Question

Ques:- A person standing between two cliffs and 580 m fromyhe nearest cliffs shouts. He hears the first echo after 4 s and he second echo after 2 s later.

Calculate :

1) The speed of sound

2)The distance of the other cliff from the person.

Answer 1

$\underline{\underline{\Huge{\textbf{Solution:}}}}$

$\textsf{Given,}$
$\sf\textsf{Distance to the Nearest Cliff for the Person = 580 m}$
$\sf\textsf{time taken by First Echo to reach the person = 4s}$
$\sf\textsf{time taken by Second Echo to reach the person = 2s}$



$\underline{\Large{\textsf{Step-1:}}}$
$\sf\textsf{Calculate the distance covered by the Echo sound}$

$\textsf{Distance Covered by the Echo Sound is 2 times the}\\ \sf\textsf{Distance to the nearest cliffs}$

$\sf\textsf{(As Echo travels to the Cliff and}\\ \sf\textsf{Comes back to the Person)}$

$\implies \sf\textsf{The Distance Travelled by the Echo = 2 \times 580 \: m}$

$\therefore \sf\textsf{The Distance Travelled by the Echo = 1160\: m}$



$\underline{\Large{\textsf{Step-2:}}}$
$\sf\textsf{Find the Speed of the sound}$

$\bigstar \: \boxed{\Large{\mathbf{Speed\: = \dfrac{Distance\: covered}{time\: taken}}}}$

$\sf\textsf{Substitute Values}$

$\implies \sf\textsf{Speed of the sound = \dfrac{1160}{4} }$

$\therefore \sf\textsf{Speed of the sound, s = 290 m/s}$



$\underline{\Large{\textsf{Step-3:}}}$
$\sf\textsf{Find the time taken by the Second Echo to reach the Person}$

$\sf\textbf{The time taken by the Second Echo to reach the Person}\\\sf \textbf{is sum of time taken by the first Echo and second echo}$

$\implies \sf\textsf{Time taken by Second Echo = 4s + 2s}$

$\therefore \sf\textsf{Time taken by Second Echo, t= 6s}$



$\underline{\Large{\textsf{Step-4:}}}$
$\sf\textsf{Calculate the Distance of other cliff from}\\\sf\textsf{the person Using Speed of the Sound}$

$\sf\textsf{We have,}$
$\sf\textsf{ s = \dfrac{d}{t} }$

$\sf\textsf{ But Echo travels to the Cliff} \\\sf\textsf{and comes back, so distance is doubled}$

$\implies \sf\textsf{s = \dfrac{2d}{t} }$

$\implies \sf \textbf{Distance of the other cliff from the person = \dfrac{s\times t}{2} \: m}$

$\implies \sf \textsf{Distance of the other cliff from the person = \dfrac{290 \times 6}{2} \: m}$

$\implies \sf \textsf{Distance of the other cliff from the person = 290 \times 3\: m}$

$\therefore \sf \textsf{Distance of the other cliff from the person = 870\: m}$


$\blacksquare\: \: \sf\textsf{Speed of the sound = \underline{\Large{\textbf{290 m/s}}}}$

$\blacksquare\: \: \sf \textsf{Distance of the other cliff from the person =\underline{\Large{\textbf{870\: m}}}}$

Answer 2

Given :- Distance of the nearest cliff = 580m

Time taken by first echo = 4s

Time taken by second echo = 2s

Calculating :-
( i ) Total distance traveled by the sound in coming back = 2 × 580m = 1160m

Speed of sound=$\frac{total \: distance \: travelled}{time \: taken} = \frac{1160}{4} = 290$

Time taken by first and second echo = 4sec + 2sec = 6sec

$\therefore \: v = \frac{2d}{t}$

$= d = \frac{vt}{2}$

( ii ) $distance = \frac{290 \times 6}{2} = 290 \times 3 = 870$