Doppler’s Effect Questions in English

(B) According to the Doppler effect,when a source of sound (train) moves towards a stationary observer,the observed frequency $f$ is given by $f = f_0 \left( \frac{v}{v - v_s} \right)$,where $v$ is the speed of sound and $v_s$ is the speed of the source. This frequency is higher than the source frequency $f_0$ and remains constant as long as the source approaches with constant velocity.
When the train passes the observer and starts moving away,the observed frequency becomes $f = f_0 \left( \frac{v}{v + v_s} \right)$,which is lower than the source frequency $f_0$ and remains constant as long as the source moves away with constant velocity.
Therefore,the frequency $f$ will be higher than $f_0$ initially and will suddenly drop to a value lower than $f_0$ at the moment the train passes the observer. This behavior is correctly represented by graph $B$.

(B) The source is moving towards point $A$ with velocity $v_s = \frac{100}{3} \, m/s$. The observer is at point $O$. The distance $AO = 4 \, m$ and the distance of the source from $A$ is $3 \, m$. The distance of the source from the observer $O$ is $r = \sqrt{3^2 + 4^2} = 5 \, m$.
The component of the source's velocity along the line joining the source and the observer is the effective velocity $v_{s, \text{eff}} = v_s \cos \theta$,where $\theta$ is the angle between the velocity vector and the line $OS$.
From the geometry,$\cos \theta = \frac{3}{5}$.
Therefore,$v_{s, \text{eff}} = \frac{100}{3} \times \frac{3}{5} = 20 \, m/s$.
Using the Doppler effect formula for a moving source and stationary observer: $n' = n \left( \frac{v}{v - v_{s, \text{eff}}} \right)$.
Substituting the values: $n' = 640 \times \left( \frac{340}{340 - 20} \right) = 640 \times \left( \frac{340}{320} \right) = 640 \times 1.0625 = 680 \, Hz$.

(C) The engine acts as both the source and the observer. The sound travels to the wall and reflects back.
Since the engine is moving towards the wall with velocity $v_s = 50\, ms^{-1}$,the frequency $f'$ received by the wall is given by the Doppler effect formula for a moving source and stationary observer:
$f' = f_0 \left( \frac{v}{v - v_s} \right) = 1.2 \left( \frac{350}{350 - 50} \right) = 1.2 \left( \frac{350}{300} \right) = 1.4\, kHz$.
Now,the wall acts as a stationary source reflecting this frequency $f'$,and the engine acts as an observer moving towards the wall with velocity $v_o = 50\, ms^{-1}$.
The frequency $f''$ heard by the driver is:
$f'' = f' \left( \frac{v + v_o}{v} \right) = 1.4 \left( \frac{350 + 50}{350} \right) = 1.4 \left( \frac{400}{350} \right) = 1.4 \times \frac{8}{7} = 1.6\, kHz$.
Alternatively,using the combined formula for reflection from a wall:
$f'' = f_0 \left( \frac{v + v_s}{v - v_s} \right) = 1.2 \left( \frac{350 + 50}{350 - 50} \right) = 1.2 \left( \frac{400}{300} \right) = 1.6\, kHz$.

(D) The apparent frequency $n'$ heard by the observer from the train approaching with speed $v_s$ is given by:
$n' = n \left[ \frac{v}{v - v_s} \right]$
The apparent frequency $n''$ heard by the observer from the train moving away with speed $v_s$ is given by:
$n'' = n \left[ \frac{v}{v + v_s} \right]$
The number of beats per second is the difference between these two frequencies:
$\Delta n = n' - n''$
$\Delta n = nv \left[ \frac{1}{v - v_s} - \frac{1}{v + v_s} \right] = nv \left[ \frac{(v + v_s) - (v - v_s)}{v^2 - v_s^2} \right] = nv \left[ \frac{2v_s}{v^2 - v_s^2} \right]$
Since $v_s^2$ is very small compared to $v^2$ ($4^2 = 16$ vs $320^2 = 102400$),we can approximate $v^2 - v_s^2 \approx v^2$:
$\Delta n \approx nv \left[ \frac{2v_s}{v^2} \right] = \frac{2nv_s}{v}$
Substituting the given values ($n = 240\,Hz$,$v_s = 4\,m/s$,$v = 320\,m/s$):
$\Delta n = \frac{2 \times 240 \times 4}{320} = \frac{1920}{320} = 6\,Hz$
Thus,the number of beats per second heard by the man is $6$.

(B) The apparent frequency $f'$ of the horn of the police van as heard by the motorcyclist is given by the Doppler effect formula:
$f' = \left( \frac{v - u_m}{v - v_s} \right) f_s$
Given $v = 330 \, m/s$,$v_s = 22 \, m/s$,and $f_s = 176 \, Hz$:
$f' = \left( \frac{330 - u_m}{330 - 22} \right) \times 176 = \left( \frac{330 - u_m}{308} \right) \times 176 \quad \dots (i)$
The apparent frequency $f''$ of the stationary siren as heard by the motorcyclist is:
$f'' = \left( \frac{v + u_m}{v} \right) f_{sireen}$
Given $f_{sireen} = 165 \, Hz$:
$f'' = \left( \frac{330 + u_m}{330} \right) \times 165 \quad \dots (ii)$
Since the motorcyclist does not observe any beats,the frequencies must be equal $(f' = f'')$:
$\left( \frac{330 - u_m}{308} \right) \times 176 = \left( \frac{330 + u_m}{330} \right) \times 165$
$\frac{330 - u_m}{308} \times 176 = \frac{330 + u_m}{2}$
$(330 - u_m) \times 0.5714 \times 176 = (330 + u_m) \times 0.5$
Simplifying the ratio: $\frac{176}{308} = \frac{4}{7}$ and $\frac{165}{330} = \frac{1}{2}$.
$\frac{4(330 - u_m)}{7} = \frac{330 + u_m}{2}$
$8(330 - u_m) = 7(330 + u_m)$
$2640 - 8u_m = 2310 + 7u_m$
$15u_m = 330$
$u_m = 22 \, m/s$.

(D) $1$. The wavelength of sound is defined as $\lambda = v/f$. Since the source determines the frequency $f$ and the medium determines the speed $v$,the wavelength is independent of the observer's motion. Thus,statement $A$ is true.
$2$. When the source moves,the distance between successive wave crests changes,which directly affects the wavelength. Thus,statement $B$ is true.
$3$. If the medium moves with velocity $u$,the speed of sound relative to the ground becomes $v' = v \pm u$. The wavelength also changes to $\lambda' = (v \pm u)/f$. Since $f$ is constant,the ratio $\lambda'/v' = 1/f$ remains constant,meaning they change in equal proportion. Thus,statement $C$ is true.
$4$. When the medium moves from source to observer,the effective speed of sound increases to $v' = v + u$. The frequency heard by the observer is $f' = v'/\lambda$. Since $\lambda = v/f$,we get $f' = (v+u)/(v/f) = f(1 + u/v)$. This is greater than $f$. However,in the Doppler effect,if the source and observer are at rest relative to the ground,the frequency heard is $f' = (v+u)/((\lambda)) = (v+u)/(v/f) = f(1+u/v)$. Wait,the frequency heard by a stationary observer from a stationary source when the medium moves is actually the same as the original frequency because the number of waves passing per second remains constant. Therefore,statement $D$ is false.

