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Question

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?

The correct answer is

34.2 m

Calculating Distance Using Sound Echo

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.

Given Information:

  • Time taken for the echo, \(t = 0.2\) s
  • Speed of sound, \(v = 342\) m/s

Concept of Echo and Distance

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:

\(d = \frac{v \times t}{2}\)

Step-by-Step Calculation:

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.

Checking the Options:

Let's compare our calculated distance with the given options:

  • Option 1: 34.2 m
  • Option 2: 68.4 m
  • Option 3: 68.8 m
  • Option 4: 34.4 m

Our calculated distance, 34.2 m, matches Option 1.

Revision Table: Sound Echo Distance Calculation

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. \(d = \frac{v \times t}{2}\)

Additional Information on Sound and Echoes

  • What is Sound? Sound is a form of energy that travels as waves through a medium (like air, water, or solids). These waves are longitudinal waves in air.
  • What is an Echo? An echo is the repetition of sound caused by the reflection of sound waves from a surface such as a wall, mountain, or cliff.
  • Conditions for Hearing a Distinct Echo: To hear a separate echo, there must be a sufficient time gap between the original sound and the reflected sound. In air, this minimum time gap is generally considered to be about 0.1 seconds. This corresponds to a minimum distance to the reflecting surface. If the speed of sound is 343 m/s, the sound travels \(343 \times 0.1 = 34.3\) meters in 0.1 s. The total distance for an echo is \(2d\), so \(2d = 34.3\) m, meaning \(d \approx 17.15\) meters. The reflecting surface must be at least this far away.
  • Factors Affecting Speed of Sound: The speed of sound in air is affected by temperature, humidity, and air pressure. It increases with increasing temperature and humidity.
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Important Questions from Wave

  1. Which one of the following statements is true for sound waves propa- gating in air?

  2. Which one of the following optical phenomena supports that the light is a transverse wave?

  3. Which of the following is the horizontal distance between two successive crests?

  4. Which of the following statements is correct?

  5. What is the correct order of radiations in descending order of frequencies?

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