Doppler’s Effect Questions in English

(D) The apparent frequency $f'$ observed due to the Doppler effect is given by the formula $f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right)$,where $f$ is the source frequency,$v$ is the speed of sound,$v_o$ is the observer velocity,and $v_s$ is the source velocity.
From this expression,it is clear that the Doppler shift depends on the source frequency,the velocity of the source,and the velocity of the observer.
However,the formula does not contain any term representing the distance between the source and the observer.
Therefore,the Doppler shift is independent of the distance between the source and the listener.

(B) The apparent frequency $n'$ heard by a stationary observer when the source is moving towards them is given by the Doppler effect formula:
$n' = n \left( \frac{v}{v - v_s} \right)$
Where:
$n = 450 \text{ cycles/sec}$ (source frequency)
$v = 340 \text{ m/sec}$ (speed of sound)
$v_s = 34 \text{ m/sec}$ (speed of source)
Substituting the values:
$n' = 450 \left( \frac{340}{340 - 34} \right)$
$n' = 450 \left( \frac{340}{306} \right)$
$n' = 450 \times 1.111... = 500 \text{ cycles/sec}$.
Thus,the apparent frequency is $500 \text{ cycles/sec}$.

(A) The wavelength $\lambda$ of a sound wave emitted by a stationary source is given by $\lambda = v/n$,where $v$ is the speed of sound and $n$ is the frequency.
When the source moves towards a stationary observer with velocity $v_s$,the frequency heard by the observer is $n' = n \left( \frac{v}{v - v_s} \right)$.
The wavelength observed by the observer is $\lambda' = \frac{v}{n'} = \frac{v}{n \left( \frac{v}{v - v_s} \right)} = \lambda \left( \frac{v - v_s}{v} \right)$.
Given $\lambda = 120 \, cm$,$v = 330 \, m/s$,and $v_s = 60 \, m/s$:
$\lambda' = 120 \left( \frac{330 - 60}{330} \right) = 120 \left( \frac{270}{330} \right) = 120 \left( \frac{9}{11} \right) \approx 98.18 \, cm$.
Rounding to the nearest integer provided in the options,the value is $98 \, cm$.

(B) According to the Doppler effect,when a source moves towards a stationary observer,the apparent frequency $n'$ is given by the formula:
$n' = n \left( \frac{v}{v - v_s} \right)$
Where:
$n = 600 \, Hz$ (actual frequency)
$v = 330 \, m/s$ (velocity of sound)
$v_s = 30 \, m/s$ (velocity of source)
Substituting the values:
$n' = 600 \left( \frac{330}{330 - 30} \right)$
$n' = 600 \left( \frac{330}{300} \right)$
$n' = 600 \times 1.1 = 660 \, Hz$
Therefore,the apparent frequency is $660 \, Hz$.

(C) According to the Doppler effect,the frequency $f'$ heard by an observer moving away from a stationary source is given by $f' = f \left( \frac{V - v_o}{V} \right)$,where $V$ is the speed of sound and $v_o$ is the speed of the observer.
Since both observers are moving away from the source with the same speed $v_o = 0.2V$,the frequency heard by both observers is $f' = f \left( \frac{V - 0.2V}{V} \right) = f \left( \frac{0.8V}{V} \right) = 0.8f$.
Therefore,the ratio of the frequencies heard by the two observers is $0.8f : 0.8f = 1:1$.

(B) According to the Doppler effect,the apparent frequency $n'$ is given by the formula:
$n' = n \left( \frac{v - v_o}{v - v_s} \right)$
Here,$v$ is the velocity of sound,$v_o$ is the velocity of the observer,and $v_s$ is the velocity of the source.
Since both the source and the observer are moving in the same direction as the sound wave propagation,their velocities are taken as positive relative to the direction of sound.
Substituting these into the general Doppler formula,we get $n' = n \frac{(v - v_o)}{(v - v_s)}$.

(C) According to the Doppler effect,when an observer moves towards a stationary source,the apparent frequency $n'$ is given by the formula:
$n' = n \left( \frac{v + v_O}{v} \right)$
where $v$ is the velocity of sound and $v_O$ is the velocity of the observer.
Given that $n' = 2n$ and $v = 332 \, m/s$,we substitute these values into the equation:
$2n = n \left( \frac{332 + v_O}{332} \right)$
Dividing both sides by $n$:
$2 = \frac{332 + v_O}{332}$
$664 = 332 + v_O$
$v_O = 664 - 332 = 332 \, m/s$.
Therefore,the velocity of the observer is $332 \, m/s$.

