Doppler’s Effect Questions in English

(C) The apparent frequency $f'$ is given by the Doppler effect formula: $f' = f \left( \frac{v \pm v_L}{v \mp v_S} \right)$.
Given that the apparent frequency $(390 \ Hz)$ is less than the actual frequency $(400 \ Hz)$,the relative distance between the source and the listener must be increasing.
This occurs if the listener moves away from the source or the source moves away from the listener.
Among the given options,the listener moving away from the source is the correct scenario.

(C) The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source.
It is applicable whenever there is relative motion between the source of the wave and the observer.
This phenomenon applies to all types of waves,including sound and light.
Therefore,the fundamental requirement for the Doppler effect is the existence of relative motion between the source and the observer.

(D) According to the Doppler effect,the apparent frequency $n'$ heard by an observer is given by the formula: $n' = n \left( \frac{v + v_o}{v - v_s} \right)$.
Here,$v$ 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 observer are moving towards each other,$v_o = \frac{v}{2}$ and $v_s = \frac{v}{2}$.
Substituting these values into the formula:
$n' = n \left( \frac{v + v/2}{v - v/2} \right)$
$n' = n \left( \frac{3v/2}{v/2} \right)$
$n' = n \times 3 = 3n$.
Therefore,the frequency heard by the observer is $3n$.

(C) When the engine approaches the observer,the apparent frequency is $n' = n \left( \frac{v}{v - v_s} \right)$.
When the engine moves away from the observer,the apparent frequency is $n'' = n \left( \frac{v}{v + v_s} \right)$.
Given the ratio of apparent frequencies is $n'/n'' = 5/3$.
Therefore,$\frac{n'}{n''} = \frac{v + v_s}{v - v_s} = \frac{5}{3}$.
Substituting $v = 340 \ m/s$:
$\frac{340 + v_s}{340 - v_s} = \frac{5}{3}$.
$3(340 + v_s) = 5(340 - v_s)$.
$1020 + 3v_s = 1700 - 5v_s$.
$8v_s = 680$.
$v_s = 85 \ m/s$.

(A) The frequency heard by an observer when moving towards a stationary source is given by the Doppler effect formula:
$n' = n \left( \frac{v + v_0}{v} \right)$
Where:
$n = 240 \text{ Hz}$ (source frequency)
$v = 330 \text{ m/s}$ (speed of sound)
$v_0 = 11 \text{ m/s}$ (speed of observer)
Substituting the values:
$n' = 240 \left( \frac{330 + 11}{330} \right)$
$n' = 240 \left( \frac{341}{330} \right)$
$n' = 240 \times 1.0333 = 248 \text{ Hz}$.

(C) The speed of the person is $v_p = 18 \text{ km/h} = 18 \times \frac{5}{18} = 5 \text{ m/s}$.
Since the person is moving towards a reflecting surface,the reflected sound acts as a source moving towards the observer.
The frequency of the sound heard directly from the whistle is $f = 272 \text{ Hz}$.
The frequency of the reflected sound heard by the person is given by the Doppler effect formula for a moving observer and a moving source (the image of the whistle):
$f' = f \left( \frac{v + v_p}{v - v_p} \right)$
Substituting the values: $f' = 272 \left( \frac{345 + 5}{345 - 5} \right) = 272 \left( \frac{350}{340} \right) = 272 \times \frac{35}{34} = 8 \times 35 = 280 \text{ Hz}$.
The number of beats heard is the difference between the reflected frequency and the original frequency:
$\text{Beats} = f' - f = 280 \text{ Hz} - 272 \text{ Hz} = 8 \text{ Hz}$.

(B) The passenger on the bus hears two sounds: the direct sound from the horn and the reflected sound from the wall.
Since the bus is moving towards the wall with velocity $v_B = 5 \ m/s$,the wall acts as a source of reflected sound.
The frequency of the direct sound heard by the passenger is $n = 165 \ Hz$.
The frequency of the reflected sound $(n')$ heard by the passenger is given by the Doppler effect formula for a moving observer and a stationary source (the wall acts as a stationary source reflecting the sound):
$n' = n \left( \frac{v + v_B}{v - v_B} \right)$
Substituting the given values ($v = 355 \ m/s$,$v_B = 5 \ m/s$,$n = 165 \ Hz$):
$n' = 165 \left( \frac{355 + 5}{355 - 5} \right) = 165 \left( \frac{360}{350} \right) = 165 \times \frac{36}{35} = 169.71 \approx 170 \ Hz$.
The number of beats heard per second is the difference between the frequencies:
$\text{Beats} = n' - n = 170 \ Hz - 165 \ Hz = 5 \ Hz$.

(A) The observer hears two sounds: one directly from the source and the other from the reflected sound (echo) from the wall.
For the direct sound,the source is moving away from the observer. The apparent frequency $n_1$ is given by:
$n_1 = n \left( \frac{v}{v + v_S} \right) = 256 \left( \frac{330}{330 + 5} \right) = 256 \left( \frac{330}{335} \right) \approx 252.18 \ Hz$.
For the reflected sound,the wall acts as a stationary source reflecting the sound. The source is moving towards the wall,so the wall receives sound at a higher frequency,which it then reflects back to the observer. The apparent frequency $n_2$ is given by:
$n_2 = n \left( \frac{v}{v - v_S} \right) = 256 \left( \frac{330}{330 - 5} \right) = 256 \left( \frac{330}{325} \right) \approx 260.00 \ Hz$.
The number of beats per second is the difference between these two frequencies:
$f_{beat} = n_2 - n_1 = n \left( \frac{v}{v - v_S} - \frac{v}{v + v_S} \right) = n \left( \frac{v(v + v_S) - v(v - v_S)}{v^2 - v_S^2} \right) = \frac{2nvv_S}{v^2 - v_S^2}$.
Substituting the values:
$f_{beat} = \frac{2 \times 256 \times 330 \times 5}{330^2 - 5^2} = \frac{844800}{108900 - 25} = \frac{844800}{108875} \approx 7.76 \ Hz \approx 7.8 \ Hz$.

