Doppler’s Effect Questions in English

(B) According to the Doppler effect,the apparent frequency $f'$ heard by an observer when the source is moving towards a stationary observer is given by:
$f' = f \left( \frac{v}{v - v_s} \right)$
Where:
$f = 320\,Hz$ (source frequency)
$v = 330\,m/s$ (speed of sound)
$v_s = 66\,m/s$ (speed of the source)
Substituting the values:
$f' = 320 \left( \frac{330}{330 - 66} \right)$
$f' = 320 \left( \frac{330}{264} \right)$
$f' = 320 \times 1.25$
$f' = 400\,Hz$
Therefore,the frequency observed by the observer is $400\,Hz$.

(B) The apparent frequency $f'$ heard by an observer is given by the Doppler effect formula: $f' = f_0 \left( \frac{v \pm v_o}{v \mp v_s} \right)$.
Here,$v = 330\,m/s$ (speed of sound),$v_o = 0$ (observer is stationary),$v_s = 30\,m/s$ (speed of source),and $f_0 = 300\,Hz$.
For train $A$ (approaching the station/observer),the source is moving towards the observer: $f_A = 300 \left( \frac{330}{330 - 30} \right) = 300 \left( \frac{330}{300} \right) = 330\,Hz$.
For train $B$ (leaving the station/observer),the source is moving away from the observer: $f_B = 300 \left( \frac{330}{330 + 30} \right) = 300 \left( \frac{330}{360} \right) = 300 \left( \frac{11}{12} \right) = 275\,Hz$.
The difference in frequencies heard is $\Delta f = f_A - f_B = 330 - 275 = 55\,Hz$.

(D) Let the frequency of the horn be $f_0$. The speed of the car is $v_c = 15\,m/s$ and the speed of sound is $v = 330\,m/s$.
When the car approaches the wall,the wall acts as a stationary observer receiving the sound,and then as a source reflecting the sound back to the driver.
The frequency of the sound heard by the wall is $f' = f_0 \left( \frac{v}{v - v_c} \right)$.
This reflected sound is then heard by the driver moving towards the wall at speed $v_c$,so the observed frequency $f$ is $f = f' \left( \frac{v + v_c}{v} \right) = f_0 \left( \frac{v + v_c}{v - v_c} \right)$.
Given the change in frequency is $\Delta f = f - f_0 = 40\,Hz$.
Substituting the values: $f = f_0 \left( \frac{330 + 15}{330 - 15} \right) = f_0 \left( \frac{345}{315} \right)$.
$\Delta f = f_0 \left( \frac{345}{315} - 1 \right) = f_0 \left( \frac{30}{315} \right) = 40$.
$f_0 = \frac{40 \times 315}{30} = 4 \times 105 = 420\,Hz$.

(B) The Doppler effect occurs due to the relative motion between the source of sound and the observer.
In this case,the passenger is inside the train,and the engine (the source of the sound) is also part of the same train.
Since both the source and the observer are moving together at the same velocity,their relative velocity is $0\,ms^{-1}$.
Because there is no relative motion between the source and the observer,there is no shift in the frequency.
Therefore,the frequency heard by the passenger is the same as the frequency produced by the whistle,which is $400\,Hz$.

(B) According to the Doppler effect,the observed frequency $f$ is given by the formula:
$f = f_0 \left( \frac{c + v_o}{c + v_s} \right)$
Where:
$f_0 = 400\,Hz$ (source frequency)
$c = 360\,ms^{-1}$ (velocity of sound)
$v_o = 40\,ms^{-1}$ (velocity of the observer,car $Q$,moving towards the source)
$v_s = 20\,ms^{-1}$ (velocity of the source,car $P$,moving away from the observer)
Substituting the values:
$f = 400 \left( \frac{360 + 40}{360 + 20} \right)$
$f = 400 \left( \frac{400}{380} \right)$
$f = 400 \times 1.0526$
$f \approx 421.05\,Hz$
Thus,the frequency heard by the passenger is approximately $421\,Hz$.

(D, A, A) $1.$ The speed of sound is relative to the medium (air). Since both trains are moving through the same air,the speed of sound relative to the passengers in train $A$ (moving at $20 \ m/s$) is $v_A = v_{sound} - v_A = 340 - 20 = 320 \ m/s$ (if moving away) or $360 \ m/s$ (if moving towards). However,for the whistle on train $A$ heard by passengers on train $A$,they are at rest relative to the source,so the speed of sound is $340 \ m/s$. For passengers in train $B$ (moving at $30 \ m/s$ away from $A$),the speed of sound relative to them is $v_B = 340 - 30 = 310 \ m/s$. Thus,option $D$ is correct.
$2.$ Since the passengers in train $A$ are at rest relative to the source (the engine of train $A$),they observe the same frequency range as emitted by the source. Thus,the intensity distribution remains unchanged,which is represented by graph $A$.
$3.$ The observed frequencies $f'_1$ and $f'_2$ for passengers in train $B$ are given by the Doppler effect formula: $f' = f \left( \frac{v - v_o}{v - v_s} \right)$. Here $v = 340 \ m/s$,$v_o = 30 \ m/s$ (moving away),$v_s = 20 \ m/s$ (moving away). So,$f' = f \left( \frac{340 - 30}{340 - 20} \right) = f \left( \frac{310}{320} \right) = f \left( \frac{31}{32} \right)$.
The new spread is $\Delta f' = f'_2 - f'_1 = (f_2 - f_1) \times \frac{31}{32} = 320 \times \frac{31}{32} = 310 \ Hz$. Thus,option $A$ is correct.