(B) According to the Doppler effect,when an observer moves towards a stationary source,the apparent frequency $n'$ is given by the formula:
$n' = n \left( \frac{v + v_O}{v} \right)$
Where $n$ is the actual frequency,$v$ is the speed of sound,and $v_O$ is the velocity of the observer.
Since the observer is moving towards the source,$v_O > 0$.
Therefore,the factor $\frac{v + v_O}{v} > 1$.
This implies that $n' > n$,meaning the apparent frequency is greater than the real frequency.

(A) The number of waves per second received by an observer is the apparent frequency $f'$.
Given the source frequency $f = 256 \text{ Hz}$.
The source velocity $v_s = \frac{v}{3}$,where $v$ is the velocity of sound.
According to the Doppler effect,when the source approaches a stationary observer,the apparent frequency is given by:
$f' = f \left( \frac{v}{v - v_s} \right)$
Substituting the values:
$f' = 256 \left( \frac{v}{v - v/3} \right)$
$f' = 256 \left( \frac{v}{2v/3} \right)$
$f' = 256 \times \frac{3}{2}$
$f' = 128 \times 3 = 384 \text{ waves per second}$.

(A) According to the Doppler effect,when the source moves towards a stationary observer,the observed frequency $n'$ is given by $n' = n \left( \frac{v}{v - v_s} \right)$,where $v$ is the speed of sound and $v_s$ is the speed of the source.
Given the frequency difference is $2.5\%$,we have $\frac{n' - n}{n} = 0.025$,which implies $\frac{n'}{n} = 1.025$.
Substituting this into the Doppler formula: $1.025 = \frac{v}{v - v_s}$.
Given $v = 320\, m/s$,we have $1.025 = \frac{320}{320 - v_s}$.
$320 - v_s = \frac{320}{1.025} \approx 312.195$.
$v_s = 320 - 312.195 = 7.805\, m/s$.
Rounding to the nearest integer,the velocity of the car is approximately $8\, m/s$.

(B) Given: Speed of both trains $v_s = v_o = 108\, km/h = 108 \times \frac{5}{18} = 30\, m/s$.
Frequency of the source $n = 750\, Hz$.
Speed of sound $v = 330\, m/s$.
After the trains cross each other,the source and the observer are moving away from each other.
Using the Doppler effect formula for this condition:
$n' = n \left( \frac{v - v_o}{v + v_s} \right)$
Substituting the values:
$n' = 750 \left( \frac{330 - 30}{330 + 30} \right)$
$n' = 750 \left( \frac{300}{360} \right)$
$n' = 750 \times \frac{5}{6} = 125 \times 5 = 625\, Hz$.

(A) The general formula for Doppler's effect is $n' = n \left( \frac{v + v_O}{v - v_S} \right)$,where $n'$ is the observed frequency,$n$ is the source frequency,$v$ is the speed of sound,$v_O$ is the velocity of the observer,and $v_S$ is the velocity of the source.
Given that the source is stationary,$v_S = 0$.
We want the observed frequency to be double the source frequency,so $n' = 2n$.
Substituting these values into the formula: $2n = n \left( \frac{v + v_O}{v - 0} \right)$.
Dividing both sides by $n$: $2 = \frac{v + v_O}{v}$.
Multiplying by $v$: $2v = v + v_O$.
Solving for $v_O$: $v_O = v$.
Since the observer is moving towards the source to hear a higher frequency,the velocity of the observer is equal to the velocity of sound directed towards the source.

(D) The Doppler effect formula for apparent frequency $f'$ is given by $f' = f \left( \frac{v_s \pm v_o}{v_s \mp v_s} \right)$,where $v_s$ is the velocity of sound,$v_o$ is the velocity of the observer,and $v_s$ is the velocity of the source.
Given that the source moves towards the observer with velocity $v = v_s$ and the observer moves away from the source with velocity $v = v_s$.
Using the sign convention: $f' = f \left( \frac{v_s - v}{v_s - v} \right)$.
Since $v = v_s$,the expression becomes $f' = f \left( \frac{v_s - v_s}{v_s - v_s} \right)$.
However,in the case where both source and observer move with the same velocity in the same direction,there is no relative motion between them.
Therefore,the relative velocity is zero,and the apparent frequency heard by the observer remains equal to the original frequency.
Thus,$f' = 200 Hz$.

(C) The Doppler effect describes the change in frequency of a wave in relation to an observer who is moving relative to the wave source. When the velocity of the sound source $(v_s)$ exceeds the velocity of sound in the medium $(v)$,the source moves faster than the waves it produces. This results in the formation of a shock wave (sonic boom),and the standard Doppler effect formulas are no longer applicable as the wave fronts overlap in a way that prevents the observer from receiving the waves in the expected sequence.