(C) Let the speed of car $B$ be $v_B$. Both cars $A$ and $B$ are moving towards the stationary siren.
$1$. Frequency heard by car $B$ from the stationary siren $(n_1)$:
Since the source is stationary and the observer (car $B$) is moving towards it,the frequency is given by:
$n_1 = n_s \left( \frac{v + v_B}{v} \right) = 170 \left( \frac{340 + v_B}{340} \right) = \frac{1}{2} (340 + v_B)$
$2$. Frequency heard by car $B$ from car $A$ $(n_2)$:
Car $A$ is the source moving towards the observer (car $B$),and car $B$ is the observer moving away from the source (car $A$).
$n_2 = n_A \left( \frac{v - v_B}{v - v_A} \right) = 180 \left( \frac{340 - v_B}{340 - 20} \right) = 180 \left( \frac{340 - v_B}{320} \right) = \frac{18}{32} (340 - v_B) = \frac{9}{16} (340 - v_B)$
$3$. Condition for no beats:
For no beats,the frequencies must be equal: $n_1 = n_2$
$\frac{1}{2} (340 + v_B) = \frac{9}{16} (340 - v_B)$
$8 (340 + v_B) = 9 (340 - v_B)$
$2720 + 8v_B = 3060 - 9v_B$
$17v_B = 340$
$v_B = 20 \, m/s$

(D) The Doppler effect for sound waves depends on the medium. The formula is given by $f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right)$,where $v$ is the speed of sound in the medium,$v_o$ is the speed of the observer,and $v_s$ is the speed of the source.
Because the medium provides a preferred frame of reference,the effect is not symmetric with respect to the source and the observer. Therefore,the Assertion is incorrect.
The motion of a source relative to a stationary observer is physically different from the motion of an observer relative to a stationary source because the medium is stationary in one case and moving relative to the observer in the other. Thus,the Reason is correct.

(D) Let $c$ be the speed of sound and $v$ be the speed of the tuning forks.
The frequency heard from the approaching tuning fork is $\nu_{1} = \left(\frac{c}{c-v}\right) \nu_{0}$.
The frequency heard from the receding tuning fork is $\nu_{2} = \left(\frac{c}{c+v}\right) \nu_{0}$.
The beat frequency is given by $\Delta \nu = \nu_{1} - \nu_{2} = 2 \; \text{Hz}$.
Substituting the expressions:
$\Delta \nu = c \nu_{0} \left(\frac{1}{c-v} - \frac{1}{c+v}\right) = c \nu_{0} \left(\frac{c+v - (c-v)}{c^{2}-v^{2}}\right) = \frac{2 c \nu_{0} v}{c^{2}-v^{2}}$.
Since $v \ll c$, we can approximate $c^{2} - v^{2} \approx c^{2}$.
Thus, $\Delta \nu \approx \frac{2 c \nu_{0} v}{c^{2}} = \frac{2 \nu_{0} v}{c} = 2$.
Substituting the given values: $\frac{2 \times 1400 \times v}{350} = 2$.
$8v = 2 \Rightarrow v = \frac{2}{8} = \frac{1}{4} \; \text{m/s}$.

(N/A) $(1)$ The observer (target) is at rest $(v_o = 0)$ and the source (rocket) is moving towards the target with speed $v_s = 200\; m s^{-1}$. The speed of sound is $v = 330\; m s^{-1}$.
Using the Doppler effect formula for a moving source and stationary observer:
$f' = f \left( \frac{v}{v - v_s} \right)$
$f' = 1000\; Hz \times \left( \frac{330}{330 - 200} \right) = 1000 \times \left( \frac{330}{130} \right) \simeq 2538.46\; Hz$.
$(2)$ Now, the target acts as the source of the reflected sound (echo) and is stationary $(v_s = 0)$. The rocket acts as the observer moving towards the source with speed $v_o = 200\; m s^{-1}$. The frequency of the source is the frequency received by the target, $f' = 2538.46\; Hz$.
Using the Doppler effect formula for a stationary source and moving observer:
$f'' = f' \left( \frac{v + v_o}{v} \right)$
$f'' = 2538.46\; Hz \times \left( \frac{330 + 200}{330} \right) = 2538.46 \times \left( \frac{530}{330} \right) \simeq 4076.16\; Hz$.

(A) Frequency of the whistle,$\nu = 400\; Hz$.
Speed of the source (train),$v_s = 10\; m s^{-1}$.
Speed of sound,$v = 340\; m s^{-1}$.
The apparent frequency $(\nu')$ of the whistle as the train approaches the platform is given by the Doppler effect formula:
$\nu' = \left( \frac{v}{v - v_s} \right) \nu = \left( \frac{340}{340 - 10} \right) \times 400 = \frac{340}{330} \times 400 \approx 412.12\; Hz$.
The apparent frequency $(\nu'')$ of the whistle as the train recedes from the platform is given by:
$\nu'' = \left( \frac{v}{v + v_s} \right) \nu = \left( \frac{340}{340 + 10} \right) \times 400 = \frac{340}{350} \times 400 \approx 388.57\; Hz$.
$(ii)$ The apparent change in the frequency of sound is caused by the relative motion between the source and the observer. This motion does not affect the speed of sound in the medium. Therefore,the speed of sound in air in both cases remains $340\; m s^{-1}$.

(D) For the stationary observer: Frequency $= 400\; Hz$,Wavelength $= 0.85\; m$,Speed of sound $= 350\; m s^{-1}$.
For the running observer: Frequency $\approx 411.76\; Hz$,Wavelength $= 0.85\; m$,Speed of sound $= 340\; m s^{-1}$.
$1$. For the stationary observer:
Since the source and observer are stationary relative to each other,the frequency remains $400\; Hz$. The wind blows towards the observer,so the effective speed of sound is $v_e = v + w = 340 + 10 = 350\; m s^{-1}$. The wavelength is $\lambda = v_e / f = 350 / 400 = 0.875\; m$.
$2$. For the running observer:
Here,the air is still,so the speed of sound is $340\; m s^{-1}$. The wavelength remains $\lambda = v / f = 340 / 400 = 0.85\; m$. The observer moves towards the source at $10\; m s^{-1}$,so the frequency heard is $f' = f(v + v_o) / v = 400(340 + 10) / 340 = 411.76\; Hz$.
Conclusion: The two situations are not identical because the wavelength and the effective speed of sound differ in both cases.

