A speaker emits a sound wave of frequency f_0 | vedclass

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Question

(B) According to the Doppler effect,when the source moves towards a stationary observer with speed $u$,the observed frequency is $f_1 = \frac{C}{C - u

Vedclass · Accepted Answer

$f_1 > f_2$

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(B) According to the Doppler effect,when the source moves towards a stationary observer with speed $u$,the observed frequency is $f_1 = \frac{C}{C - u} f_0$,where $C$ is the speed of sound.When the observer moves towards a stationary source with speed $u$,the observed frequency is $f_2 = \frac{C + u}{C} f_0$.Comparing the two expressions,we analyze the ratio $\frac{f_1}{f_2} = \frac{C}{C - u} \cdot \frac{C}{C + u} = \frac{C^2}{C^2 - u^2}$.Since $C^2 > C^2 - u^2$,it follows that $\frac{C^2}{C^2 - u^2} > 1$.Therefore,$f_1 > f_2$.

$A$ 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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