Question #339ae

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Question #339ae

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Answer 1 · 2016-12-10T04:35:44

In classical physics, where the speeds of source and the receiver relative to the medium are lower than the velocity of waves in the medium, the relationship between observed frequency $f$ $f$ and emitted frequency $f_{0}$ $f_0$ is given by Doppler's effect as

$f = (\frac{c + v_{r}}{c + v_{s}}) f_{0}$ $f=((c+v_r)/(c+v_s))f_0$

where

$c \to the velocity of waves in the medium$ $c->"the velocity of waves in the medium"$

$v_{r} \to the velocity of the receiver relative to the medium$ $v_ r-> "the velocity of the receiver relative to the medium"$ (positive if the receiver is moving towards the source (and negative in the other direction)

$v_{s} \to the velocity of the source relative to the medium$ $v_s ->"the velocity of the source relative to the medium"$
(positive if the source is moving away from the receiver (and negative in the other direction )

In our problem

$c = 343.2 m s^{- 1}$ $c=343.2ms^-1$

$v_{r} = - v_{r}$ $v_r=-v_r$

$v_{s} = + v_{r}$ $v_s=+v_r$

$f_{0} = 440 H z$ $f_0=440Hz$

$f = 438 H z$ $f=438Hz$

So

$f = (\frac{c + v_{r}}{c + v_{s}}) f_{0}$ $f=((c+v_r)/(c+v_s))f_0$

$\Rightarrow 438 = (\frac{343.2 - v_{r}}{343.2 + v_{r}}) \times 440$ $=>438=((343.2-v_r)/(343.2+v_r))xx440$

$878 v_{r} = (440 - 438) \times 343.2$ $878v_r=(440-438)xx343.2$

$\Rightarrow v_{r} = \frac{686.4}{878} \approx 0.782 m/s$ $=>v_r=686.4/878~~0.782" m/s"$

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Question #339ae

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