(D) The operating frequency of the $SONAR$ is $f = 40.0 \; kHz = 40000 \; Hz$.
Speed of the enemy submarine (observer),$v_o = 360 \; km/h = 360 \times (5/18) \; m/s = 100 \; m/s$.
Speed of sound in water,$v = 1450 \; m/s$.
First,the enemy submarine acts as an observer moving towards the stationary $SONAR$ source. The frequency $f'$ received by the submarine is:
$f' = f \left( \frac{v + v_o}{v} \right) = 40 \left( \frac{1450 + 100}{1450} \right) = 40 \left( \frac{1550}{1450} \right) \approx 42.76 \; kHz$.
Next,the submarine acts as a source reflecting this frequency back to the $SONAR$. Since the submarine is moving towards the $SONAR$,the frequency $f''$ received by the $SONAR$ is:
$f'' = f' \left( \frac{v}{v - v_o} \right) = 42.76 \left( \frac{1450}{1450 - 100} \right) = 42.76 \left( \frac{1450}{1350} \right) \approx 45.93 \; kHz$.
Rounding to the nearest integer,the frequency is approximately $46 \; kHz$.

(B) Let $v$ be the speed of sound in air and $f = 40\; kHz$ be the frequency emitted by the bat.
The bat moves toward the wall with speed $v_b = 0.03v$.
First,the wall acts as a stationary observer receiving the sound. The frequency $f'$ heard by the wall is:
$f' = f \left( \frac{v}{v - v_b} \right) = 40 \left( \frac{v}{v - 0.03v} \right) = \frac{40}{0.97}\; kHz$.
Next,the wall reflects this sound,acting as a stationary source,and the bat acts as an observer moving toward the source with speed $v_b = 0.03v$. The frequency $f''$ heard by the bat is:
$f'' = f' \left( \frac{v + v_b}{v} \right) = \left( \frac{40}{0.97} \right) \left( \frac{v + 0.03v}{v} \right) = \frac{40 \times 1.03}{0.97} \approx 42.47\; kHz$.
Rounding to one decimal place,we get $42.5\; kHz$.

(N/A) The Doppler effect is the phenomenon where the observed frequency of a wave changes due to the relative motion between the source of the wave and the observer.
Example: The pitch (or frequency) of the whistle of a fast-moving train appears to decrease as it recedes away from an observer. Conversely,when an observer approaches a stationary source of sound at high speed,the pitch of the sound heard appears to be higher than the actual frequency of the source.
Analysis: The Doppler effect is a wave phenomenon applicable to both sound and electromagnetic waves. We analyze the frequency changes under three primary conditions:
$(1)$ The observer is stationary,but the source is moving.
$(2)$ The observer is moving,but the source is stationary.
$(3)$ Both the observer and the source are moving relative to the medium.

(D) The Doppler effect describes the change in frequency of a wave in relation to an observer who is moving relative to the wave source. The standard pre-assumptions (conditions) for the derivation of the Doppler effect formula are:
$(1)$ The medium through which the sound travels is stationary.
$(2)$ The velocities of the source and the observer are significantly less than the velocity of sound in the medium $(v_s, v_o < v)$.
$(3)$ The motion of the source and the observer is along the straight line joining them.
$(4)$ The observer is capable of hearing the sound.

(N/A) For this,we use the following sign convention: the direction from the observer to the source is taken as the positive direction of velocity.
Consider a source $S$ moving with velocity $v_{s}$ and an observer who is stationary in a frame where the medium is also at rest.
Let the speed of a wave of angular frequency $\omega$ and period $T_{0}$,both measured by an observer at rest with respect to the medium,be $v$.
As shown in the figure,at time $t=0$,the source is at point $S_{1}$,located at a distance $L$ from the observer,and emits a crest. This reaches the observer at time $t_{1} = \frac{L}{v} \quad \dots (1)$
At time $t=T_{0}$,the source has moved a distance $v_{s}T_{0}$ and is at point $S_{2}$,located at a distance $L + v_{s}T_{0}$ from the observer. At $S_{2}$,the source emits a second crest. This reaches the observer at time $t_{2} = T_{0} + \frac{L + v_{s}T_{0}}{v}$.
At time $n T_{0}$,the source emits its $(n+1)^{th}$ crest,which reaches the observer at time $t_{n+1} = n T_{0} + \frac{L + n v_{s}T_{0}}{v} \quad \dots (2)$.
The time interval between the arrival of the $(n+1)^{th}$ crest and the first crest at the observer is $\Delta t = t_{n+1} - t_{1} = n T_{0} + \frac{L + n v_{s}T_{0}}{v} - \frac{L}{v} = n T_{0} + \frac{n v_{s}T_{0}}{v} = n T_{0} \left( 1 + \frac{v_{s}}{v} \right)$.
The observed period $T$ is the time interval between consecutive crests,so $T = \frac{\Delta t}{n} = T_{0} \left( 1 + \frac{v_{s}}{v} \right) = T_{0} \left( \frac{v + v_{s}}{v} \right)$.
Since frequency $\nu = \frac{1}{T}$ and $\nu_{0} = \frac{1}{T_{0}}$,the observed frequency is $\nu = \nu_{0} \left( \frac{v}{v + v_{s}} \right)$.

(N/A) When the observer moves with velocity $v_{0}$ towards a stationary source,we analyze the situation in the reference frame of the moving observer. In this frame,the source and the medium approach the observer at speed $v_{0}$,and the speed at which the wave crests approach the observer is $v + v_{0}$.
The time interval between the arrival of the first and the $(n+1)^{\text{th}}$ crests is given by:
$t_{n+1} - t_{n} = n T_{0} - \frac{n v_{0} T_{0}}{v + v_{0}}$
The observer measures the period of the wave as:
$T = \frac{t_{n+1} - t_{1}}{n} = T_{0} \left( 1 - \frac{v_{0}}{v + v_{0}} \right) = T_{0} \left( \frac{v}{v + v_{0}} \right)$
Since frequency $\nu = \frac{1}{T}$ and original frequency $\nu_{0} = \frac{1}{T_{0}}$,the observed frequency is:
$\nu = \nu_{0} \left( \frac{v + v_{0}}{v} \right) = \nu_{0} \left( 1 + \frac{v_{0}}{v} \right)$
When the observer moves away from the stationary source,we replace $v_{0}$ with $-v_{0}$ in the equation:
$\nu = \nu_{0} \left( 1 - \frac{v_{0}}{v} \right)$

