Question

A man is standing between source and cliff. When he start moving along line joining him and source, he hears 10 beats per second. Velocity of man is : (Frequency of source = 60 Hz, Velocity of sound = 330 m/s)

Answer 1

Answer:

Given:

Man starts moving along a line joining the source and cliff. He gets to hear beat of 10 Hz. Frequency of sound = 60 Hz and velocity of sound = 330 m/s

To find:

Velocity of man

Concept :

We shall apply Doppler's Effect equation and beats equations simultaneously .

Calculation:

Let's assume that the person is moving towards the cliff.

So apparent frequency heard directly from source shall be f2.

$f2 = \bigg \{\dfrac{330 -v}{330 - 0} \bigg \}60$

Similarly , apparent frequency heard by observer after Reflection from the cliff be f3 :

$f3= \bigg \{\dfrac{330 + v}{330 - 0} \bigg \}60$

Now these 2 frequencies shall produce beats of 10 Hz.

$f3 - f2 = 10$

$= >\bigg \{\dfrac{330 + v}{330 - 0} \bigg \}60 - \bigg \{\dfrac{330 - v}{330 - 0} \bigg \}60= 10$

$= > \dfrac{330 + v - 330 + v}{330} = \dfrac{10}{60}$

$= > \dfrac{2v}{330} = \dfrac{1}{6}$

$= > v = \dfrac{330}{12}$

$= > v = 27.5 \: m {s}^{ - 1}$

So final answer is :

Velocity of man = 27.5 m/s

Answer 2

Velocity of man = 27.5 m/s

Explanation :

Given :

The hears 10 beats per seconds.

Velocity of man is : (Frequency of source = 60 Hz, Velocity of sound = 330 m/s.

To Find :

The Velocity of man .

Solution :

We know about Doppler's Effect equation.

In which,

For determining the frequency :

• ƒ = observed frequency,
• c = speed of sound.
• Vs = velocity of source

[ It will be negative if it's moving toward the observer. ]

• ƒ₀ = emitted frequency of source.

So,

Consider the :

The person is moving towards -  The bluff.

The frequency that can be heard directly from the source as - f₂

he frequency heard by observer after omitting from the bluff as - f₃

After finding f₂ & f₃ , These two frequencies shall produce beats of 10 Hz.

So,

• f₂ :

$\implies \sf \bigg\{\dfrac{330 -v}{330 - 0} \bigg\}\;60$

$\rule{300}{1.5}$

• f₃ :

$\implies \sf \bigg \{\dfrac{330 + v}{330 - 0} \bigg \}\;60$

Now :

• Beats of 10 Hz :

$\implies \sf f_3 - f_2 = 10\;Beats.\\ \\ \\\implies \sf \dfrac{330 + v - 330 + v}{330} = \dfrac{10}{60} \\ \\ \\\implies \sf \dfrac{2v}{330} = \dfrac{1}{6} \\ \\ \\\implies \sf v = \dfrac{330}{12} \\ \\ \\\implies \sf v = 27.5 \: m {s}^{ - 1}$

Hence,

The velocity of man will be 27.5 m/s .