- #1

songoku

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- Homework Statement
- Please see below

- Relevant Equations
- ##f_2=\frac{v\pm v_o}{v\pm v_s}f_1##

My answer is (A) but the correct answer is (B).

My attempt:

$$f_2=\frac{v\pm v_o}{v\pm v_s}f_1$$

$$=\frac{v+0.1v}{v}f_o$$

$$=1.1f_o$$

If we consider the observer to move pass through the sound source and now is moving away from the stationary source, then:

$$f_2=\frac{v-0.1v}{v}f_o$$

$$=0.9f_o$$

My questions:

1) why the answer is not ##0.9f_o\leq f\leq1.1f_o##?

2) based on the doppler effect formula, the frequency heard by observer will be constant (higher) when the observer is approaching the source (or source approaching the observer) and will also be constant (lower) when observer is moving away from source (or source is moving away from observer). But in real life, when ambulance is approaching me, the sound heard will be louder and louder (not constant) as it is approaching. This seems to contradict the formula. Or maybe the sound is louder and louder due to amplitude is increasing but the frequency stays constant?

Thanks