echo to be heard, the reflecting surface (like a mountain or a wall) must be far enough away so that the reflected sound reaches the listener after the original sound has faded.
Given Information:
- Time taken for the echo to be heard (total time): \(t_{\text{total}} = 5\) seconds
- Speed of sound in air: \(v = 340\) m/s
Step-by-Step Calculation:
The total time \(t_{\text{total}}\) is the time for the sound to travel from the source to the mountain (let's call this distance \(d\)) and then back from the mountain to the source (which is also distance \(d\)). So, the total distance covered by the sound is \(d + d = 2d\).
The relationship between distance, speed, and time is given by:
\(\text{Distance} = \text{Speed} \times \text{Time}\)
Using the total distance (\(2d\)) and the total time (\(t_{\text{total}}\)):
\(2d = v \times t_{\text{total}}\)
We want to find the distance \(d\) of the mountain from the source. We can rearrange the formula to solve for \(d\):
\(d = \frac{v \times t_{\text{total}}}{2}\)
Now, let's substitute the given values into the formula:
\(d = \frac{340 \, \text{m/s} \times 5 \, \text{s}}{2}\)
\(d = \frac{1700 \, \text{m}}{2}\)
\(d = 850 \, \text{m}\)
The distance of the mountain from the source is 850 meters.
Converting Distance to Kilometers
The options are given in kilometers. We need to convert meters to kilometers. There are 1000 meters in 1 kilometer.
\(1 \, \text{km} = 1000 \, \text{m}\)
So, to convert meters to kilometers, we divide the distance in meters by 1000.
\(d_{\text{km}} = \frac{d_{\text{m}}}{1000}\)
\(d_{\text{km}} = \frac{850 \, \text{m}}{1000 \, \text{m/km}}\)
\(d_{\text{km}} = 0.85 \, \text{km}\)
Therefore, the distance of the mountain from the source of sound is 0.85 km.
Summary of Calculation
| Quantity | Symbol | Value |
|---|---|---|
| Total time for echo | \(t_{\text{total}}\) | 5 s |
| Speed of sound | \(v\) | 340 m/s |
| Total distance covered by sound (to and fro) | \(2d\) | \(v \times t_{\text{total}} = 340 \times 5 = 1700\) m |
| Distance to the mountain | \(d\) | \(\frac{1700}{2} = 850\) m |
| Distance to the mountain in km | \(d_{\text{km}}\) | \(\frac{850}{1000} = 0.85\) km |
Final Answer Determination
The calculated distance of the mountain is 0.85 km, which matches one of the provided options.
Revision Table: Key Concepts
| Concept | Description | Formula/Relation |
|---|---|---|
| Echo | Reflection of sound waves from a surface. | Requires a reflecting surface and sufficient distance. |
| Speed | Distance traveled per unit time. | \(v = \frac{d}{t}\) |
| Distance Calculation (Echo) | Half the total distance covered by sound round trip. | \(d = \frac{v \times t_{\text{total}}}{2}\) |
| Unit Conversion | Changing from one unit of measurement to another (e.g., m to km). | \(1 \, \text{km} = 1000 \, \text{m}\) |
Additional Information: Factors Affecting Speed of Sound
The speed of sound is not constant; it depends on several factors:
- Medium: Sound travels fastest in solids, slower in liquids, and slowest in gases. This is because particles are closer together in solids, allowing vibrations to transmit more quickly.
- Temperature: In gases, the speed of sound increases with increasing temperature. At higher temperatures, molecules move faster and collide more frequently, speeding up the transfer of sound energy. The approximate speed of sound in air at temperature \(T\) (in Celsius) can be given by \(v \approx 331.4 + 0.6T\) m/s.
- Humidity: The presence of water vapor (humidity) in air slightly increases the speed of sound.
- Pressure: For an ideal gas, the speed of sound is generally independent of pressure at a constant temperature because density also changes proportionally with pressure, and these effects cancel out.
In this problem, a constant speed of sound (340 m/s) is given, which is a typical value for the speed of sound in dry air at around 15-20°C.
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