(A) Let $v_0$ be the speed of the car and $v$ be the speed of sound. The observed frequencies from $M$ and $N$ by the car moving towards $Q$ are $f_M = f_M^0 \left( \frac{v + v_0 \cos \theta}{v} \right)$ and $f_N = f_N^0 \left( \frac{v + v_0 \cos \theta}{v} \right)$,where $\theta$ is the angle between the velocity vector of the car and the line connecting the car to the speakers.
The beat frequency $v(t) = |f_N - f_M| = (121 - 118) \left( \frac{v + v_0 \cos \theta}{v} \right) = 3 \left( 1 + \frac{v_0}{v} \cos \theta \right)$.
At point $P$,$\theta$ is acute,so $\cos \theta > 0$,thus $v_P > 3$. At point $Q$,$\theta = 90^{\circ}$,so $\cos \theta = 0$,thus $v_Q = 3$. At point $R$,$\theta$ is obtuse,so $\cos \theta < 0$,thus $v_R < 3$.
Since $v_P = 3(1 + k)$ and $v_R = 3(1 - k)$ where $k = \frac{v_0}{v} \cos \theta_P$,we have $v_P + v_R = 6 = 2v_Q$. Thus,$(A)$ is true.
The rate of change of beat frequency is $\frac{dv}{dt} = 3 \frac{v_0}{v} (-\sin \theta) \frac{d\theta}{dt}$. This is maximum when $\sin \theta$ is maximum,which occurs at $Q$ where $\theta = 90^{\circ}$. Thus,$(B)$ is true.
The variation of $\cos \theta$ with time as the car moves from $P$ to $R$ follows a sigmoid-like curve,which is represented by the left plot. Thus,$(C)$ is true.

(D) The frequency $f_1$ received by the car approaching the stationary source is given by the Doppler effect formula:
$f_1 = f_0 \left( \frac{v + v_c}{v} \right) = 492 \left( \frac{330 + 2}{330} \right) = 492 \left( \frac{332}{330} \right) \,Hz$.
The car acts as a moving source reflecting this frequency back to the stationary observer (the source). The frequency $f_2$ received by the source is:
$f_2 = f_1 \left( \frac{v}{v - v_c} \right) = 492 \left( \frac{332}{330} \right) \left( \frac{330}{330 - 2} \right) = 492 \left( \frac{332}{328} \right) \,Hz$.
Calculating $f_2$:
$f_2 = 492 \times 1.012195 \approx 498 \,Hz$.
The beat frequency $f_B$ is the difference between the reflected frequency and the original frequency:
$f_B = f_2 - f_0 = 498 - 492 = 6 \,Hz$.

(A) Let the stationary man be at point $C$ and the horizontal line be $AB$. Let $O$ be the point on the line $AB$ directly below $C$. Given $CO = 12 \,m$. The men are at $A$ and $B$ such that $AO = OB = 5 \,m$. The distance $AC = BC = \sqrt{12^2 + 5^2} = 13 \,m$. The angle $\theta$ between the line $AC$ (or $BC$) and the vertical $CO$ is given by $\cos \theta = \frac{12}{13}$ and $\sin \theta = \frac{5}{13}$.
The man at $A$ (behind) is moving towards $O$ with velocity $v_A = 2.0 \,m \,s^{-1}$. The component of his velocity along the line $AC$ is $v_{A, \text{eff}} = v_A \sin \theta = 2.0 \times \frac{5}{13} \,m \,s^{-1}$. Since he is moving towards $C$,the observed frequency is $f_A = f \left( \frac{v}{v - v_A \sin \theta} \right) = 1430 \left( \frac{330}{330 - 2 \times \frac{5}{13}} \right) \approx 1430 \left( 1 + \frac{10}{13 \times 330} \right)$.
The man at $B$ (front) is moving away from $O$ with velocity $v_B = 1.0 \,m \,s^{-1}$. The component of his velocity along the line $BC$ is $v_{B, \text{eff}} = v_B \sin \theta = 1.0 \times \frac{5}{13} \,m \,s^{-1}$. Since he is moving away from $C$,the observed frequency is $f_B = f \left( \frac{v}{v + v_B \sin \theta} \right) = 1430 \left( \frac{330}{330 + 1 \times \frac{5}{13}} \right) \approx 1430 \left( 1 - \frac{5}{13 \times 330} \right)$.
The beat frequency is $\Delta f = |f_A - f_B| = 1430 \left( \frac{10}{13 \times 330} + \frac{5}{13 \times 330} \right) = 1430 \left( \frac{15}{13 \times 330} \right) = \frac{1430}{330} \times \frac{15}{13} = \frac{13}{3} \times \frac{15}{13} = 5 \,Hz$.

(A) When a source is stationary and a reflector (car) approaches it with speed $v$,the frequency $f'$ received by the car is $f' = f_0 \frac{c+v}{c}$.
This frequency is then reflected back to the source. The car acts as a moving source,so the frequency $f''$ heard by the stationary source is $f'' = f' \frac{c}{c-v} = f_0 \frac{c+v}{c-v}$.
Since $v \ll c$,we use the binomial approximation: $f'' = f_0 (1 + v/c)(1 - v/c)^{-1} \approx f_0 (1 + v/c)(1 + v/c) \approx f_0 (1 + 2v/c)$.
The frequency shift for a single car is $\Delta f = f'' - f_0 = f_0 (2v/c)$.
For two cars with speeds $v_1$ and $v_2$,the difference in reflected frequencies is $\Delta f_{diff} = |f''_1 - f''_2| = f_0 \frac{2(v_1 - v_2)}{c}$.
Given $\Delta f_{diff} = 0.012 f_0$,we have $0.012 f_0 = f_0 \frac{2 \Delta v}{c}$.
$\Delta v = 0.006 \times c = 0.006 \times 330 \ m/s = 1.98 \ m/s$.
Converting to $km/h$: $\Delta v = 1.98 \times 3.6 \ km/h \approx 7.128 \ km/h$.
The nearest integer is $7 \ km/h$.

