Two vehicles, each moving with speed u on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity w. One of these vehicles blows a whistle of frequency f1. An observer in the other vehicle hears the frequency of the whistle to be f2. The speed of sound in still air is V. The correct statement(s) is(are)

If the wind blows from the observer to the source, f2 > f1., If the wind blows from the source to the observer, f2 > f1.

If the wind blows from the observer to the source, f2 > f1.

If the wind blows from the source to the observer, f2 > f1.

If the wind blows from the observer to the source, f2 1

If the wind blows from the source to the observer, f2 1.

Waves PYQ — JEE Advanced 2013 · Shift 2 · Q49 with Solution

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Correct answer: A, B

Choose a positive direction

Let the positive direction be from the source vehicle to the observer vehicle at the instant considered.

Since the two vehicles are approaching each other with equal speed $u$ :

source moves toward observer: $v_s = +u$
observer moves toward source: $v_o = -u$

Wind speed relative to ground is $w$ along the road.

If wind blows from source to observer, then $w=+w$ .
If wind blows from observer to source, then $w=-w$ .

The speed of sound relative to ground becomes: $v_{\,\text{sound, ground}}=V+w$ for propagation from source to observer.

General Doppler formula in moving medium

In a medium moving with wind, use velocities relative to the medium.

Frequency heard is $f_2=f_1\frac{(V-v_o')}{(V-v_s')}$ where $v_o'$ and $v_s'$ are observer and source velocities relative to the medium, taken positive in the direction from source to observer.

Since medium moves with velocity $w$ , $v_s'=v_s-w=u-w$ $v_o'=v_o-w=-u-w$

Therefore,

=f_1\frac{V+u+w}{V-u+w}$$ This is for wind from source to observer ($w$ positive). If wind is opposite, substitute $w\to -w$: $$f_2=f_1\frac{V+u-w}{V-u-w}$$ This is for wind from observer to source. 3. **Case 1: wind from source to observer** Then $$\frac{f_2}{f_1}=\frac{V+u+w}{V-u+w}$$ Compare numerator and denominator: $$V+u+w > V-u+w$$ so $$\frac{f_2}{f_1}>1 \Rightarrow f_2>f_1$$ Hence **B is correct**. 4. **Case 2: wind from observer to source** Then $$\frac{f_2}{f_1}=\frac{V+u-w}{V-u-w}$$ Again, $$V+u-w > V-u-w$$ so $$\frac{f_2}{f_1}>1 \Rightarrow f_2>f_1$$ Hence **A is correct**. 5. **Check remaining options** Since in both wind directions we get $$f_2>f_1,$$ any option claiming $f_2<f_1$ (or equivalently $f_2/f_1<1$) is false. Thus: - A: correct - B: correct - C: false - D: false 6. **Final answer** The correct options are: $$\boxed{A,\ B}$$ This matches the stored correct answer.

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View written solutionFree

Correct answer: A, B

Choose a positive direction

Let the positive direction be from the source vehicle to the observer vehicle at the instant considered.

Since the two vehicles are approaching each other with equal speed $u$ :

source moves toward observer: $v_s = +u$
observer moves toward source: $v_o = -u$

Wind speed relative to ground is $w$ along the road.

If wind blows from source to observer, then $w=+w$ .
If wind blows from observer to source, then $w=-w$ .

The speed of sound relative to ground becomes: $v_{\,\text{sound, ground}}=V+w$ for propagation from source to observer.

General Doppler formula in moving medium

In a medium moving with wind, use velocities relative to the medium.

Frequency heard is $f_2=f_1\frac{(V-v_o')}{(V-v_s')}$ where $v_o'$ and $v_s'$ are observer and source velocities relative to the medium, taken positive in the direction from source to observer.

Since medium moves with velocity $w$ , $v_s'=v_s-w=u-w$ $v_o'=v_o-w=-u-w$

Therefore,

=f_1\frac{V+u+w}{V-u+w}$$ This is for wind from source to observer ($w$ positive). If wind is opposite, substitute $w\to -w$: $$f_2=f_1\frac{V+u-w}{V-u-w}$$ This is for wind from observer to source. 3. **Case 1: wind from source to observer** Then $$\frac{f_2}{f_1}=\frac{V+u+w}{V-u+w}$$ Compare numerator and denominator: $$V+u+w > V-u+w$$ so $$\frac{f_2}{f_1}>1 \Rightarrow f_2>f_1$$ Hence **B is correct**. 4. **Case 2: wind from observer to source** Then $$\frac{f_2}{f_1}=\frac{V+u-w}{V-u-w}$$ Again, $$V+u-w > V-u-w$$ so $$\frac{f_2}{f_1}>1 \Rightarrow f_2>f_1$$ Hence **A is correct**. 5. **Check remaining options** Since in both wind directions we get $$f_2>f_1,$$ any option claiming $f_2<f_1$ (or equivalently $f_2/f_1<1$) is false. Thus: - A: correct - B: correct - C: false - D: false 6. **Final answer** The correct options are: $$\boxed{A,\ B}$$ This matches the stored correct answer.

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