(A) According to the Doppler effect,when a source of sound moves towards a stationary observer,the observed frequency $n'$ is given by the formula:
$n' = n \left( \frac{v}{v - v_s} \right)$
where $n$ is the original frequency,$v$ is the velocity of sound,and $v_s$ is the velocity of the source.
Given that the observed frequency is three times the original frequency,we have $n' = 3n$.
Substituting this into the formula:
$3n = n \left( \frac{v}{v - v_s} \right)$
$3 = \frac{v}{v - v_s}$
$3(v - v_s) = v$
$3v - 3v_s = v$
$2v = 3v_s$
$v_s = \frac{2}{3}v$
Therefore,the speed of the source is $\frac{2}{3}v$.

(C) According to the Doppler effect,when a source of sound approaches a stationary observer,the apparent frequency $n'$ is given by the formula:
$n' = n \left( \frac{v}{v - v_s} \right)$
Where:
$n = 1200 \, cps$ (actual frequency)
$v = 350 \, m/s$ (speed of sound)
$v_s = 50 \, m/s$ (speed of the source)
Substituting the values:
$n' = 1200 \times \left( \frac{350}{350 - 50} \right)$
$n' = 1200 \times \left( \frac{350}{300} \right)$
$n' = 1200 \times \frac{7}{6} = 200 \times 7 = 1400 \, cps$
Therefore,the apparent frequency heard by the observer is $1400 \, cps$.

(D) According to the Doppler effect,when a source of sound approaches a stationary observer,the apparent frequency $n'$ is given by the formula:
$n' = n \left( \frac{v}{v - v_s} \right)$
Where:
$n = 1200 \ Hz$ (source frequency)
$v = 400 \ m/s$ (speed of sound)
$v_s = 100 \ m/s$ (speed of the source)
Substituting the values into the formula:
$n' = 1200 \left( \frac{400}{400 - 100} \right)$
$n' = 1200 \left( \frac{400}{300} \right)$
$n' = 1200 \times \frac{4}{3}$
$n' = 400 \times 4 = 1600 \ Hz$
Therefore,the apparent frequency heard by the observer is $1600 \ Hz$.

(A) According to the Doppler effect,when a source moves towards a stationary observer,the apparent frequency $n'$ is given by the formula:
$n' = n \times \left( \frac{v}{v - v_s} \right)$
Given:
Frequency of source $n = 150 \, Hz$
Velocity of sound $v = 330 \, m/s$
Velocity of source $v_s = 110 \, m/s$
Substituting the values:
$n' = 150 \times \left( \frac{330}{330 - 110} \right)$
$n' = 150 \times \left( \frac{330}{220} \right)$
$n' = 150 \times 1.5 = 225 \, Hz$
Therefore,the frequency heard by the person is $225 \, Hz$.

(A) When the source is approaching the stationary observer,the observed frequency $n_a$ is given by the Doppler effect formula:
$n_a = \left( \frac{v}{v - v_s} \right) n$
Substituting the values $v = 340\, m/s$,$v_s = 20\, m/s$,and $n = 1000\, Hz$:
$n_a = \left( \frac{340}{340 - 20} \right) \times 1000 = \frac{340}{320} \times 1000 = \frac{17}{16} \times 1000 = 1062.5\, Hz$
When the source is receding from the observer,the observed frequency $n_r$ is:
$n_r = \left( \frac{v}{v + v_s} \right) n$
$n_r = \left( \frac{340}{340 + 20} \right) \times 1000 = \frac{340}{360} \times 1000 = \frac{17}{18} \times 1000 = 944.4\, Hz$
The ratio of the frequencies is:
$\frac{n_a}{n_r} = \frac{\frac{v}{v - v_s} n}{\frac{v}{v + v_s} n} = \frac{v + v_s}{v - v_s}$
$\frac{n_a}{n_r} = \frac{340 + 20}{340 - 20} = \frac{360}{320} = \frac{9}{8}$
Thus,the ratio is $9:8$.

(A) The apparent frequency $n'$ heard by a stationary observer when the source is moving with velocity $v_s$ is given by the Doppler effect formula: $n' = n \left( \frac{v}{v \mp v_s} \right)$.
When the source is approaching the observer,the apparent frequency is $n_a = n \left( \frac{v}{v - v_s} \right) = 1000 \ Hz$.
When the source is moving away from the observer,the apparent frequency is $n_r = n \left( \frac{v}{v + v_s} \right)$.
Dividing the two expressions: $\frac{n_a}{n_r} = \frac{v + v_s}{v - v_s}$.
Substituting the given values: $\frac{1000}{n_r} = \frac{350 + 50}{350 - 50} = \frac{400}{300} = \frac{4}{3}$.
Therefore,$n_r = 1000 \times \frac{3}{4} = 750 \ Hz$.