(N/A) Let us take the direction from the observer to the source as the positive direction.
Let the source and the observer be moving with velocities $v_{s}$ and $v_{0}$ respectively as shown in the figure.
Suppose at time $t=0$,the observer is at $O_{1}$ and the source is at $S_{1}$,with $O_{1}$ being to the left of $S_{1}$.
The source emits a wave of velocity $v$,frequency $\nu$,and period $T_{0}$,all measured by an observer at rest with respect to the medium.
Let $L$ be the distance between $O_{1}$ and $S_{1}$ at $t=0$,when the source emits the first crest. Since the observer is moving,the velocity of the wave relative to the observer is $v+v_{0}$. Therefore,the first crest reaches the observer at time $t_{1}=\frac{L}{v+v_{0}}$.
At time $t=T_{0}$,both the observer and the source have moved to their new positions $O_{2}$ and $S_{2}$ respectively. The new distance between the observer and the source $O_{2}S_{2}$ would be $L+(v_{s}-v_{0})T_{0}$.
At $S_{2}$,the source emits a second crest. This reaches the observer at time $t_{2}=T_{0}+\frac{L+(v_{s}-v_{0})T_{0}}{v+v_{0}}$.
At time $nT_{0}$,the source emits its $(n+1)^{th}$ crest,and this reaches the observer at time $t_{n+1}=nT_{0}+\frac{L+n(v_{s}-v_{0})T_{0}}{v+v_{0}}$.
The time interval between the arrival of the first and $(n+1)^{th}$ crest is $\Delta t = t_{n+1} - t_{1} = nT_{0} + \frac{n(v_{s}-v_{0})T_{0}}{v+v_{0}} = nT_{0} \left(1 + \frac{v_{s}-v_{0}}{v+v_{0}}\right) = nT_{0} \left(\frac{v+v_{0}+v_{s}-v_{0}}{v+v_{0}}\right) = nT_{0} \left(\frac{v+v_{s}}{v+v_{0}}\right)$.
The observed period is $T' = \frac{\Delta t}{n} = T_{0} \left(\frac{v+v_{s}}{v+v_{0}}\right)$.
Thus,the observed frequency is $\nu' = \frac{1}{T'} = \nu \left(\frac{v+v_{0}}{v+v_{s}}\right)$.

(N/A) The general formula for the Doppler effect is given by $\frac{f_{L}}{v + v_{L}} = \frac{f_{S}}{v + v_{S}}$,where $v$ is the speed of sound,$v_{L}$ is the velocity of the listener,and $v_{S}$ is the velocity of the source.
Given that the listener is stationary,$v_{L} = 0$.
When the source moves away from the listener,the velocity of the source $v_{S}$ is taken as positive in the direction of sound propagation (from listener to source),so the denominator becomes $(v + v_{S})$.
Substituting these values,we get $\frac{f_{L}}{v + 0} = \frac{f_{S}}{v + v_{S}}$.
Therefore,$f_{L} = \left( \frac{v}{v + v_{S}} \right) f_{S}$.
Since $v + v_{S} > v$,the fraction $\frac{v}{v + v_{S}} < 1$. Consequently,$f_{L} < f_{S}$.

(A) For $(a)$,when both the source and the listener move towards a wall,the sound reflects off the wall. The wall acts as a secondary source. Since both are moving towards the reflecting surface,the frequency of the reflected sound (echo) heard by the listener increases due to the Doppler effect. Thus,$(a-iii)$.
For $(b)$,when the source and the listener move away from each other,the relative velocity between them increases,causing the observed frequency to decrease according to the Doppler effect formula $f' = f \left( \frac{v - v_L}{v + v_S} \right)$. Thus,$(b-ii)$.

(B) According to the Doppler effect,the apparent frequency $f_{L}$ heard by an observer is given by the formula:
$f_{L} = f_{S} \left( \frac{v + v_{L}}{v + v_{S}} \right)$
Here,$f_{S} = 400 \ Hz$ is the source frequency,$v = 330 \ ms^{-1}$ is the speed of sound in air,$v_{L} = 0 \ ms^{-1}$ is the velocity of the observer (standing on the platform),and $v_{S} = -10 \ ms^{-1}$ is the velocity of the source (moving towards the observer).
Substituting the values:
$f_{L} = 400 \times \left( \frac{330 + 0}{330 - 10} \right)$
$f_{L} = 400 \times \left( \frac{330}{320} \right)$
$f_{L} = 400 \times 1.03125 = 412.5 \ Hz$
Thus,the frequency of the sound heard by the observer is $412.5 \ Hz$.

(A) Let $v_B$ be the speed of the bus and $v = 330\, ms^{-1}$ be the speed of sound.
First,the sound from the horn reaches the wall. The wall acts as a stationary observer receiving sound from a moving source (the bus). The frequency $f'$ received by the wall is:
$f' = f_0 \left( \frac{v}{v - v_B} \right) = 420 \left( \frac{330}{330 - v_B} \right)$
Next,the wall reflects this sound,acting as a stationary source of frequency $f'$. The driver (moving source) acts as an observer moving towards this source. The frequency $f''$ heard by the driver is:
$f'' = f' \left( \frac{v + v_B}{v} \right) = 420 \left( \frac{330}{330 - v_B} \right) \left( \frac{330 + v_B}{330} \right)$
Given $f'' = 490\, Hz$,we have:
$490 = 420 \left( \frac{330 + v_B}{330 - v_B} \right)$
Dividing both sides by $70$:
$7 = 6 \left( \frac{330 + v_B}{330 - v_B} \right)$
$7(330 - v_B) = 6(330 + v_B)$
$2310 - 7v_B = 1980 + 6v_B$
$13v_B = 330$
$v_B = \frac{330}{13} \approx 25.38\, ms^{-1}$
Converting to $kmh^{-1}$:
$v_B = \frac{330}{13} \times \frac{18}{5} = \frac{66 \times 18}{13} = \frac{1188}{13} \approx 91.38\, kmh^{-1}$
Rounding to the nearest integer,the speed is $91\, kmh^{-1}$.

(D) Let $f_0 = 440 \, Hz$ be the original frequency and $f_2 = 480 \, Hz$ be the reflected frequency heard by the driver.
First,the wall acts as a stationary observer receiving the sound from the moving car. The frequency $f_1$ received by the wall is given by the Doppler effect formula for a moving source and stationary observer:
$f_1 = f_0 \left( \frac{v}{v - v_c} \right)$
where $v = 345 \, m/s$ is the speed of sound and $v_c$ is the speed of the car.
Next,the wall acts as a stationary source reflecting the sound back to the moving driver (who is now the observer). The frequency $f_2$ heard by the driver is:
$f_2 = f_1 \left( \frac{v + v_c}{v} \right)$
Substituting $f_1$ into the equation for $f_2$:
$f_2 = f_0 \left( \frac{v}{v - v_c} \right) \left( \frac{v + v_c}{v} \right) = f_0 \left( \frac{v + v_c}{v - v_c} \right)$
Given $f_2 / f_0 = 480 / 440 = 48 / 44 = 12 / 11$:
$12 / 11 = (345 + v_c) / (345 - v_c)$
$12(345 - v_c) = 11(345 + v_c)$
$4140 - 12v_c = 3795 + 11v_c$
$23v_c = 345$
$v_c = 345 / 23 = 15 \, m/s$
Converting to $km/hr$:
$v_c = 15 \times (18 / 5) = 54 \, km/hr$.