(A) The frequency of the sound reflected by the building and heard by the driver is given by the Doppler effect formula for a moving source and a moving observer.
Given:
Frequency of source $f_0 = 8 \ kHz = 8000 \ Hz$
Velocity of source (car) $v_s = 36 \ km/h = 36 \times \frac{5}{18} \ m/s = 10 \ m/s$
Velocity of observer (driver) $v_o = 10 \ m/s$
Speed of sound $v = 320 \ m/s$
First,the building acts as an observer receiving sound from the car:
$f' = f_0 \left( \frac{v}{v - v_s} \right) = 8000 \left( \frac{320}{320 - 10} \right) = 8000 \left( \frac{320}{310} \right)$
Then,the building acts as a source reflecting the sound to the driver:
$f'' = f' \left( \frac{v + v_o}{v} \right) = 8000 \left( \frac{320}{310} \right) \left( \frac{320 + 10}{320} \right)$
$f'' = 8000 \left( \frac{330}{310} \right) = 8000 \times 1.0645 \approx 8516 \ Hz = 8.516 \ kHz \approx 8.50 \ kHz$.

(B) Let the stationary man be at point $C$ and the two moving men be at $A$ and $B$. The distance $CO = 12 \ m$. Since the men are $10 \ m$ apart and the stationary man is equidistant,$AO = OB = 5 \ m$.
The distance $AC = BC = \sqrt{12^2 + 5^2} = 13 \ m$.
The angle $\theta$ between the line of motion and the line joining the observer to the source is given by $\cos \theta = \frac{5}{13}$.
For the man behind $(A)$ moving towards $O$ with speed $v_s = 2.0 \ m \ s^{-1}$,the component of velocity along the line $AC$ is $v_s \cos \theta$ towards $C$. The observed frequency is $f_A = f \left( \frac{v}{v - v_s \cos \theta} \right) = 1430 \left( \frac{330}{330 - 2 \cos \theta} \right) \approx 1430 \left( 1 + \frac{2 \cos \theta}{330} \right)$.
For the man in front $(B)$ moving away from $O$ with speed $v_s = 1.0 \ m \ s^{-1}$,the component of velocity along the line $BC$ is $v_s \cos \theta$ away from $C$. The observed frequency is $f_B = f \left( \frac{v}{v + v_s \cos \theta} \right) = 1430 \left( \frac{330}{330 + 1 \cos \theta} \right) \approx 1430 \left( 1 - \frac{\cos \theta}{330} \right)$.
The beat frequency is $\Delta f = f_A - f_B = 1430 \left( \frac{2 \cos \theta + \cos \theta}{330} \right) = 1430 \left( \frac{3 \cos \theta}{330} \right) = 13 \cos \theta$.
Substituting $\cos \theta = \frac{5}{13}$,we get $\Delta f = 13 \times \frac{5}{13} = 5 \ Hz$.

(B) First,convert the velocities from $km/h$ to $m/s$:
$v_{S1} = 108 \times \frac{5}{18} = 30 \ m/s$
$v_O = 36 \times \frac{5}{18} = 10 \ m/s$
For observer $O$ and source $S_2$:
The source $S_2$ is stationary $(v_s = 0)$. The observer $O$ moves towards $S_2$ with a velocity component along the line joining them. Since $O$ moves directly towards $S_2$,the velocity of the observer is $v_o = 10 \ m/s$.
Using the Doppler effect formula: $f_2 = f_0 \left( \frac{v + v_o}{v} \right) = 120 \left( \frac{330 + 10}{330} \right) = 120 \left( \frac{340}{330} \right) \approx 123.64 \ Hz$.
For observer $O$ and source $S_1$:
The distance between $S_1$ and $S_2$ is $800 \ m$,and $O$ is $600 \ m$ from $S_2$. The distance $OS_1 = \sqrt{800^2 + 600^2} = 1000 \ m$.
The angle $\theta$ such that $\cos \theta = \frac{800}{1000} = 0.8$ and $\sin \theta = \frac{600}{1000} = 0.6$.
The velocity component of $S_1$ towards $O$ is $v_{s1} = v_{S1} \cos \theta = 30 \times 0.8 = 24 \ m/s$.
The velocity component of $O$ towards $S_1$ is $v_{o1} = v_O \sin \theta = 10 \times 0.6 = 6 \ m/s$.
Using the Doppler effect formula: $f_1 = f_0 \left( \frac{v + v_{o1}}{v - v_{s1}} \right) = 120 \left( \frac{330 + 6}{330 - 24} \right) = 120 \left( \frac{336}{306} \right) \approx 131.76 \ Hz$.
Beat frequency $= |f_1 - f_2| = 131.76 - 123.64 = 8.12 \ Hz$.
The number of beats heard is approximately $8$.

(A) For a stationary tuning fork,the resonance frequency is $f = \frac{v}{4\ell_1}$ (for a closed pipe),so $f \propto \frac{1}{\ell_1}$.
When the tuning fork moves parallel to the open end,the frequency perceived by the pipe is the Doppler-shifted frequency. Since the fork moves perpendicular to the line joining the source and the observer (the pipe opening),the frequency remains $f$. However,the question implies the effective frequency changes due to the relative motion or the setup. Given the standard interpretation of this problem,the frequency $f'$ perceived at the pipe is $f' = f \left( \frac{v}{v - v_T} \right)$ where $v_T$ is the speed of the fork.
For resonance,$f' = \frac{v}{4\ell_2}$.
Thus,$\frac{v}{4\ell_1} \left( \frac{v}{v - v_T} \right) = \frac{v}{4\ell_2}$.
This simplifies to $\frac{\ell_2}{\ell_1} = \frac{v - v_T}{v} = 1 - \frac{v_T}{v}$.
Therefore,$\frac{\ell_2 - \ell_1}{\ell_1} = -\frac{v_T}{v}$.
Substituting the values: $\frac{\Delta \ell}{\ell_1} \times 100 = -\frac{2}{320} \times 100 = -0.625 \%$.
The magnitude of the percentage change is $0.625 \%$,which is approximately $0.63 \%$.