(B) According to the Doppler effect,when both the source and the listener are moving towards each other,the apparent frequency $f'$ heard by the listener is given by the formula:
$f' = f \left( \frac{v + v_O}{v - v_S} \right)$
Here,$v_O$ is the velocity of the observer (listener) and $v_S$ is the velocity of the source.
Given that $v_O = v/10$ and $v_S = v/10$,we substitute these values into the formula:
$f' = f \left( \frac{v + v/10}{v - v/10} \right)$
$f' = f \left( \frac{1.1v}{0.9v} \right)$
$f' = f \left( \frac{11}{9} \right)$
$f' \approx 1.22 f$
Thus,the frequency heard by the listener is approximately $1.22 f$.

(C) The source is moving in a circular path. The linear speed of the source is given by $v_S = r\omega$.
Given $r = 70 \text{ cm} = 0.70 \text{ m}$ and frequency of revolution $f_{rev} = 5 \text{ rev/s}$.
Angular velocity $\omega = 2\pi f_{rev} = 2 \times \pi \times 5 = 10\pi \text{ rad/s}$.
Thus,$v_S = 0.70 \times 10\pi = 7\pi \approx 7 \times 3.1416 = 21.99 \approx 22 \text{ m/s}$.
The minimum frequency is heard when the source is moving directly away from the listener.
Using the Doppler effect formula for a receding source: $n_{\min} = n \left( \frac{v}{v + v_S} \right)$.
Substituting the values: $n_{\min} = 1000 \times \left( \frac{352}{352 + 22} \right) = 1000 \times \left( \frac{352}{374} \right) \approx 941.17 \text{ Hz}$.
Rounding to the nearest integer,the minimum frequency is $941 \text{ Hz}$.

(B) The observer hears two sounds: one directly from the source and one reflected from the wall (echo).
$1$. For the direct sound,the source is moving away from the stationary observer. The frequency heard $(n_1)$ is given by the Doppler effect formula:
$n_1 = n \left( \frac{v}{v + v_s} \right) = 500 \left( \frac{332}{332 + 2} \right) = 500 \left( \frac{332}{334} \right) \ Hz$
$2$. For the reflected sound (echo),the wall acts as a stationary source reflecting the sound. Effectively,the image of the source is moving towards the observer at a speed of $v_s = 2 \ m/s$. The frequency heard $(n_2)$ is:
$n_2 = n \left( \frac{v}{v - v_s} \right) = 500 \left( \frac{332}{332 - 2} \right) = 500 \left( \frac{332}{330} \right) \ Hz$
$3$. The beat frequency is the difference between these two frequencies:
$\text{Beat frequency} = n_2 - n_1 = 500 \times 332 \left( \frac{1}{330} - \frac{1}{334} \right)$
$= 500 \times 332 \left( \frac{334 - 330}{330 \times 334} \right) = 500 \times 332 \left( \frac{4}{110220} \right) \approx 6.02 \ Hz$
Thus,the number of beats per second is approximately $6$.

(C) The frequency of the reflected sound heard by the driver is given by the Doppler effect formula for a moving source and a moving observer.
Here,the car acts as both the source and the observer.
The velocity of the car $v_s = v_o = 72\,km/hr = 72 \times \frac{5}{18} = 20\,m/s$.
The velocity of sound $v = 330\,m/s$.
The frequency of the horn $n = 124\,vib/sec$.
The formula for the reflected frequency $n'$ is:
$n' = n \left( \frac{v + v_o}{v - v_s} \right)$
Substituting the values:
$n' = 124 \left( \frac{330 + 20}{330 - 20} \right)$
$n' = 124 \left( \frac{350}{310} \right)$
$n' = 124 \times \frac{35}{31} = 4 \times 35 = 140\,vib/sec$.

(D) According to the Doppler effect,when a source of sound moves towards a stationary observer,the apparent frequency $n_1$ is given by the formula:
$n_1 = n \left( \frac{V}{V - S} \right)$
where $n$ is the actual frequency,$V$ is the speed of sound in air,and $S$ is the speed of the source.
Dividing both sides by $n$,we get:
$\frac{n_1}{n} = \frac{V}{V - S}$

(B) The Doppler effect occurs when there is a relative velocity between the source and the observer along the line joining them.
In this problem,the vehicle is moving in a direction perpendicular to the line joining the observer and the vehicle.
Therefore,the component of the velocity of the source along the line joining the observer and the source is $v_s \cos(90^{\circ}) = 0$.
Since there is no relative motion along the line of sight,the frequency perceived by the observer remains the same as the source frequency.
Thus,the observed frequency is $n' = n$.
Given that the observed frequency is $n + n_1$,we have $n + n_1 = n$,which implies $n_1 = 0$.

