A speaker emits a sound wave of frequency f_0 | vedclass

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(B) When the observer is stationary and the source is moving towards the observer,the observed frequency is given by the Doppler effect formula:$f_1 =

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$f_1 > f_2$

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(B) When the observer is stationary and the source is moving towards the observer,the observed frequency is given by the Doppler effect formula:$f_1 = f_0 \left( \frac{v}{v - u} \right)$where $v$ is the speed of sound and $u$ is the speed of the source.When the source is stationary and the observer is moving towards the source,the measured frequency is:$f_2 = f_0 \left( \frac{v + u}{v} \right)$To compare $f_1$ and $f_2$,we look at the ratio:$f_1 = f_0 \left( 1 - \frac{u}{v} \right)^{-1} \approx f_0 \left( 1 + \frac{u}{v} + \frac{u^2}{v^2} + \dots \right)$$f_2 = f_0 \left( 1 + \frac{u}{v} \right)$Comparing the two,$f_1 = f_0 \left( \frac{v}{v-u} \right)$ and $f_2 = f_0 \left( \frac{v+u}{v} \right) = f_0 \left( 1 + \frac{u}{v} \right)$.Since $\frac{v}{v-u} > 1 + \frac{u}{v}$ (because $\frac{v^2}{v(v-u)} > \frac{v^2-u^2}{v(v-u)}$),it follows that $f_1 > f_2$.

$A$ speaker emits a sound wave of frequency $f_0$. When it moves towards a stationary observer with speed $u$,the observer measures a frequency $f_1$. If the speaker is stationary and the observer moves towards it with speed $u$,the measured frequency is $f_2$. Then,

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