SATHEE: Wave Motion 4 Question 3

Wave Motion 4 Question 3

3. A source of sound $S$ is moving with a velocity of $50 m / s$ towards a stationary observer. The observer measures the frequency of the source as $1000 H z$ . What will be the apparent frequency of the source when it is moving away from the observer after crossing him? (Take, velocity of sound in air is $350 m / s$ )

(2019 Main, 10 April II)

(a) $807 Hz$

(b) $1143 Hz$

(d) $857 Hz$

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Answer:

Correct Answer: 3. (c)

Solution:

Initially,

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When source is moving towards stationary observer, frequency observed is more than source frequency due to Doppler’s effect, it is given by

$f_{observed} = f (\frac{v}{v - v_{s}})$

where, $f =$ source frequency,

$f_{o} = observed frequency = 1000 H z,$

$v =$ speed of sound in air $= 350 m s^{- 1}$

and $v_{s} =$ speed of source $= 50 m s^{- 1}$

So, $f = \frac{f_{o b s} (v - v_{s})}{v} = \frac{1000 (350 - 50)}{350} = \frac{6000}{7} H z$

When source moves away from stationary observer, observed frequency will be lower due to Doppler’s effect and it is given by

$\begin{aligned} f_{0} & = f (\frac{v}{v + v_{s}}) = \frac{6000 \times 350}{7 \times (350 + 50)} \\ = \frac{6000 \times 350}{7 \times 400} = 750 H z \end{aligned}$

Wave Motion 4 Question 3

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Wave Motion 4 Question 3

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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