(D) The apparent frequency $n'$ heard by the observer is given by the Doppler effect formula for a source moving towards a stationary observer:
$n' = n \left( \frac{v}{v - v_s} \right)$
Given:
Source frequency $n = 450 Hz$
Speed of sound $v = 330 m/s$
Speed of source $v_s = 33 m/s$
Substituting the values:
$n' = 450 \times \left( \frac{330}{330 - 33} \right)$
$n' = 450 \times \left( \frac{330}{297} \right)$
$n' = 450 \times \frac{10}{9}$
$n' = 50 \times 10 = 500 Hz$
Therefore,the frequency heard by the observer is $500 Hz$.

(A) The Doppler effect formula for an observer moving away from a stationary source is given by:
$n' = n \left( \frac{v - v_O}{v} \right)$
Where:
$n = 100 \, Hz$ (source frequency)
$v = 330 \, m/s$ (speed of sound)
$v_O = 33 \, m/s$ (speed of observer moving away)
Substituting the values:
$n' = 100 \left( \frac{330 - 33}{330} \right)$
$n' = 100 \left( \frac{297}{330} \right)$
$n' = 100 \times 0.9 = 90 \, Hz$
Thus,the observed frequency is $90 \, Hz$.

(C) When the train is approaching,the frequency heard by the observer is given by the Doppler effect formula: $n_a = n \left( \frac{v}{v - v_s} \right)$.
Substituting the values: $219 = n \left( \frac{340}{340 - v_s} \right)$ ... $(i)$.
When the train is receding,the frequency heard by the observer is: $n_r = n \left( \frac{v}{v + v_s} \right)$.
Substituting the values: $184 = n \left( \frac{340}{340 + v_s} \right)$ ... $(ii)$.
Dividing equation $(i)$ by $(ii)$:
$\frac{219}{184} = \frac{340 + v_s}{340 - v_s}$.
$219(340 - v_s) = 184(340 + v_s)$.
$74460 - 219v_s = 62560 + 184v_s$.
$11900 = 403v_s$.
$v_s \approx 29.53 \, m/s \approx 29.5 \, m/s$.
Substituting $v_s$ back into equation $(i)$:
$219 = n \left( \frac{340}{340 - 29.5} \right) = n \left( \frac{340}{310.5} \right)$.
$n = \frac{219 \times 310.5}{340} \approx 200 \, Hz$.

(D) The observer hears two sounds: one directly from the source and one reflected from the wall.
Let $v = 340\, m/s$ be the speed of sound and $v_s = 1\, m/s$ be the speed of the source (the boy).
$1$. Frequency heard directly from the source $(n_1)$: Since the source is moving towards the observer,the frequency is given by $n_1 = n_0 \left( \frac{v}{v - v_s} \right) = 680 \left( \frac{340}{340 - 1} \right) = 680 \left( \frac{340}{339} \right) \approx 682\, Hz$.
$2$. Frequency heard from the reflection off the wall $(n_2)$: The wall acts as a stationary source moving away from the observer. The frequency reflected by the wall is $n' = n_0 \left( \frac{v}{v + v_s} \right)$. This reflected sound then travels to the observer. Thus,$n_2 = 680 \left( \frac{340}{340 + 1} \right) = 680 \left( \frac{340}{341} \right) \approx 678\, Hz$.
$3$. The number of beats per second is the difference between the two frequencies: $n_1 - n_2 = 680 \left( \frac{340}{339} - \frac{340}{341} \right) = 680 \times 340 \left( \frac{341 - 339}{339 \times 341} \right) = 680 \times 340 \left( \frac{2}{115600 - 1} \right) \approx 680 \times 340 \times \frac{2}{115600} = 680 \times \frac{680}{115600} = 4\, Hz$.

