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Chapter 2: Problem 45

A man beats drum at a certain distance from a large mountain. He slowly increases the rate of beating and finds that the echo is not heard distinctly, when the drum beating is at the rate of 40 per minute. He moves by 80 meter towards the mountain and finds that the echo is again not heard distinctly, when the rate of beating of the drum is 1 per second. What is the initial distance of the man from the mountain ? (1) \(120 \mathrm{~m}\) (2) \(240 \mathrm{~m}\) (3) \(270 \mathrm{~m}\) (4) \(340 \mathrm{~m}\)

Short Answer

Expert verified
The initial distance is 255 meters.

Step by step solution

01

Understanding the problem

When the man beats the drum at 40 beats per minute, he does not hear the echo distinctly. This implies that the echoes overlap with the subsequent beats.
02

Converting beats per minute to beats per second

Convert the rate of beating from 40 beats per minute to beats per second: \ \( \text{Rate} = \frac{40}{60} \text{ beats/second} = \frac{2}{3} \text{ beats/second} \)
03

Echo overlap condition at initial position

For the man not to hear an echo distinctly, the time taken for a sound to travel to the mountain and back must be equal to the interval between beats: \ \( \text{Time interval} = \frac{3}{2} \text{ seconds} \)
04

Calculating the distance using speed

Let the initial distance to the mountain be \( d \). The speed of sound \( v \) is assumed to be \( 340 \text{ m/s} \). The round trip time is given by: \ \( \text{Time} = \frac{2d}{340} \text{ seconds} \) \( \text{where the round trip time} \) results in \( \frac{2d}{340} = \frac{3}{2}\), solve this for the initial distance \( d \).
05

Solving for the initial distance \( d \)

Rearrange the equation \( \frac{2d}{340} = \frac{3}{2} \) to solve for \( d \): \ \( 2d = 340 \times \frac{3}{2} \) \ \( 2d = 510 \) \ \( d = 255 \text{ meters} \)
06

Determining distance with new conditions

When the man moves 80 meters closer, the new distance is \ \( 255 - 80 = 175 \text{ meters}. \) Repeating drum rate is now 1 per second (or 60 beats per minute), thus the time interval is 1 second.
07

Calculate the new round trip time

Calculate using the new distance: \( \frac{2 \times 175}{340} = 1 \text{ second}\), which checks out correctly. Therefore, our initial calculation is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

echo in physics
An echo is a reflection of sound that arrives at the listener after a delay. The delay is caused by the time it takes for the sound wave to travel to a reflective surface and back. In this exercise, the drum beats create sound waves that travel to a mountain and then reflect back to the man. If the time interval between the drum beats is shorter than the round trip travel time of the sound, the echoes will overlap with the subsequent beats, making the echo indistinct. This means the visibility of the echo depends on the rate at which the sound is produced and the distance to the reflective surface.
speed of sound calculation
The speed of sound in air is typically around 340 meters per second. This value can change slightly depending on factors such as temperature and humidity, but 340 m/s is a good average estimation. To calculate the distance based on the speed of sound, we need to know the time it takes for the sound wave to travel the desired distance. The formula to use is:


  • \frac{\text{speed of sound} (\text{m/s}) \times \text{time} (\text{s})}{2} = \text{distance (m)},
  • In this scenario, the sound travels to the mountain and back, doubling the travel time in the formula. If we take the initial condition where the rate is 40 beats per minute, the interval between beats is 1.5 seconds. Using
    \(\frac{1.5 (\text{s})\times 340 (\text{m/s})}{2} = 255\) meters,
    we can solve for the initial distance \(d\) using the measured time and given speed.
time interval and distance math problem
This exercise is essentially a problem of calculating the distance using time intervals and the speed of sound. When the man does not hear a distinct echo at 40 beats per minute (or \(\frac{2}{3}\) beats per second), it means the round trip time for the sound equals the interval between beats. We set up the equation for this condition:


  • \frac{2d}{340} = \frac{3}{2} This simplifies to:

      2d = 510,
      d = \frac{510}{2},
      d= 255 meters.


    Then, the man moves 80 meters closer and repeats the experiment at 1 beat per second. Now the time interval between beats is 1 second, and he finds the echo indistinct again. The total distance now is:

      d = 175 meters (\(255 - 80\) is calculated with the same factor).
      The round trip time condition again checks out as:
      175 \frac{2 \times 175}{ 340} = \frac(1 second)
      Therefore, the calculations confirm the correct initial distance.

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