(B) The frequency $f'$ of the sound received by the pipe from a source moving towards it with speed $u$ is given by the Doppler effect formula: $f' = f_s \left( \frac{v}{v - u} \right)$.
$A$ closed pipe of length $L$ resonates at odd harmonics of its fundamental frequency $f_0$,i.e.,$f' = n f_0$,where $n = 1, 3, 5, 7, \dots$.
$(A)$ For $u = 0.8 v$ and $f_s = f_0$: $f' = f_0 \left( \frac{v}{v - 0.8 v} \right) = f_0 \left( \frac{v}{0.2 v} \right) = 5 f_0$. Since $5$ is an odd integer,this leads to resonance.
$(B)$ For $u = 0.8 v$ and $f_s = 2 f_0$: $f' = 2 f_0 \left( \frac{v}{v - 0.8 v} \right) = 10 f_0$. Since $10$ is an even integer,this does not lead to resonance.
$(C)$ For $u = 0.8 v$ and $f_s = 0.5 f_0$: $f' = 0.5 f_0 \left( \frac{v}{v - 0.8 v} \right) = 2.5 f_0$. This is not an odd harmonic.
$(D)$ For $u = 0.5 v$ and $f_s = 1.5 f_0$: $f' = 1.5 f_0 \left( \frac{v}{v - 0.5 v} \right) = 1.5 f_0 \left( \frac{v}{0.5 v} \right) = 3 f_0$. Since $3$ is an odd integer,this leads to resonance.
Thus,combinations $(A)$ and $(D)$ lead to resonance.

(B) The general formula for the Doppler effect with wind velocity $w$ is given by $f_2 = f_1 \left( \frac{V + w - v_o}{V + w - v_s} \right)$,where $v_o$ and $v_s$ are the velocities of the observer and source relative to the ground,respectively.
Case $1$: Wind blows from source to observer.
The effective speed of sound is $(V + w)$. The observer moves towards the source with speed $u$,and the source moves towards the observer with speed $u$.
Using the sign convention (positive direction from source to observer): $v_o = -u$ and $v_s = u$.
$f_2 = f_1 \left( \frac{(V + w) - (-u)}{(V + w) - u} \right) = f_1 \left( \frac{V + w + u}{V + w - u} \right)$.
Since $(V + w + u) > (V + w - u)$,we have $f_2 > f_1$.
Case $2$: Wind blows from observer to source.
The effective speed of sound is $(V - w)$.
Using the sign convention: $v_o = -u$ and $v_s = u$.
$f_2 = f_1 \left( \frac{(V - w) - (-u)}{(V - w) - u} \right) = f_1 \left( \frac{V - w + u}{V - w - u} \right)$.
Since $(V - w + u) > (V - w - u)$,we have $f_2 > f_1$.
In both cases,the observed frequency $f_2$ is greater than the source frequency $f_1$. Thus,statements $(A)$ and $(B)$ are correct.

(A) The frequency $f_0$ received by an observer is given by the Doppler effect formula: $f_0 = \left( \frac{C \pm V_0}{C \mp V_s} \right) f_s$,where $C$ is the speed of sound,$V_0$ is the speed of the observer,and $V_s$ is the speed of the source.
Case $1$: Source and observer move towards each other.
$f_1 = \left( \frac{C + v}{C - v} \right) f_s$
$288 = \left( \frac{C + v}{C - v} \right) 240$
$\frac{C + v}{C - v} = \frac{288}{240} = \frac{6}{5} \quad \dots (1)$
Case $2$: Source and observer move away from each other.
$f_2 = n = \left( \frac{C - v}{C + v} \right) f_s$
$n = \left( \frac{C - v}{C + v} \right) 240 \quad \dots (2)$
From equation $(1)$,we have $\frac{C - v}{C + v} = \frac{5}{6}$.
Substituting this into equation $(2)$:
$n = \left( \frac{5}{6} \right) 240$
$n = 5 \times 40 = 200 \ Hz$.

(A) $(1)$ The frequency $f'$ measured by the detector when $S_2$ is at $Q$ is given by the Doppler effect formula: $f' = \frac{C}{C + v \cos \theta} f$,where $C = 324 \ ms^{-1}$ is the speed of sound,$v = 4 \sqrt{2} \ ms^{-1}$ is the speed of the source,and $\theta = 45^{\circ}$ is the angle between the velocity vector of $S_2$ and the line joining $Q$ to $P$.
$f' = \frac{324}{324 + 4 \sqrt{2} \cos 45^{\circ}} \times 656 = \frac{324}{324 + 4 \sqrt{2} \times \frac{1}{\sqrt{2}}} \times 656 = \frac{324}{328} \times 656 = 648 \ Hz$.
$(2)$ When $S_2$ is at $R$,its velocity is perpendicular to the line $RP$,so there is no Doppler shift for $S_2$. Thus,$f_{P, S_2} = 656 \ Hz$.
For $S_1$ moving towards the detector at $P$ with speed $v_s = 4 \ ms^{-1}$,the observed frequency is $f_{P, S_1} = \frac{C}{C - v_s} f = \frac{324}{324 - 4} \times 656 = \frac{324}{320} \times 656 = 664.2 \ Hz$.
The beat frequency is $\Delta f = |f_{P, S_1} - f_{P, S_2}| = 664.2 - 656 = 8.2 \ Hz$.

(D) The angular amplitude is $\theta_0 = \cos^{-1}(0.9)$. Since $\theta_0$ is small,$\cos \theta_0 \approx 1 - \frac{\theta_0^2}{2} = 0.9$,which gives $\frac{\theta_0^2}{2} = 0.1$,so $\theta_0^2 = 0.2$ and $\theta_0 = \sqrt{0.2} = \sqrt{\frac{1}{5}}$.
For a simple pendulum,the maximum velocity is $v' = \ell \omega \theta_0 = \ell \sqrt{\frac{g}{\ell}} \theta_0 = \theta_0 \sqrt{g\ell}$.
Substituting the values: $v' = \sqrt{\frac{1}{5}} \sqrt{10 \times 8} = \sqrt{\frac{80}{5}} = \sqrt{16} = 4 \ m/s$.
The Doppler shift formula for source and observer moving towards each other is $f_{max} = \left( \frac{v + v'}{v - v'} \right) f$ and moving away is $f_{min} = \left( \frac{v - v'}{v + v'} \right) f$.
The maximum variation in frequency is $\Delta f = f_{max} - f_{min} = f \left( \frac{v + v'}{v - v'} - \frac{v - v'}{v + v'} \right) = f \left( \frac{(v + v')^2 - (v - v')^2}{v^2 - v'^2} \right) = f \left( \frac{4vv'}{v^2 - v'^2} \right)$.
Substituting $v = 330 \ m/s$,$v' = 4 \ m/s$,and $f = 660 \ Hz$:
$\Delta f = 660 \times \left( \frac{4 \times 330 \times 4}{330^2 - 4^2} \right) = 660 \times \left( \frac{5280}{108900 - 16} \right) \approx 660 \times \left( \frac{5280}{108884} \right) \approx 660 \times 0.04849 \approx 32.003 \ Hz$.
Thus,the value lies in the range $(31.19$ to $32.27)$. Therefore,the correct option is $D$.