(D) The frequency $f$ heard by the observer is given by the Doppler effect formula: $f = f_{0} \left( \frac{v_{s}}{v_{s} - v \cos \theta} \right)$,where $v_{s}$ is the speed of sound,$v$ is the speed of the source,and $\theta$ is the angle between the velocity vector of the source and the line joining the source to the observer.
Before the source reaches the point of closest approach $(t < t_{0})$,the source is approaching the observer,so $\theta$ is acute,$\cos \theta > 0$,and the observed frequency $f > f_{0}$. As the source approaches the point of closest approach,$\theta$ increases towards $90^{\circ}$,so $\cos \theta$ decreases,and $f$ decreases towards $f_{0}$.
At the point of closest approach $(t = t_{0})$,$\theta = 90^{\circ}$,$\cos \theta = 0$,and $f = f_{0}$.
After the source passes the point of closest approach $(t > t_{0})$,the source is moving away from the observer,so $\theta$ is obtuse,$\cos \theta < 0$,and the observed frequency $f < f_{0}$. As the source moves further away,$\theta$ increases,$\cos \theta$ becomes more negative,and $f$ decreases further below $f_{0}$.
Thus,the frequency $f$ starts above $f_{0}$,decreases to $f_{0}$ at $t = t_{0}$,and continues to decrease below $f_{0}$ for $t > t_{0}$. This behavior is best represented by Graph $D$.

(A) The apparent frequency $f'$ heard by a stationary observer from a moving source is given by $f' = f \left( \frac{v}{v \mp v_s} \right)$,where $v$ is the speed of sound and $v_s$ is the speed of the source.
For source $S_1$ moving towards the observer,the apparent frequency is $f_1' = f \left( \frac{v}{v - V} \right)$.
For source $S_2$ moving away from the observer,the apparent frequency is $f_2' = f \left( \frac{v}{v + V} \right)$.
The beat frequency is the difference between these two frequencies:
$|f_1' - f_2'| = 3$
$f \left( \frac{v}{v - V} - \frac{v}{v + V} \right) = 3$
Substituting the given values $f = 500 \, Hz$ and $v = 330 \, m/s$:
$500 \left( \frac{330(v + V) - 330(v - V)}{v^2 - V^2} \right) = 3$
$500 \left( \frac{660V}{330^2 - V^2} \right) = 3$
Since $V^2 \ll 330^2$,we can approximate $330^2 - V^2 \approx 330^2$:
$500 \left( \frac{660V}{330^2} \right) = 3$
$500 \left( \frac{2V}{330} \right) = 3$
$V = \frac{3 \times 330}{1000} = \frac{990}{1000} = 0.99 \, m/s \approx 1 \, m/s$.

(A) Given: Speed of each car $v_s = v_o = 7.2\, km/hr = 7.2 \times \frac{5}{18} = 2\, m/s$. Frequency of horn $f_0 = 676\, Hz$. Velocity of sound $v = 340\, m/s$.
Each driver hears two sounds: one directly from the other car and one reflected from the other car.
$1$. Frequency of sound heard directly from the other car $(f_d)$:
$f_d = f_0 \left( \frac{v + v_o}{v - v_s} \right) = 676 \left( \frac{340 + 2}{340 - 2} \right) = 676 \left( \frac{342}{338} \right) = 684\, Hz$.
$2$. Frequency of sound reflected from the other car $(f_r)$:
The sound from the other car travels to the driver's car,reflects,and returns. Effectively,the other car acts as a source moving towards the driver,and the driver acts as a source reflecting the sound back. The frequency heard after reflection is:
$f_r = f_0 \left( \frac{v + v_s}{v - v_s} \right) \left( \frac{v + v_o}{v - v_o} \right) = 676 \left( \frac{342}{338} \right) \left( \frac{342}{338} \right) = 676 \left( \frac{342}{338} \right)^2 \approx 692\, Hz$.
Beat frequency = $f_r - f_d = 692 - 684 = 8\, Hz$.

(C) Given:
Speed of source $V_S = 20 \, m/s$
Speed of observer/detector $V_O = 20 \, m/s$
Speed of sound $V = 340 \, m/s$
Observed frequency $f' = 1800 \, Hz$
According to the Doppler effect,when the source and observer move away from each other,the observed frequency is given by:
$f' = f \left( \frac{V - V_O}{V + V_S} \right)$
Substituting the values:
$1800 = f \left( \frac{340 - 20}{340 + 20} \right)$
$1800 = f \left( \frac{320}{360} \right)$
$1800 = f \left( \frac{8}{9} \right)$
$f = \frac{1800 \times 9}{8}$
$f = 225 \times 9 = 2025 \, Hz$
Therefore,the original frequency of the source is $2025 \, Hz$.

(A) Given:
Velocity of source (car $X$),$V_s = 36 \; km/h = 36 \times \frac{5}{18} = 10 \; m/s$.
Velocity of observer (car $Y$),$V_o = 72 \; km/h = 72 \times \frac{5}{18} = 20 \; m/s$.
Velocity of sound,$V = 340 \; m/s$.
Apparent frequency,$f' = 1320 \; Hz$.
Using the Doppler effect formula for source and observer approaching each other:
$f' = f_0 \left( \frac{V + V_o}{V - V_s} \right)$
Substituting the values:
$1320 = f_0 \left( \frac{340 + 20}{340 - 10} \right)$
$1320 = f_0 \left( \frac{360}{330} \right)$
$1320 = f_0 \left( \frac{36}{33} \right)$
$f_0 = 1320 \times \frac{33}{36}$
$f_0 = 1210 \; Hz$.
Therefore,the actual frequency of the whistle is $1210 \; Hz$.

(B) Let $f_0 = 400\, Hz$ be the source frequency,$f_2 = 500\, Hz$ be the reflected frequency heard by the driver,$C = 330\, m/s$ be the speed of sound,and $V$ be the speed of the car.
First,the wall acts as an observer receiving the sound from the moving car:
$f_1 = f_0 \left( \frac{C}{C - V} \right)$
Next,the wall acts as a stationary source reflecting the sound back to the driver (observer) who is moving towards the wall:
$f_2 = f_1 \left( \frac{C + V}{C} \right)$
Substituting $f_1$ into the equation for $f_2$:
$f_2 = f_0 \left( \frac{C}{C - V} \right) \left( \frac{C + V}{C} \right) = f_0 \left( \frac{C + V}{C - V} \right)$
Given $f_2 = 500\, Hz$ and $f_0 = 400\, Hz$:
$500 = 400 \left( \frac{330 + V}{330 - V} \right)$
$\frac{5}{4} = \frac{330 + V}{330 - V}$
$5(330 - V) = 4(330 + V)$
$1650 - 5V = 1320 + 4V$
$9V = 330$
$V = \frac{330}{9} = \frac{110}{3}\, m/s$
Converting the speed to $km/h$:
$V = \frac{110}{3} \times \frac{18}{5} = 22 \times 6 = 132\, km/h$.