(A) The problem involves the Doppler effect with a moving source and a moving observer (the driver).
$1$. First,the sound from the horn travels to the hill. The hill acts as a stationary observer receiving sound from a source (the car) moving towards it at $v_s = 30 \ m/s$. The frequency $n_1$ received by the hill is:
$n_1 = n \left( \frac{v}{v - v_s} \right) = 600 \left( \frac{330}{330 - 30} \right) = 600 \left( \frac{330}{300} \right) = 660 \ Hz$.
$2$. The hill reflects this sound,acting as a stationary source of frequency $n_1 = 660 \ Hz$. The driver (observer) is moving towards the hill at $v_o = 30 \ m/s$. The frequency $n'$ heard by the driver is:
$n' = n_1 \left( \frac{v + v_o}{v} \right) = 660 \left( \frac{330 + 30}{330} \right) = 660 \left( \frac{360}{330} \right) = 2 \times 360 = 720 \ Hz$.
Alternatively,using the combined formula for a moving observer and source towards a stationary reflector:
$n' = n \left( \frac{v + v_o}{v - v_s} \right) = 600 \left( \frac{330 + 30}{330 - 30} \right) = 600 \left( \frac{360}{300} \right) = 720 \ Hz$.

(B) The observer is moving away from siren $1$ and towards siren $2$.
The frequency of sound heard from siren $1$ is given by the Doppler effect formula for a moving observer moving away from a stationary source:
$n_1 = n \left( \frac{v - v_0}{v} \right) = 330 \left( \frac{330 - 2}{330} \right) = 328 \ Hz$
The frequency of sound heard from siren $2$ is given by the Doppler effect formula for a moving observer moving towards a stationary source:
$n_2 = n \left( \frac{v + v_0}{v} \right) = 330 \left( \frac{330 + 2}{330} \right) = 332 \ Hz$
The beat frequency is the difference between the two observed frequencies:
$\text{Beat frequency} = n_2 - n_1 = 332 - 328 = 4 \ Hz$.

(D) The formula for the apparent frequency $(n')$ when the source and listener are approaching each other is given by:
$n' = n \left( \frac{v + v_O}{v - v_S} \right)$
Where:
$n'$ = apparent frequency = $400 \, cps$
$n$ = true frequency
$v$ = velocity of sound = $360 \, m/s$
$v_O$ = velocity of observer = $40 \, m/s$
$v_S$ = velocity of source = $40 \, m/s$
Substituting the values into the formula:
$400 = n \left( \frac{360 + 40}{360 - 40} \right)$
$400 = n \left( \frac{400}{320} \right)$
$400 = n \left( \frac{5}{4} \right)$
$n = 400 \times \frac{4}{5} = 320 \, cps$
Therefore,the true frequency is $320 \, cps$.

(A) According to the Doppler effect,when the source is moving away from a stationary observer,the observed frequency $f'$ is given by the formula:
$f' = f \left( \frac{v}{v + v_s} \right)$
Where:
$f = 500 \; Hz$ (source frequency)
$v = 330 \; m/s$ (speed of sound)
$v_s = 50 \; m/s$ (speed of the source)
Substituting the values:
$f' = 500 \times \left( \frac{330}{330 + 50} \right)$
$f' = 500 \times \left( \frac{330}{380} \right)$
$f' = 500 \times 0.8684$
$f' \approx 434.2 \; Hz$

(C) According to the Doppler effect,the apparent frequency heard by an observer depends on the relative motion between the source of sound and the observer.
In this scenario,the man (observer) and the engine (source) are both part of the same moving train.
Since they are moving together at the same velocity,there is no relative motion between the source and the observer.
Therefore,the frequency heard by the man remains equal to the actual frequency of the whistle,which is $600 \ Hz$.

(A) According to the Doppler effect,when a source moves towards a stationary observer,the apparent frequency $n'$ is given by the formula:
$n' = n \left( \frac{v}{v - v_S} \right)$
Where:
$n = 500 Hz$ (source frequency)
$v = 330 m/s$ (speed of sound)
$v_S = 30 m/s$ (velocity of source)
Substituting the values:
$n' = 500 \left( \frac{330}{330 - 30} \right)$
$n' = 500 \left( \frac{330}{300} \right)$
$n' = 500 \times 1.1 = 550 Hz$
Therefore,the frequency heard by the observer is $550 Hz$.

(C) According to the Doppler effect,when a source of sound approaches a stationary observer,the observed frequency $n'$ is given by the formula:
$n' = n \left( \frac{v}{v - v_s} \right)$
Where:
$n = 90 \text{ vibrations/sec}$ (source frequency)
$v = \text{speed of sound}$
$v_s = \frac{v}{10}$ (speed of the source)
Substituting the values:
$n' = 90 \left( \frac{v}{v - \frac{v}{10}} \right)$
$n' = 90 \left( \frac{v}{\frac{9v}{10}} \right)$
$n' = 90 \times \frac{10}{9} = 100 \text{ vibrations/sec}$.

