A man standing between two cliffs produces a sound and hears two successive echoes at intervals of 3s and 4s respectively. Calculate the distance between two cliffs. The speed of sound in the air is 330m/s
Answer
608.7k+ views
Hint: It is given in the question the question that the man is standing somewhere in between the cliff. Therefore sound created by him will hit the either cliffs and reach back to him and this is basically what an echo is. In the question it is given that he hears a successive echo at an interval of 3s and 4s. Hence we will use the distance formula to calculate the distance between the man and the cliffs as time taken for sound in air to reflect and come back to the man is given.
Complete answer:
To begin with let us first define what an echo means.
An echo is a repeated sound in a regular interval of time. In the above question a man hears two successive echo’s such that the interval between them is one sec. Given below is a diagram depicting the following.
In the above diagram we can see that the two cliffs are at a distance D apart. Let us say the man is standing at a distance x from the cliff A and at a distance y from cliff B. Hence we can write $D=x+y...(1)$. The sound produced by the man hits the cliff A and reverts back to him in a time interval of 2secs. The relation between distance (d), time(t) and speed (v)is given as $v=\dfrac{d}{t}$.
The sound travels twice the distance twice of x as it hits the cliff and comes back. Hence distance x is equal to,
$\begin{align}
& v=\dfrac{2x}{t} \\
& 330=\dfrac{2x}{3} \\
& x=\dfrac{990}{2}=495m \\
\end{align}$
Similarly the sound takes 4secs to hit cliff b and reach back to the man. Hence distance y is equal to,
$\begin{align}
& v=\dfrac{2y}{t} \\
& 330=\dfrac{2y}{4} \\
& y=\dfrac{1320}{2}=660m \\
\end{align}$
Hence substituting x and y in equation 1 we get,
$\begin{align}
& D=x+y \\
& D=495+660 \\
& D=1155m \\
\end{align}$
Hence the distance between the two cliffs is 1155 meters.
Note:
Echo’s are a result of reflection of sound. In caves and closed surfaces echoes are more prominently heard as the reflection of sound takes place frequently. In the above question a man is standing somewhere in between the two cliffs, if the man had to stand at the midpoints between the two cliffs then a single echo would have been heard by the man.
Complete answer:
To begin with let us first define what an echo means.
An echo is a repeated sound in a regular interval of time. In the above question a man hears two successive echo’s such that the interval between them is one sec. Given below is a diagram depicting the following.
In the above diagram we can see that the two cliffs are at a distance D apart. Let us say the man is standing at a distance x from the cliff A and at a distance y from cliff B. Hence we can write $D=x+y...(1)$. The sound produced by the man hits the cliff A and reverts back to him in a time interval of 2secs. The relation between distance (d), time(t) and speed (v)is given as $v=\dfrac{d}{t}$.
The sound travels twice the distance twice of x as it hits the cliff and comes back. Hence distance x is equal to,
$\begin{align}
& v=\dfrac{2x}{t} \\
& 330=\dfrac{2x}{3} \\
& x=\dfrac{990}{2}=495m \\
\end{align}$
Similarly the sound takes 4secs to hit cliff b and reach back to the man. Hence distance y is equal to,
$\begin{align}
& v=\dfrac{2y}{t} \\
& 330=\dfrac{2y}{4} \\
& y=\dfrac{1320}{2}=660m \\
\end{align}$
Hence substituting x and y in equation 1 we get,
$\begin{align}
& D=x+y \\
& D=495+660 \\
& D=1155m \\
\end{align}$
Hence the distance between the two cliffs is 1155 meters.
Note:
Echo’s are a result of reflection of sound. In caves and closed surfaces echoes are more prominently heard as the reflection of sound takes place frequently. In the above question a man is standing somewhere in between the two cliffs, if the man had to stand at the midpoints between the two cliffs then a single echo would have been heard by the man.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What are the major means of transport Explain each class 12 social science CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

Why should a magnesium ribbon be cleaned before burning class 12 chemistry CBSE