(C) The Doppler effect formula for sound is given by $f' = f \left( \frac{v - v_o}{v - v_s} \right)$,where $v$ is the speed of sound,$v_o$ is the velocity of the observer,and $v_s$ is the velocity of the source.
In this case,the source moves towards the observer,so $v_s = v/4$.
The observer moves away from the source,so $v_o = v/6$.
Substituting these values into the formula:
$f' = f \left( \frac{v - v/6}{v - v/4} \right)$
$f' = f \left( \frac{5v/6}{3v/4} \right)$
$f' = f \left( \frac{5}{6} \times \frac{4}{3} \right)$
$f' = f \left( \frac{20}{18} \right) = \frac{10}{9} f$.

(B) The frequency of the sound heard by the wall (which acts as a stationary observer) is given by $f' = f \left( \frac{v}{v - v_s} \right)$, where $v = 335 \,m/s$ and $v_s = 5 \,m/s$.
$f' = 165 \left( \frac{335}{335 - 5} \right) = 165 \left( \frac{335}{330} \right) = 165 \times \frac{335}{330} = \frac{335}{2} = 167.5 \,Hz$.
Now, the wall acts as a source reflecting this sound back to the bus (which acts as a moving observer). The frequency $f''$ heard by the passengers is $f'' = f' \left( \frac{v + v_o}{v} \right)$, where $v_o = 5 \,m/s$.
$f'' = 167.5 \left( \frac{335 + 5}{335} \right) = 167.5 \left( \frac{340}{335} \right) = 167.5 \times \frac{340}{335} = 167.5 \times 1.0149 \approx 170 \,Hz$.
The number of beats per second is the difference between the reflected frequency and the original frequency: $f_{beats} = f'' - f = 170 \,Hz - 165 \,Hz = 5 \,Hz$.

(B) According to the Doppler effect,when a source of sound moves towards a stationary observer with velocity $v_s$,the apparent frequency $f'$ is given by $f' = f \left( \frac{v}{v - v_s} \right)$,where $v$ is the speed of sound and $f$ is the actual frequency.
Since the denominator $(v - v_s)$ is less than $v$,the apparent frequency $f'$ is greater than the actual frequency $f$,meaning the frequency increases.
The wavelength $\lambda'$ observed by the stationary observer is given by $\lambda' = \frac{v - v_s}{f}$. Since $(v - v_s) < v$,the apparent wavelength $\lambda'$ is less than the actual wavelength $\lambda = v/f$. Thus,the wavelength decreases.

(C) According to the Doppler effect,when an observer moves with velocity $V_1$ towards a stationary source,the apparent frequency $F_1$ is given by: $F_1 = F_0 \left( \frac{v + V_1}{v} \right)$,where $F_0$ is the actual frequency of the source.
When the observer moves away from the stationary source with velocity $V_1$,the apparent frequency $F_2$ is given by: $F_2 = F_0 \left( \frac{v - V_1}{v} \right)$.
Given the ratio $\frac{F_1}{F_2} = 2$,we have: $\frac{F_0 \left( \frac{v + V_1}{v} \right)}{F_0 \left( \frac{v - V_1}{v} \right)} = 2$.
This simplifies to: $\frac{v + V_1}{v - V_1} = 2$.
Cross-multiplying gives: $v + V_1 = 2(v - V_1) = 2v - 2V_1$.
Rearranging the terms: $3V_1 = v$.
Therefore,$\frac{v}{V_1} = 3$.

(C) According to the Doppler effect,when the source of sound and the observer move towards each other,the apparent frequency $f'$ is given by $f' = f \left( \frac{v + v_o}{v - v_s} \right)$,where $v$ is the speed of sound,$v_o$ is the speed of the observer,and $v_s$ is the speed of the source.
Since the numerator $(v + v_o)$ is greater than $v$ and the denominator $(v - v_s)$ is less than $v$,the apparent frequency $f'$ will be greater than the actual frequency $f$.
Since the speed of sound $v$ in the medium remains constant,the relationship between frequency $f$ and wavelength $\lambda$ is given by $v = f \lambda$,or $\lambda = v/f$.
As the apparent frequency $f'$ increases,the apparent wavelength $\lambda'$ must decrease to keep the speed of sound constant.
Therefore,the observer hears a higher frequency and perceives a lower wavelength.

(C) Let the velocity of sound be $v$ and the frequency of the source be $f$. The observer moves towards the stationary source with a velocity $v_o = v/5$.
According to the Doppler effect,the apparent frequency $f'$ heard by the observer is given by the formula: $f' = f \left( \frac{v + v_o}{v} \right)$.
Substituting the given values: $f' = f \left( \frac{v + v/5}{v} \right) = f \left( \frac{6v/5}{v} \right) = 1.2f$.
The increase in frequency is $\Delta f = f' - f = 1.2f - f = 0.2f$.
The percentage increase is given by $\left( \frac{\Delta f}{f} \right) \times 100 = \left( \frac{0.2f}{f} \right) \times 100 = 20 \%$.
Therefore,the correct option is $C$.