(A) The linear velocity of the whistle is given by $v_S = r\omega$.
Given $r = 1.2 \, m$ and angular velocity $\omega = 2\pi \times \frac{400}{60} \, rad/s$.
$v_S = 1.2 \times 2\pi \times \frac{400}{60} = 1.2 \times 2\pi \times \frac{20}{3} = 16\pi \approx 50.26 \, m/s$. Using $v_S \approx 50 \, m/s$ as per standard approximation.
When the whistle approaches the listener,the observed frequency is maximum: $n_{max} = n \left( \frac{v}{v - v_S} \right) = 500 \left( \frac{340}{340 - 50} \right) = 500 \left( \frac{340}{290} \right) \approx 586 \, Hz$.
When the whistle moves away from the listener,the observed frequency is minimum: $n_{min} = n \left( \frac{v}{v + v_S} \right) = 500 \left( \frac{340}{340 + 50} \right) = 500 \left( \frac{340}{390} \right) \approx 436 \, Hz$.
Thus,the range of frequencies heard is $436 \, Hz$ to $586 \, Hz$.

(D) According to the Doppler effect,when a source moves towards a stationary observer,the observed frequency $f'$ is given by $f' = f \left( \frac{v}{v - v_s} \right)$,where $v$ is the speed of sound and $v_s$ is the speed of the source.
For the first case,$v_s = 34 \ m/s$:
$f_1 = f \left( \frac{340}{340 - 34} \right) = f \left( \frac{340}{306} \right)$
For the second case,$v_s = 17 \ m/s$:
$f_2 = f \left( \frac{340}{340 - 17} \right) = f \left( \frac{340}{323} \right)$
Now,calculating the ratio $f_1/f_2$:
$\frac{f_1}{f_2} = \frac{f \left( \frac{340}{306} \right)}{f \left( \frac{340}{323} \right)} = \frac{323}{306}$
Dividing both numerator and denominator by $17$:
$\frac{323 \div 17}{306 \div 17} = \frac{19}{18}$
Thus,the ratio $f_1/f_2$ is $19/18$.

(D) The frequency heard by an observer is given by the Doppler effect formula: $f' = f \left( \frac{v + v_o}{v - v_s} \right)$,where $v$ is the speed of sound,$v_o$ is the velocity of the observer,and $v_s$ is the velocity of the source.
Since the source and the observer are both at rest,$v_o = 0$ and $v_s = 0$.
Substituting these values into the formula,we get $f' = f \left( \frac{v + 0}{v - 0} \right) = f \left( \frac{v}{v} \right) = f$.
Therefore,the observed frequency $f'$ is equal to the source frequency $f$,regardless of the speed of sound $v$.
Thus,the frequency heard by the observer will not change.

(B) According to the Doppler effect,when the source and observer move away from each other,the observed frequency $n'$ is given by the formula:
$n' = n \left( \frac{v - v_O}{v + v_S} \right)$
Here,$v = 340\; m/s$ is the speed of sound,$v_O = 10\; m/s$ is the velocity of the observer,and $v_S = 10\; m/s$ is the velocity of the source.
Given $n' = 1950\; Hz$,we substitute the values:
$1950 = n \left( \frac{340 - 10}{340 + 10} \right)$
$1950 = n \left( \frac{330}{350} \right)$
$1950 = n \left( \frac{33}{35} \right)$
$n = 1950 \times \frac{35}{33} \approx 2068.18\; Hz$
Rounding to the nearest whole number,the actual frequency is $2068\; Hz$.

(B) According to the Doppler effect,the apparent frequency $n'$ is given by the formula:
$n' = n \left( \frac{v + v_O}{v - v_S} \right)$
Where:
$n = 240 \ Hz$ (source frequency)
$v = 340 \ m/s$ (velocity of sound)
$v_O = 20 \ m/s$ (velocity of observer moving towards the source)
$v_S = 20 \ m/s$ (velocity of source moving towards the observer)
Substituting the values:
$n' = 240 \left( \frac{340 + 20}{340 - 20} \right)$
$n' = 240 \left( \frac{360}{320} \right)$
$n' = 240 \times 1.125 = 270 \ Hz$.

(B) According to the Doppler effect,when an observer moves towards a stationary source,the observed frequency $n'$ is given by:
$n' = n \left( \frac{v + v_0}{v} \right)$
where $n$ is the source frequency,$v$ is the speed of sound,and $v_0$ is the speed of the observer.
For train $A$:
$5.5 = 5 \left( \frac{v + v_A}{v} \right) \implies 1.1 = 1 + \frac{v_A}{v} \implies \frac{v_A}{v} = 0.1$ ... $(i)$
For train $B$:
$6.0 = 5 \left( \frac{v + v_B}{v} \right) \implies 1.2 = 1 + \frac{v_B}{v} \implies \frac{v_B}{v} = 0.2$ ... $(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{v_B / v}{v_A / v} = \frac{0.2}{0.1} = 2$
Therefore,the ratio of the velocity of train $B$ to that of train $A$ is $2$.

