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A man standing unsymmetrically between two parallel cliffs, claps his hand and starts hearing a series of echoes at intervals of $1{\text{ s}}$. If the speed of sound in air is \[340{\text{ m}}{{\text{s}}^{ - 1}}\], then the distance between the two parallel cliffs is
A. $170{\text{ m}}$
B. $340{\text{ m}}$
C. $510{\text{ m}}$
D. $680{\text{ m}}$

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Last updated date: 11th Jun 2024
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Answer
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Hint Sound wave is a longitudinal wave and hence travel in straight line motion. So, time taken by a sound to travel a distance $d$ can be given by $t = \dfrac{d}{v}$ where $v$ is the velocity of sound waves in a particular medium.

Complete step by step answer
As given in the question that the man is standing unsymmetrically between the cliffs that means he can be anywhere between the cliffs.
Let the man is standing at a distance ${x_1}{\text{ m}}$ from the nearest cliff and ${x_2}{\text{ m}}$ from the farther cliff so that the distance between the cliffs will be $\left( {{x_1} + {x_2}} \right){\text{ m}}$
Now, as we know that the sound wave is a longitudinal wave and hence travels in straight line motion. So, time taken by a sound to travel a distance $d$ can be given by $t = \dfrac{d}{v}$ where $v$ is the velocity of sound waves in a particular medium.
As given in the question that the man hears the echoes at an interval of $1{\text{ s}}$
So, the first sound he will hear after $1{\text{ s}}$ which was get reflected by the nearest cliff after covering a distance of $2{x_1}$
Therefore from the above equation we have
$1 = \dfrac{{2{x_1}}}{{340}}$
On solving we get ${x_1} = 170{\text{ m}}$
Now, the second echo he will hear after ${\text{2 s}}$ which was get reflected by the farther cliff after covering a distance of $2{x_2}$
Therefore from the above equation we have
$2 = \dfrac{{2{x_2}}}{{340}}$
On solving we get ${x_2} = 340{\text{ m}}$
So, the distance between the cliffs is $\left( {{x_1} + {x_2}} \right) = 170 + 340 = 510{\text{ m}}$
Hence, option C is correct.

Note As sound wave is a longitudinal wave it has compressions and rarefactions while moving. The compressions and rarefactions after striking a rigid wall or cliff, reflects back as compressions and rarefaction respectively as there is phase difference of $180^\circ $ . This phenomenon is known as reflection of sound waves.