(A) The observed frequency $f_{0}$ when an observer moves towards a stationary source is given by the Doppler effect formula: $f_{0} = \left(\frac{v + v_{0}}{v}\right) f_{s}$,where $v$ is the velocity of sound and $v_{0}$ is the velocity of the observer.
Given that $v_{0} = \frac{v}{5}$,we substitute this into the formula:
$f_{0} = \left(\frac{v + \frac{v}{5}}{v}\right) f_{s} = \left(\frac{\frac{6v}{5}}{v}\right) f_{s} = \frac{6}{5} f_{s} = 1.2 f_{s}$.
The percentage change in frequency is calculated as:
$\% \text{ change} = \frac{f_{0} - f_{s}}{f_{s}} \times 100$.
Substituting $f_{0} = 1.2 f_{s}$:
$\% \text{ change} = \frac{1.2 f_{s} - f_{s}}{f_{s}} \times 100 = 0.2 \times 100 = 20 \%$.

(C) The velocity of the observer is $V_{o} = 18\,km/h = 18 \times \frac{5}{18} = 5\,m/s$. The source is stationary,so $V_{s} = 0$.
For the direct sound,the observer is moving away from the source. The apparent frequency is given by:
$f_{\text{direct}} = f_{0} \left( \frac{v - V_{o}}{v} \right) = 640 \left( \frac{320 - 5}{320} \right) = 640 \left( \frac{315}{320} \right) = 2 \times 315 = 630\,Hz$.
For the reflected sound,the hill acts as a stationary source of frequency $f_{0} = 640\,Hz$. The observer is moving towards the hill (towards the reflected source). The apparent frequency is given by:
$f_{\text{reflected}} = f_{0} \left( \frac{v + V_{o}}{v} \right) = 640 \left( \frac{320 + 5}{320} \right) = 640 \left( \frac{325}{320} \right) = 2 \times 325 = 650\,Hz$.
The beat frequency is the difference between the two frequencies:
$f_{\text{beat}} = f_{\text{reflected}} - f_{\text{direct}} = 650\,Hz - 630\,Hz = 20\,Hz$.

(D) Let $f_0$ be the actual frequency of the horn,$C$ be the speed of sound,and $V_s$ be the speed of the car.
When the car approaches the observer: $f_1 = f_0 \left( \frac{C}{C - V_s} \right) = 100 \, Hz$.
When the car moves away from the observer: $f_2 = f_0 \left( \frac{C}{C + V_s} \right) = 50 \, Hz$.
Dividing the two equations: $\frac{f_1}{f_2} = \frac{100}{50} = 2 = \frac{C + V_s}{C - V_s}$.
Solving for $V_s$: $2(C - V_s) = C + V_s \implies 2C - 2V_s = C + V_s \implies C = 3V_s \implies V_s = \frac{C}{3}$.
Substituting $V_s$ into the first equation: $100 = f_0 \left( \frac{C}{C - C/3} \right) = f_0 \left( \frac{C}{2C/3} \right) = f_0 \left( \frac{3}{2} \right)$.
Thus,$f_0 = 100 \times \frac{2}{3} = \frac{200}{3} \, Hz$.
If the observer moves with the car,the relative velocity between the source and the observer is zero,so the observed frequency is equal to the actual frequency $f_0$.
Therefore,$f = f_0 = \frac{200}{3} \, Hz$.
Comparing this with $\frac{x}{3} \, Hz$,we get $x = 200$.

(B) The hill acts as a secondary source of sound.
Given:
Frequency of whistle,$f = 320\,Hz$
Velocity of car,$v_s = v_o = 36\,km/h = 36 \times \frac{5}{18} = 10\,m/s$
Velocity of sound,$v = 330\,m/s$
Step $1$: Calculate the frequency received by the hill $(f_1)$.
Since the source (car) is moving towards the stationary observer (hill),the frequency received by the hill is:
$f_1 = f \left( \frac{v}{v - v_s} \right) = 320 \left( \frac{330}{330 - 10} \right) = 320 \left( \frac{330}{320} \right) = 330\,Hz$
Step $2$: Calculate the frequency of the echo heard by the driver $(f_2)$.
The hill reflects this frequency $f_1$. Now,the hill acts as a stationary source and the car acts as a moving observer moving towards the source.
$f_2 = f_1 \left( \frac{v + v_o}{v} \right) = 330 \left( \frac{330 + 10}{330} \right) = 340\,Hz$

(C) The source of sound is moving in a circle,so its linear speed $v_s$ is given by $v_s = r\omega$.
Given $r = 2 \,m$ and $\omega = 15 \,rad/s$,we have $v_s = 2 \times 15 = 30 \,m/s$.
The speed of sound is $v = 330 \,m/s$.
According to the Doppler effect,when the source moves towards the stationary observer,the observed frequency is $f_{\max} = \left(\frac{v}{v - v_s}\right) f$.
When the source moves away from the stationary observer,the observed frequency is $f_{\min} = \left(\frac{v}{v + v_s}\right) f$.
The ratio of the highest to the lowest frequency is:
$\frac{f_{\max}}{f_{\min}} = \frac{\left(\frac{v}{v - v_s}\right) f}{\left(\frac{v}{v + v_s}\right) f} = \frac{v + v_s}{v - v_s}$.
Substituting the values:
$\frac{f_{\max}}{f_{\min}} = \frac{330 + 30}{330 - 30} = \frac{360}{300} = 1.2$.

(A) Let $v$ be the speed of sound in air. The speed of the engine is $v_s = 0.005v$.
First,the sound travels from the moving engine to the stationary cliff. The frequency $f_1$ received by the cliff is given by the Doppler effect formula for a moving source and stationary observer:
$f_1 = f \left( \frac{v}{v + v_s} \right) = f \left( \frac{v}{v + 0.005v} \right) = f \left( \frac{1}{1.005} \right)$
Next,the cliff acts as a stationary source reflecting the sound of frequency $f_1$ back to the moving engine (observer). The frequency $f_2$ received by the engine is given by the Doppler effect formula for a stationary source and moving observer:
$f_2 = f_1 \left( \frac{v - v_o}{v} \right)$
Since the engine is moving away from the cliff,the observer speed $v_o = v_s = 0.005v$ is taken as negative relative to the direction of the reflected sound:
$f_2 = f_1 \left( \frac{v - 0.005v}{v} \right) = f_1 \left( \frac{0.995v}{v} \right) = 0.995 f_1$
Substituting $f_1$ into the equation for $f_2$:
$f_2 = 0.995 \times f \left( \frac{1}{1.005} \right) \approx 0.995 \times 0.995 f \approx 0.990 f$
Thus,the correct option is $A$.