(B) The linear speed of the whistle is given by $v_s = r\omega.$
Given $r = 50 \; cm = 0.5 \; m$ and $\omega = 20 \; rad/s.$
So,$v_s = 0.5 \times 20 = 10 \; m/s.$
The minimum frequency is heard when the source is moving directly away from the observer.
The formula for the observed frequency is $n_{\min} = n \left( \frac{v}{v + v_s} \right).$
Substituting the values: $n_{\min} = 385 \left( \frac{340}{340 + 10} \right) = 385 \left( \frac{340}{350} \right) = 385 \times \frac{34}{35} = 11 \times 34 = 374 \; Hz.$

(A) According to the Doppler effect,when the source is moving away from a stationary observer,the observed frequency $n'$ is given by the formula:
$n' = n \left( \frac{v}{v + v_S} \right)$
Where:
$n = 800 \; Hz$ (source frequency)
$v = 330 \; m/s$ (velocity of sound)
$v_S = 30 \; m/s$ (velocity of source)
Substituting the values:
$n' = 800 \left( \frac{330}{330 + 30} \right)$
$n' = 800 \left( \frac{330}{360} \right)$
$n' = 800 \left( \frac{11}{12} \right)$
$n' = \frac{8800}{12} \approx 733.33 \; Hz$.

(A) According to the Doppler effect,the frequency $n'$ heard by a stationary observer when the source moves with speed $v_c$ is given by $n' = n \left( \frac{v}{v \mp v_c} \right)$.
Before passing,the source approaches the observer: $n_{Before} = n \left( \frac{v}{v - v_c} \right)$.
After passing,the source moves away from the observer: $n_{After} = n \left( \frac{v}{v + v_c} \right)$.
The ratio of the frequencies is given as $\frac{n_{Before}}{n_{After}} = \frac{11}{9}$.
Substituting the expressions:
$\frac{n \left( \frac{v}{v - v_c} \right)}{n \left( \frac{v}{v + v_c} \right)} = \frac{v + v_c}{v - v_c} = \frac{11}{9}$.
Cross-multiplying gives:
$9(v + v_c) = 11(v - v_c)$
$9v + 9v_c = 11v - 11v_c$
$20v_c = 2v$
$v_c = \frac{2v}{20} = \frac{1}{10}v$.

(C) The formula for the apparent frequency $n'$ heard by a stationary observer when the source moves towards them with velocity $v_S$ is given by:
$n' = n \left( \frac{v}{v - v_S} \right)$
Given that the apparent frequency is double the actual frequency,$n' = 2n$.
Substituting this into the formula:
$2n = n \left( \frac{v}{v - v_S} \right)$
$2 = \frac{v}{v - v_S}$
$2(v - v_S) = v$
$2v - 2v_S = v$
$v = 2v_S$
$v_S = \frac{v}{2}$
Therefore,the velocity of the source should be $\frac{v}{2}$.

(D) According to the Doppler effect,when the source and the observer are moving towards each other,the observed frequency $n'$ is given by:
$n' = n \left( \frac{v + v_o}{v - v_s} \right)$
Here,$n = 600 \, Hz$ is the source frequency,$v = 340 \, m/s$ is the speed of sound,$v_s = 20 \, m/s$ is the speed of the source (first train),and $v_o = 15 \, m/s$ is the speed of the observer (second train).
Substituting the values:
$n' = 600 \left( \frac{340 + 15}{340 - 20} \right)$
$n' = 600 \left( \frac{355}{320} \right)$
$n' = 600 \times 1.109375 = 665.625 \, Hz \approx 666 \, Hz$.

(A) The formula for the apparent frequency $n'$ when the source and observer approach each other is given by:
$n' = n \left[ \frac{v + v_O}{v - v_S} \right]$
Given:
Apparent frequency $n' = 435 \, s^{-1}$
Speed of sound $v = 332 \, m/s$
Velocity of observer $v_O = 50 \, m/s$
Velocity of source $v_S = 50 \, m/s$
Substituting the values into the formula:
$435 = n \left[ \frac{332 + 50}{332 - 50} \right]$
$435 = n \left[ \frac{382}{282} \right]$
$435 = n \times 1.3546$
$n = \frac{435}{1.3546} \approx 321.12 \, s^{-1}$
Rounding to the nearest given option,the real frequency is $320 \, s^{-1}$.