# Question #339ae

##### 1 Answer
Dec 10, 2016

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 $f$ and emitted frequency ${f}_{0}$ is given by Doppler's effect as

$f = \left(\frac{c + {v}_{r}}{c + {v}_{s}}\right) {f}_{0}$

$$where


$c \to \text{the velocity of waves in the medium}$

${v}_{r} \to \text{the velocity of the receiver relative to the medium}$ (positive if the receiver is moving towards the source (and negative in the other direction)

${v}_{s} \to \text{the velocity of the source relative to the medium}$
(positive if the source is moving away from the receiver (and negative in the other direction )

In our problem

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

${v}_{r} = - {v}_{r}$

${v}_{s} = + {v}_{r}$

${f}_{0} = 440 H z$

$f = 438 H z$

So

$f = \left(\frac{c + {v}_{r}}{c + {v}_{s}}\right) {f}_{0}$

$\implies 438 = \left(\frac{343.2 - {v}_{r}}{343.2 + {v}_{r}}\right) \times 440$

$878 {v}_{r} = \left(440 - 438\right) \times 343.2$

$\implies {v}_{r} = \frac{686.4}{878} \approx 0.782 \text{ m/s}$