(A) The general formula for the Doppler effect is given by $f' = f \left( \frac{V \pm v_o}{V \mp v_s} \right)$,where $V$ is the speed of sound,$v_o$ is the speed of the observer,and $v_s$ is the speed of the source.
Given:
Source velocity $v_s = V/3$ (moving towards the observer,so the denominator decreases).
Observer velocity $v_o = V/4$ (moving away from the source,so the numerator decreases).
Substituting these values into the formula:
$f' = F \left( \frac{V - V/4}{V - V/3} \right)$
$f' = F \left( \frac{3V/4}{2V/3} \right)$
$f' = F \left( \frac{3}{4} \times \frac{3}{2} \right)$
$f' = \frac{9}{8} F$
Therefore,the apparent frequency is $9/8 F$.

(A) According to the Doppler effect,when a source moves towards a stationary observer,the apparent frequency $f'$ is given by the formula:
$f' = f \left( \frac{v}{v - v_s} \right)$
where $f$ is the real frequency,$v$ is the speed of sound,and $v_s$ is the speed of the source.
Given that $v_s = \frac{1}{10} v$,we substitute this into the formula:
$f' = f \left( \frac{v}{v - \frac{v}{10}} \right)$
$f' = f \left( \frac{v}{\frac{9v}{10}} \right)$
$f' = f \left( \frac{10}{9} \right)$
Therefore,the ratio of apparent frequency to real frequency is $\frac{f'}{f} = \frac{10}{9}$.

(D) The formula for the apparent frequency $f'$ heard by a stationary observer when the source moves away with velocity $v_s$ is given by: $f' = f \left( \frac{V}{V + v_s} \right)$,where $V$ is the speed of sound and $f$ is the original frequency.
Given that the apparent frequency is half the original frequency,we have $f' = \frac{f}{2}$.
Substituting this into the equation: $\frac{f}{2} = f \left( \frac{V}{V + v_s} \right)$.
Canceling $f$ from both sides: $\frac{1}{2} = \frac{V}{V + v_s}$.
Cross-multiplying gives: $V + v_s = 2V$.
Solving for $v_s$: $v_s = 2V - V = V$.
Therefore,the source must move at a speed equal to the speed of sound,$V$.

(C) According to the Doppler effect,the apparent frequency $f'$ is given by the formula: $f' = f \left( \frac{V \pm v_o}{V \mp v_s} \right)$.
Here,$V$ is the velocity of sound,$v_s = \frac{V}{3}$ is the velocity of the source moving towards the observer,and $v_o = \frac{V}{5}$ is the velocity of the observer moving away from the source.
Since the source moves towards the observer,the denominator becomes $(V - v_s) = (V - \frac{V}{3}) = \frac{2V}{3}$.
Since the observer moves away from the source,the numerator becomes $(V - v_o) = (V - \frac{V}{5}) = \frac{4V}{5}$.
Substituting these values into the formula:
$f' = f \left( \frac{4V/5}{2V/3} \right) = f \left( \frac{4}{5} \times \frac{3}{2} \right) = f \left( \frac{12}{10} \right) = \frac{6}{5} f$.
Thus,the apparent frequency is $\frac{6}{5} f$.

(A) According to the Doppler effect,when the source is stationary and the observer moves with velocity $V_1$,the apparent frequency $F$ is given by $F = F_0 \left( \frac{V \pm V_1}{V} \right)$,where $F_0$ is the actual frequency and $V$ is the speed of sound.
Case $1$: Observer moves towards the source.
The apparent frequency is $F_1 = F_0 \left( \frac{V + V_1}{V} \right)$.
Case $2$: Observer moves away from the source.
The apparent frequency is $F_2 = F_0 \left( \frac{V - V_1}{V} \right)$.
Given the ratio $F_1 / F_2 = 2$,we have:
$\frac{F_0 (V + V_1) / V}{F_0 (V - V_1) / V} = 2$
$\frac{V + V_1}{V - V_1} = 2$
$V + V_1 = 2(V - V_1)$
$V + V_1 = 2V - 2V_1$
$3V_1 = V$
$\frac{V}{V_1} = 3$.
Thus,the correct option is $A$.

(A) The apparent frequency $f'$ heard by a stationary observer when the source is moving with speed $v_s$ is given by the Doppler effect formula: $f' = f \left( \frac{V}{V \mp v_s} \right)$.
Before passing,the source approaches the observer: $f_1 = f \left( \frac{V}{V - v_s} \right) = 1200 \left( \frac{V}{V - v_s} \right)$.
After passing,the source moves away from the observer: $f_2 = f \left( \frac{V}{V + v_s} \right) = 1200 \left( \frac{V}{V + v_s} \right)$.
The ratio is given as $f_1 / f_2 = 7/5$.
Substituting the expressions: $\frac{V + v_s}{V - v_s} = \frac{7}{5}$.
Cross-multiplying: $5(V + v_s) = 7(V - v_s) \implies 5V + 5v_s = 7V - 7v_s$.
Rearranging terms: $12v_s = 2V \implies v_s = \frac{2V}{12} = \frac{V}{6}$.

(C) The frequency heard by an observer due to the Doppler effect is given by $f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right)$. Here,$v = 330 \ m/s$,$v_s = 30 \ m/s$,$v_o = 0$,and $f = 300 \ Hz$.
For the train approaching the station (first train),the apparent frequency is $f_1 = f \left( \frac{v}{v - v_s} \right) = 300 \left( \frac{330}{330 - 30} \right) = 300 \left( \frac{330}{300} \right) = 330 \ Hz$.
For the train leaving the station (second train),the apparent frequency is $f_2 = f \left( \frac{v}{v + v_s} \right) = 300 \left( \frac{330}{330 + 30} \right) = 300 \left( \frac{330}{360} \right) = 300 \left( \frac{11}{12} \right) = 275 \ Hz$.
The difference in frequencies heard by the person is $\Delta f = f_1 - f_2 = 330 \ Hz - 275 \ Hz = 55 \ Hz$.

