Range of frequency heard by observer related to Doppler effect

[songoku]

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

 

[haruspex]

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

It is not made clear, but it seems you are to consider the instaneous position shown, not the whole transit.

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).

Only if on a line through the source.

maybe the sound is louder and louder due to amplitude is increasing but the frequency stays constant?

Amplitude is independent of frequency, but loudness is a matter of perception. The human auditory system is more sensitive to some frequencies than others.

 

[Orodruin]

The image shows the obsever travelling towards, but not straight at, the source. I would assume the question intends to ask about this particular setup.

This means that the observer approaches the source with a speed lower than $0.1v$. Since it us an approach, the frequency cannog be lowered.

 

[songoku]

It is not made clear, but it seems you are to consider the instaneous position shown, not the whole transit.

The image shows the obsever travelling towards, but not straight at, the source. I would assume the question intends to ask about this particular setup.

This means that the observer approaches the source with a speed lower than $0.1v$. Since it us an approach, the frequency cannog be lowered.

I understand.

Only if on a line through the source.

Do you mean in real life there can be some sound wave that are not on a line through the source so adding to the frequency heard by the observer?

Thanks

 

[Orodruin]

Do you mean in real life there can be some sound wave that are not on a line through the source so adding to the frequency heard by the observer?

He means the motion of the observer.

 

[songoku]

He means the motion of the observer.

If I imagine observer approaching source in a straight line in real life, the sound heard will be louder, not constant (not really sure whether the pitch will be higher too). Is this related to Doppler effect or is it something else?

Thanks

 

[haruspex]

(not really sure whether the pitch will be higher too).

The pitch only changes if the approach speed changes.

Is this related to Doppler effect

no

or is it something else?

The emitted sound spreads out in concentric spherical shells, so when a wave has reached distance r it is spread out over a shell of area $4\pi r^2$. (corrected)

 

[Steve4Physics]

The emitted sound spreads out in concentric spherical shells, so when a wave has reached distance r it is spread out over a shell of area $\pi r^2$.

I know it's only a typo' but (just in case it confuses the OP) that should be an area of $4\pi r^2$, the surface area of a sphere.

 

[songoku]

The pitch only changes if the approach speed changes.

no

The emitted sound spreads out in concentric spherical shells, so when a wave has reached distance r it is spread out over a shell of area $4\pi r^2$. (corrected)

I know it's only a typo' but (just in case it confuses the OP) that should be an area of $4\pi r^2$, the surface area of a sphere.

I get it. It is actually related to the intensity of the sound wave, not to the frequency. The frequency is still consistent with Doppler effect formula.

Thank you very much for the help and explanation haruspex, Orodruin, Steve4Physics