(B) When the observer is stationary and the source is moving towards the observer,the observed frequency is given by the Doppler effect formula:
$f_1 = f_0 \left( \frac{v}{v - u} \right)$
where $v$ is the speed of sound and $u$ is the speed of the source.
When the source is stationary and the observer is moving towards the source,the measured frequency is:
$f_2 = f_0 \left( \frac{v + u}{v} \right)$
To compare $f_1$ and $f_2$,we look at the ratio:
$f_1 = f_0 \left( 1 - \frac{u}{v} \right)^{-1} \approx f_0 \left( 1 + \frac{u}{v} + \frac{u^2}{v^2} + \dots \right)$
$f_2 = f_0 \left( 1 + \frac{u}{v} \right)$
Comparing the two,$f_1 = f_0 \left( \frac{v}{v-u} \right)$ and $f_2 = f_0 \left( \frac{v+u}{v} \right) = f_0 \left( 1 + \frac{u}{v} \right)$.
Since $\frac{v}{v-u} > 1 + \frac{u}{v}$ (because $\frac{v^2}{v(v-u)} > \frac{v^2-u^2}{v(v-u)}$),it follows that $f_1 > f_2$.

(B) The source (bus) is moving towards a stationary observer (person at the bus stop).
Given:
Velocity of source,$v_s = 39.6 \,km/h = 39.6 \times \frac{5}{18} = 11 \,ms^{-1}$.
Velocity of sound,$v = 330 \,ms^{-1}$.
Time interval of source,$T = 30 \,s$.
When the source moves towards a stationary observer,the apparent time interval $T^{\prime}$ between two consecutive pulses is given by the formula:
$T^{\prime} = T \left( \frac{v - v_s}{v} \right)$
Substituting the values:
$T^{\prime} = 30 \left( \frac{330 - 11}{330} \right)$
$T^{\prime} = 30 \left( \frac{319}{330} \right)$
$T^{\prime} = \frac{319}{11} = 29 \,s$.
Therefore,the person will hear the horn at an interval of $29 \,s$.

(A) The apparent frequency $\nu'$ is $10 \%$ less than the original frequency $\nu$,so $\nu' = 0.9 \nu$.
Using the Doppler effect formula for a source moving away from a stationary observer: $\nu' = \nu \left( \frac{v}{v + v_s} \right)$,where $v = 330 \, m/s$ is the speed of sound and $v_s$ is the speed of the car.
Substituting the values: $0.9 \nu = \nu \left( \frac{330}{330 + v_s} \right)$.
$0.9 = \frac{330}{330 + v_s} \Rightarrow 0.9(330 + v_s) = 330$.
$297 + 0.9 v_s = 330$.
$0.9 v_s = 330 - 297 = 33$.
$v_s = \frac{33}{0.9} = \frac{330}{9} \approx 36.7 \, m/s$.

(C) The observer is at point $O$ and the locomotive is at point $S$. The crossing is at point $C$. The distance $SC = 1 \, km$ and $OC = \sqrt{3} \, km$.
In the right-angled triangle $SCO$,the angle $\theta$ at $S$ is given by $\tan \theta = \frac{OC}{SC} = \frac{\sqrt{3}}{1} = \sqrt{3}$.
Therefore,$\theta = 60^{\circ}$.
According to the Doppler effect for a moving source and a stationary observer,the apparent frequency $f'$ is given by:
$f' = f \left( \frac{c}{c - v \cos \theta} \right)$
where $c = 330 \, m/s$ is the speed of sound,$v = 20 \, m/s$ is the speed of the source,and $f = 640 \, Hz$ is the source frequency.
Substituting the values:
$f' = 640 \left( \frac{330}{330 - 20 \cos 60^{\circ}} \right)$
Since $\cos 60^{\circ} = 0.5$:
$f' = 640 \left( \frac{330}{330 - 20 \times 0.5} \right) = 640 \left( \frac{330}{330 - 10} \right) = 640 \left( \frac{330}{320} \right)$
$f' = 640 \times \frac{33}{32} = 20 \times 33 = 660 \, Hz$.

(B) Let $v = 330 \, m/s$ be the speed of sound and $n = 660 \, Hz$ be the frequency of the sources.
When the listener moves from $A$ to $B$ with velocity $u$,they move away from source $A$ and towards source $B$.
The apparent frequency heard from source $A$ is $n_1 = n \left( \frac{v - u}{v} \right)$.
The apparent frequency heard from source $B$ is $n_2 = n \left( \frac{v + u}{v} \right)$.
The number of beats per second is the difference between these frequencies: $n_2 - n_1 = 8$.
Substituting the expressions: $n \left( \frac{v + u}{v} \right) - n \left( \frac{v - u}{v} \right) = 8$.
$\frac{n}{v} (v + u - v + u) = 8$.
$\frac{n}{v} (2u) = 8$.
Substituting the values: $\frac{660}{330} (2u) = 8$.
$2(2u) = 8$.
$4u = 8$.
$u = 2 \, m/s$.

(B) Let the original frequency of the whistle be $f_0$.
When the train approaches the observer,the observed frequency is $f_1 = f_0 \left( \frac{v}{v - v_s} \right)$.
When the train moves away from the observer,the observed frequency is $f_2 = f_0 \left( \frac{v}{v + v_s} \right)$.
However,the problem states the frequency drops by $140 \, Hz$ as the train passes. This implies the change is from the frequency when approaching to the frequency when receding: $f_1 - f_2 = 140 \, Hz$.
Given $v = 330 \, m/s$ and $v_s = 70 \, m/s$:
$f_0 \left( \frac{330}{330 - 70} - \frac{330}{330 + 70} \right) = 140$
$f_0 \left( \frac{330}{260} - \frac{330}{400} \right) = 140$
$f_0 \left( 1.269 - 0.825 \right) = 140$
$f_0 (0.444) = 140 \implies f_0 \approx 315 \, Hz$.
Wait,re-evaluating the standard interpretation: If the train is moving away,the drop is $f_0 - f_2 = 140$.
$f_0 \left( 1 - \frac{330}{330 + 70} \right) = 140$
$f_0 \left( 1 - \frac{330}{400} \right) = 140$
$f_0 \left( \frac{70}{400} \right) = 140$
$f_0 = 140 \times \frac{400}{70} = 800 \, Hz$.