(C) Step $1$: The sound from the siren travels towards the wall. The wall acts as a stationary observer. The frequency $n_1$ received by the wall is given by the Doppler effect formula for a moving source and stationary observer: $n_1 = n \left( \frac{V}{V - V_1} \right)$.
Step $2$: The wall reflects this sound. Now,the wall acts as a stationary source of frequency $n_1$,and the driver acts as an observer moving towards the wall with speed $V_1$. The frequency $n_2$ heard by the driver is given by: $n_2 = n_1 \left( \frac{V + V_1}{V} \right)$.
Step $3$: Substituting the value of $n_1$ into the equation for $n_2$: $n_2 = \left( n \frac{V}{V - V_1} \right) \left( \frac{V + V_1}{V} \right) = n \left( \frac{V + V_1}{V - V_1} \right)$.

(D) Let the original frequency of the whistle be $n$ and the observed frequency be $n^{\prime}$.
Given that the frequency drops by $30 \%$,the observed frequency is $n^{\prime} = n - 0.30n = 0.7n$.
According to the Doppler effect,when a source moves away from a stationary observer,the observed frequency is given by $n^{\prime} = n \left( \frac{v}{v + v_s} \right)$,where $v = 350 \ m/s$ is the speed of sound and $v_s$ is the speed of the engine.
Substituting the values: $0.7n = n \left( \frac{350}{350 + v_s} \right)$.
Dividing both sides by $n$: $0.7 = \frac{350}{350 + v_s}$.
$0.7(350 + v_s) = 350$.
$245 + 0.7v_s = 350$.
$0.7v_s = 350 - 245 = 105$.
$v_s = \frac{105}{0.7} = 150 \ m/s$.

(A) Given:
Frequency of source $(n_0) = 510 \,Hz$.
Velocity of source $(v_s) = 72 \,km/hr = 72 \times \frac{5}{18} = 20 \,m/s$.
Velocity of sound in air $(v) = 320 \,m/s$.
$(1)$ When the source is approaching a stationary listener,the apparent frequency $(n_1)$ is given by:
$n_1 = n_0 \left( \frac{v}{v - v_s} \right)$
$n_1 = 510 \times \left( \frac{320}{320 - 20} \right) = 510 \times \left( \frac{320}{300} \right) = 544 \,Hz$.
$(2)$ When the source is moving away from a stationary listener,the apparent frequency $(n_2)$ is given by:
$n_2 = n_0 \left( \frac{v}{v + v_s} \right)$
$n_2 = 510 \times \left( \frac{320}{320 + 20} \right) = 510 \times \left( \frac{320}{340} \right) = 480 \,Hz$.
Thus,the frequencies are $544 \,Hz$ and $480 \,Hz$.

(B) According to the Doppler effect,when the listener moves towards a stationary source with velocity $V_1$,the apparent frequency $F_1$ is given by:
$F_1 = F \left( \frac{V + V_1}{V} \right)$
When the listener moves away from the stationary source with velocity $V_1$,the apparent frequency $F_2$ is given by:
$F_2 = F \left( \frac{V - V_1}{V} \right)$
Given the ratio $\frac{F_1}{F_2} = 2$,we substitute the expressions:
$\frac{F_1}{F_2} = \frac{V + V_1}{V - V_1} = 2$
$V + V_1 = 2(V - V_1)$
$V + V_1 = 2V - 2V_1$
$3V_1 = V$
Therefore,$\frac{V}{V_1} = 3$.

(D) Applying Doppler's effect to sound waves,the observed frequency $n^{\prime}$ is given by $n^{\prime} = n \left( \frac{v + v_0}{v} \right) = n \left( 1 + \frac{v_0}{v} \right)$,where $v$ is the velocity of sound and $v_0$ is the velocity of the observer moving towards the source.
Given that the observed frequency is triple the source frequency,$n^{\prime} = 3n$.
Substituting this into the equation: $3n = n \left( 1 + \frac{v_0}{v} \right)$.
Dividing both sides by $n$: $3 = 1 + \frac{v_0}{v}$.
Solving for $v_0$: $\frac{v_0}{v} = 2$,which implies $v_0 = 2v$.
Therefore,the observer must move with twice the velocity of sound towards the source.

(A) The wall acts as a stationary source of reflected sound. The frequency of sound received by the wall is $n_w = n \left(\frac{V}{V-V_1}\right)$.
This reflected sound acts as a source for the driver. Since the driver is moving towards the wall (source),the frequency heard by the driver is $n' = n_w \left(\frac{V+V_1}{V}\right)$.
Substituting $n_w$,we get $n' = n \left(\frac{V}{V-V_1}\right) \left(\frac{V+V_1}{V}\right) = n \left(\frac{V+V_1}{V-V_1}\right)$.

(B) According to the Doppler effect,the observed frequency $n^{\prime}$ is given by the formula:
$n^{\prime} = \left( \frac{V + V_{L}}{V - V_{S}} \right) n$
Here,the speed of the listener $V_{L} = \frac{V}{10}$ and the speed of the source $V_{S} = \frac{V}{10}$.
Substituting these values into the formula:
$n^{\prime} = \left( \frac{V + \frac{V}{10}}{V - \frac{V}{10}} \right) n$
$n^{\prime} = \left( \frac{\frac{11V}{10}}{\frac{9V}{10}} \right) n$
$n^{\prime} = \frac{11}{9} n$
$n^{\prime} \approx 1.22 n$

(A) According to the Doppler effect,when the listener moves towards the source,the apparent frequency increases,so the numerator is $(V+V_L)$.
When the source moves towards the listener,the apparent wavelength decreases,which increases the frequency,so the denominator is $(V-V_S)$.
Therefore,the observed frequency $n$ is given by the formula:
$n=n_0\left[\frac{V+V_{L}}{V-V_{s}}\right]$

(B) For an observer moving towards a stationary source,the observed frequency is given by $n_1 = n_0 \left[ \frac{v + v_L}{v} \right]$.
For a source moving towards a stationary observer,the observed frequency is given by $n_2 = n_0 \left[ \frac{v}{v - v_s} \right]$.
Substituting the values $v = 330 \,m/s$,$v_L = 50 \,m/s$,and $v_s = 50 \,m/s$:
$n_1 = n_0 \left[ \frac{330 + 50}{330} \right] = n_0 \left[ \frac{380}{330} \right] \approx 1.151 n_0$.
$n_2 = n_0 \left[ \frac{330}{330 - 50} \right] = n_0 \left[ \frac{330}{280} \right] \approx 1.178 n_0$.
Comparing the two results,we find that $n_2 > n_1$.
Therefore,the frequency heard will be more in the second case than in the first case.

