The echo of the sound of a boy standing in front of a mountain is heard after 0.2 s. What is the distance of the mountain from the boy when speed of the sound is 342 m/s?
34.2 m
Understanding how echoes work is key to solving this problem. An echo is simply the reflection of sound waves off a surface, like a mountain or a wall, that returns to the listener. When a boy makes a sound, it travels to the mountain and then reflects back to him. The time given (0.2 s) is the total time taken for the sound to travel from the boy to the mountain and back to the boy.
The total distance covered by the sound wave is twice the distance between the boy and the mountain because the sound travels to the mountain and then returns. Let the distance between the boy and the mountain be \(d\).
Total distance traveled by sound \( = d (\text{to mountain}) + d (\text{back from mountain}) = 2d\).
The relationship between distance, speed, and time is given by the formula:
\(\text{Distance} = \text{Speed} \times \text{Time}\)
In this case, the total distance is \(2d\), the speed is \(v\), and the time is \(t\). So, we have:
\(2d = v \times t\)
To find the distance \(d\) of the mountain from the boy, we need to rearrange the formula:
\(d = \frac{v \times t}{2}\)
Now, we substitute the given values into the formula:
\(d = \frac{342 \, \text{m/s} \times 0.2 \, \text{s}}{2}\)
First, calculate the total distance traveled by the sound:
\(\text{Total distance} = 342 \, \text{m/s} \times 0.2 \, \text{s} = 68.4 \, \text{m}\)
This total distance is \(2d\). So, the distance to the mountain \(d\) is half of this value:
\(d = \frac{68.4 \, \text{m}}{2}\)
\(d = 34.2 \, \text{m}\)
So, the distance of the mountain from the boy is 34.2 meters.
Let's compare our calculated distance with the given options:
Our calculated distance, 34.2 m, matches Option 1.
| Concept | Explanation | Formula Used |
|---|---|---|
| Echo Time | Total time for sound to travel to reflecting surface and back. | \(t\) |
| Speed of Sound | How fast sound travels through the medium (air). | \(v\) |
| Total Distance Traveled | \(2 \times\) distance to reflecting surface. | \(v \times t\) |
| Distance to Surface | Half of the total distance traveled by sound. | \(d = \frac{v \times t}{2}\) |
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I. Sound is a mechanical wave
II. Sound wave does not need any medium to propagate
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a. Television waves
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c. X-rays
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