(D) According to the Doppler effect,when a source moves towards a stationary observer,the observed frequency $f$ is given by $f = f_0 \left( \frac{v}{v - v_s} \right)$,where $f_0$ is the source frequency,$v$ is the speed of sound,and $v_s$ is the speed of the source.
For the first case,$v_s = 34 \, m/s$:
$f_1 = f_0 \left( \frac{340}{340 - 34} \right) = f_0 \left( \frac{340}{306} \right) = f_0 \left( \frac{10}{9} \right)$.
For the second case,$v_s = 17 \, m/s$:
$f_2 = f_0 \left( \frac{340}{340 - 17} \right) = f_0 \left( \frac{340}{323} \right) = f_0 \left( \frac{20}{19} \right)$.
Now,calculating the ratio $\frac{f_1}{f_2}$:
$\frac{f_1}{f_2} = \frac{f_0 \left( \frac{10}{9} \right)}{f_0 \left( \frac{20}{19} \right)} = \frac{10}{9} \times \frac{19}{20} = \frac{19}{18}$.

(D) In the Doppler effect,when an observer moves towards a stationary source,the frequency of the sound detected by the observer increases because the observer encounters more wave crests per unit time.
However,the wavelength of the sound waves in the medium is determined solely by the source and the properties of the medium (speed of sound $c$ and frequency $f_0$).
Since the source is stationary and the medium is stationary,the wavelength $\lambda_0$ remains unchanged for the observer.
Therefore,the wavelength shift detected by the observer is zero.

(C) The apparent frequency is maximum when the component of the velocity of the source along the line joining the source and the observer is maximum towards the observer.
Let $O$ be the position of the observer and $C$ be the center of the circular path. The distance from the observer to the center is $OC = 9 + 8 = 17 \, m$.
Let $P$ be the position of the tuning fork on the circle. The velocity vector $V_s$ of the tuning fork is tangential to the circle at $P$. The component of $V_s$ along $OP$ is maximum when the angle between $OP$ and the velocity vector $V_s$ is zero,or more simply,when the line $OP$ is tangent to the circle at $P$.
In this configuration,$\triangle OPC$ is a right-angled triangle with the right angle at $P$ (since the radius $CP$ is perpendicular to the tangent $OP$).
Using the Pythagorean theorem in $\triangle OPC$:
$OP^2 + CP^2 = OC^2$
$OP^2 + 8^2 = 17^2$
$OP^2 = 289 - 64 = 225$
$OP = \sqrt{225} = 15 \, m$.

(B) When the source approaches the stationary observer,the observed frequency $f'$ is given by $f' = \frac{v}{v - V_s} f_0$.
Given that the frequency increases by $10 \%$,we have $f' = 1.1 f_0$.
Thus,$\frac{v}{v - V_s} = 1.1 = \frac{11}{10}$.
$10v = 11v - 11V_s \implies v = 11V_s$.
When the source recedes from the observer,the observed frequency $f''$ is given by $f'' = \frac{v}{v + V_s} f_0$.
Substituting $v = 11V_s$,we get $f'' = \frac{11V_s}{11V_s + V_s} f_0 = \frac{11}{12} f_0$.
$f'' \approx 0.9166 f_0 = 91.66 \% f_0$.
The percentage change in frequency is $100 \% - 91.66 \% = 8.34 \% \approx 8.5 \%$.

(B) Let $f_0$ be the actual frequency,$v$ be the speed of sound,and $v_s$ be the speed of the train.
When the train approaches the observer,the apparent frequency $f_A$ is given by: $f_A = \frac{v}{v - v_s} f_0$.
The difference in frequency is $\Delta f_A = f_A - f_0 = \left( \frac{v}{v - v_s} - 1 \right) f_0 = \frac{v_s}{v - v_s} f_0$.
When the train recedes from the observer,the apparent frequency $f_R$ is given by: $f_R = \frac{v}{v + v_s} f_0$.
The difference in frequency is $\Delta f_R = f_0 - f_R = \left( 1 - \frac{v}{v + v_s} \right) f_0 = \frac{v_s}{v + v_s} f_0$.
Given the ratio $\frac{\Delta f_A}{\Delta f_R} = \frac{3}{2}$,we have:
$\frac{\frac{v_s}{v - v_s} f_0}{\frac{v_s}{v + v_s} f_0} = \frac{v + v_s}{v - v_s} = \frac{3}{2}$.
Cross-multiplying gives: $2(v + v_s) = 3(v - v_s)$.
$2v + 2v_s = 3v - 3v_s$.
$5v_s = v$.
$v_s = \frac{v}{5}$.

(A) Let $c$ be the speed of sound. The detector $D$ is at a distance $2R$ from the center of the circle. The source $S$ moves in a circle of radius $R$ with speed $v$. The component of the source's velocity along the line joining the source and the detector determines the Doppler shift.
The maximum frequency is detected when the source is moving directly towards the detector. At this point,the velocity component along the line $SD$ is $v$. The observed frequency is $f_{max} = f \left( \frac{c}{c-v} \right)$.
The minimum frequency is detected when the source is moving directly away from the detector. At this point,the velocity component along the line $SD$ is $v$. The observed frequency is $f_{min} = f \left( \frac{c}{c+v} \right)$.
Therefore,the ratio of maximum to minimum frequency is:
$\frac{f_{max}}{f_{min}} = \frac{f \left( \frac{c}{c-v} \right)}{f \left( \frac{c}{c+v} \right)} = \frac{c+v}{c-v}$.

(D) According to the Doppler effect,when a source of sound moves with a constant velocity $v_s$ towards a stationary observer,the observed frequency $n'$ is given by $n' = n_0 \left( \frac{v}{v - v_s} \right)$,where $n_0$ is the source frequency and $v$ is the speed of sound. This frequency is constant and higher than $n_0$.
As the source crosses the observer and moves away,the observed frequency becomes $n'' = n_0 \left( \frac{v}{v + v_s} \right)$,which is also constant but lower than $n_0$.
Since the source moves with a constant speed,the frequency remains constant at a higher value before the observer and drops abruptly to a lower constant value after passing the observer. This behavior is represented by the graph in option $D$.

(A) Let $n$ be the actual frequency of the source,$V$ be the speed of sound,and $V_s$ be the speed of the source.
When the source moves towards the stationary observer,the apparent frequency is $n_1 = n \left( \frac{V}{V - V_s} \right)$.
When the source moves away from the stationary observer,the apparent frequency is $n_2 = n \left( \frac{V}{V + V_s} \right)$.
The change in apparent frequency is $\Delta n = n_1 - n_2$.
$\Delta n = n \left( \frac{V}{V - V_s} - \frac{V}{V + V_s} \right) = n V \left( \frac{V + V_s - (V - V_s)}{V^2 - V_s^2} \right) = n V \left( \frac{2 V_s}{V^2 - V_s^2} \right)$.
Since $V_s << V$,we can approximate $V^2 - V_s^2 \approx V^2$.
Therefore,$\Delta n \approx n V \left( \frac{2 V_s}{V^2} \right) = \frac{2 n V_s}{V}$.