(D) The frequency heard by a stationary observer when the source is approaching is given by $n_b = \left( \frac{v}{v - v_s} \right) n$.
When the source is moving away from the stationary observer,the frequency heard is $n_a = \left( \frac{v}{v + v_s} \right) n$.
Taking the ratio of these two frequencies:
$\frac{n_b}{n_a} = \frac{v + v_s}{v - v_s}$.
Given the ratio $\frac{n_b}{n_a} = \frac{11}{9}$,we have:
$\frac{11}{9} = \frac{v + v_s}{v - v_s}$.
Cross-multiplying gives:
$11(v - v_s) = 9(v + v_s)$
$11v - 11v_s = 9v + 9v_s$
$2v = 20v_s$
$v_s = \frac{2v}{20} = \frac{v}{10}$.

(A) The Doppler effect depends on the relative motion between the source of sound and the observer.
In this case, the passenger (observer) and the engine (source) are both in the same train.
Since they are moving together at the same velocity, there is no relative motion between the source and the listener.
Therefore, the apparent frequency $f$ heard by the passenger is equal to the actual frequency $n$ of the whistle.
Thus, $f = n$.

(A) The apparent frequency $n$ heard by a stationary observer when the source moves towards them with speed $v_S$ is given by the Doppler effect formula: $n = n_0 \left( \frac{v}{v - v_S} \right)$,where $n_0$ is the real frequency and $v$ is the speed of sound.
Given that $v_S = \frac{v}{10}$,we substitute this into the formula:
$\frac{n}{n_0} = \frac{v}{v - \frac{v}{10}}$
$\frac{n}{n_0} = \frac{v}{\frac{9v}{10}}$
$\frac{n}{n_0} = \frac{10}{9}$
Thus,the ratio of apparent to real frequency is $10:9$.

(D) Let the original frequency be $f$. The apparent frequency $f^{\prime}$ drops by $20 \%$,so $f^{\prime} = f - 0.2f = 0.8f = \frac{4}{5}f$.
Since the frequency decreases,the source is moving away from the stationary observer.
Using the Doppler effect formula for a source moving away from a stationary observer:
$f^{\prime} = f \left[ \frac{V}{V + V_s} \right]$
where $V = 350 \ m/s$ is the speed of sound and $V_s$ is the speed of the engine.
Substituting the values:
$\frac{4}{5}f = f \left[ \frac{350}{350 + V_s} \right]$
$\frac{4}{5} = \frac{350}{350 + V_s}$
$4(350 + V_s) = 5(350)$
$1400 + 4V_s = 1750$
$4V_s = 350$
$V_s = 87.5 \ m/s$.

(B) According to the Doppler effect,the frequency heard by an observer when the source is approaching is $n_{\text{before}} = \left(\frac{V}{V - v_c}\right) n$.
When the source is receding,the frequency heard is $n_{\text{after}} = \left(\frac{V}{V + v_c}\right) n$.
Given the ratio $\frac{n_{\text{before}}}{n_{\text{after}}} = \frac{11}{9}$.
Substituting the expressions: $\frac{\frac{V}{V - v_c} n}{\frac{V}{V + v_c} n} = \frac{V + v_c}{V - v_c} = \frac{11}{9}$.
Cross-multiplying: $9(V + v_c) = 11(V - v_c)$.
$9V + 9v_c = 11V - 11v_c$.
$20v_c = 2V$.
$v_c = \frac{2V}{20} = \frac{V}{10}$.

(B) The apparent frequency $f$ heard by a stationary observer when the source moves towards them is given by the Doppler effect formula:
$f = f_0 \left( \frac{v}{v - v_s} \right)$
Given:
$v = 340 \,m/s$ (speed of sound)
$f_0 = 90 \,Hz$ (source frequency)
$v_s = \frac{v}{10} = \frac{340}{10} = 34 \,m/s$ (speed of the source)
Substituting the values into the formula:
$f = 90 \left( \frac{340}{340 - 34} \right)$
$f = 90 \left( \frac{340}{306} \right)$
$f = 90 \left( \frac{10}{9} \right)$
$f = 100 \,Hz$

(B) Concept: Doppler effect for a source moving away from a stationary observer.
Formula: $f^{\prime} = f \left( \frac{v}{v + v_s} \right)$
Where:
$f$ is the source frequency = $1152 \ Hz$
$v$ is the speed of sound = $340 \ m/s$
$v_s$ is the speed of the source.
First,convert the speed of the source from $km/h$ to $m/s$:
$v_s = 72 \times \frac{5}{18} \ m/s = 20 \ m/s$
Now,substitute the values into the formula:
$f^{\prime} = 1152 \times \left( \frac{340}{340 + 20} \right)$
$f^{\prime} = 1152 \times \left( \frac{340}{360} \right)$
$f^{\prime} = 1152 \times \left( \frac{17}{18} \right)$
$f^{\prime} = 64 \times 17 = 1088 \ Hz$
Therefore,the apparent frequency heard by the observer is $1088 \ Hz$.

(D) The Doppler effect formula for an observer moving towards a stationary source is given by $n^{\prime} = n \left( \frac{v + v_{o}}{v} \right)$,where $n^{\prime}$ is the observed frequency,$n$ is the source frequency,$v$ is the speed of sound,and $v_{o}$ is the velocity of the observer.
Given that the observed frequency is double the source frequency,we have $n^{\prime} = 2n$.
Substituting this into the formula: $2n = n \left( \frac{v + v_{o}}{v} \right)$.
Dividing both sides by $n$: $2 = \frac{v + v_{o}}{v}$.
Multiplying by $v$: $2v = v + v_{o}$.
Solving for $v_{o}$: $v_{o} = v$.
Therefore,the observer must move towards the source with a velocity equal to the speed of sound.