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Mechanical Properties Of Fluids

Question

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A man standing between two parallel cliffs clap and hears two echoes, one 2-4s and second at 3.6s after clapping. If the velocity of sound is 340m/s, find the distance between the two cliffs.

Solution

Let the man be at a distance of d₁ from one cliff and d₂ from the other cliff.
The first echo is heard after reflection from the first cliff and the second reflection is heard after reflection from the second cliff.

As the first echo is heard at time t₁ = 2.4s,

∴

2 straight d subscript 1 equals vt subscript 1

rightwards double arrow

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Second echo is heard at time

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∴

2 straight d subscript 2 equals vt subscript 2

space space space straight d subscript 2 equals vt subscript 2 over 2 equals fraction numerator 340 cross times 3.6 over denominator 2 end fraction equals 612 straight m

Thus, the distance between two cliffs is given by,

straight d equals straight d subscript 1 plus straight d subscript 2 equals 408 plus 612 equals 1020 straight m

Some More Questions From Mechanical Properties of Fluids Chapter

What is wave?

Is it logical to say that wave motion is a process of propagation of mass?

What is common property of all types of waves?

What are different types of wave motions?

Define transverse wave motion.

Define longitudinal wave motion.

Which of the two waves - elastic waves or electromagnetic waves does need any material media for its propagation?

What is sound?

How sound is produced?

Can sound travel in vacuum?

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