(A) Let $v$ be the velocity of sound and $v_0 = 40 \ m/s$ be the velocity of the listener.
When the listener moves towards a stationary source,the apparent frequency $n'$ is given by:
$n' = \left( \frac{v + v_0}{v} \right) n = 200 \dots (i)$
When the listener moves away from the same source,the apparent frequency $n''$ is given by:
$n'' = \left( \frac{v - v_0}{v} \right) n = 160 \dots (ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{v + v_0}{v - v_0} = \frac{200}{160} = \frac{5}{4}$
$4(v + 40) = 5(v - 40)$
$4v + 160 = 5v - 200$
$v = 360 \ m/s$
The velocity of sound in air is $360 \ m/s$.

(B) When an observer moves towards a stationary source,the apparent frequency $n'$ is given by the Doppler effect formula:
$n' = \left( \frac{v + v_O}{v} \right) n$
Given that the velocity of the observer $v_O = \frac{v}{5}$,where $v$ is the velocity of sound.
Substituting the value of $v_O$ into the formula:
$n' = \left( \frac{v + v/5}{v} \right) n = \left( \frac{6v/5}{v} \right) n = 1.2n$
The increase in frequency is $\Delta n = n' - n = 1.2n - n = 0.2n$.
The percentage increase in frequency is given by:
$\text{Percentage increase} = \left( \frac{\Delta n}{n} \right) \times 100 = \left( \frac{0.2n}{n} \right) \times 100 = 20\%$.

(C) The situation is shown in the figure. Both the source (engine) and the observer (passenger in the middle of the train) have the same speed $v = 30 \ m/s$.
The engine is at the start of the semicircle,and the observer is at the midpoint of the semicircle. The line joining the engine and the observer makes an angle of $45^\circ$ with the velocity vector of both the engine and the observer.
The component of the velocity of the observer towards the source is $v \cos 45^\circ$.
The component of the velocity of the source towards the observer is $v \cos 45^\circ$.
Since both the source and the observer are moving with the same velocity component along the line joining them,there is no relative motion between them along the line of sight.
Therefore,the apparent frequency heard by the observer is the same as the source frequency,which is $200 \ Hz$.

(B) When the source approaches the observer,the apparent frequency is $n' = \frac{v}{v - v_s} n = n \left( 1 - \frac{v_s}{v} \right)^{-1} \approx n \left( 1 + \frac{v_s}{v} \right)$.
When the source recedes from the observer,the apparent frequency is $n'' = \frac{v}{v + v_s} n = n \left( 1 + \frac{v_s}{v} \right)^{-1} \approx n \left( 1 - \frac{v_s}{v} \right)$.
Given that the difference $n' - n'' = \frac{2}{100} n$.
Substituting the expressions: $n \left( 1 + \frac{v_s}{v} \right) - n \left( 1 - \frac{v_s}{v} \right) = \frac{2}{100} n$.
$n \left( \frac{2 v_s}{v} \right) = \frac{2}{100} n$.
$\frac{2 v_s}{v} = \frac{2}{100} \Rightarrow v_s = \frac{v}{100}$.
Given $v = 300 \, m/s$,so $v_s = \frac{300}{100} = 3 \, m/s$.

(D) $1$. The number of waves striking the surface per second (frequency of waves reaching the moving target) is given by $n' = \frac{c + v}{\lambda} = \frac{\nu (c + v)}{c}$. Thus,option $(c)$ is correct.
$2$. The moving target acts as a source for the reflected waves. The apparent frequency $n''$ of the reflected wave is given by the Doppler effect formula for a moving source: $n'' = n' \left( \frac{c}{c - v} \right) = \nu \left( \frac{c + v}{c} \right) \left( \frac{c}{c - v} \right) = \nu \left( \frac{c + v}{c - v} \right)$. Thus,option $(a)$ is correct.
$3$. The wavelength $\lambda'$ of the reflected wave is $\lambda' = \frac{c}{n''} = \frac{c}{\nu \left( \frac{c + v}{c - v} \right)} = \frac{c(c - v)}{\nu (c + v)}$. Thus,option $(b)$ is correct.
$4$. Since $(a)$,$(b)$,and $(c)$ are all correct,the correct choice is $(d)$.

(B) Given:
Speed of car $A$ $(v_A)$ = $72\; km/hr = 72 \times \frac{5}{18} = 20\; m/s$.
Speed of car $B$ $(v_B)$ = $36\; km/hr = 36 \times \frac{5}{18} = 10\; m/s$.
Frequency of source $(n)$ = $280\; Hz$.
Speed of sound $(v)$ = $340\; m/s$ (assumed).
The component of velocity of source $(A)$ along the line joining the cars is $v_A \cos 45^o$ (moving towards the observer).
The component of velocity of observer $(B)$ along the line joining the cars is $v_B \cos 45^o$ (moving towards the source).
Using the Doppler effect formula:
$n' = n \left( \frac{v + v_B \cos 45^o}{v - v_A \cos 45^o} \right)$
$n' = 280 \left( \frac{340 + 10 \cos 45^o}{340 - 20 \cos 45^o} \right)$
$n' = 280 \left( \frac{340 + 10/\sqrt{2}}{340 - 20/\sqrt{2}} \right)$
$n' = 280 \left( \frac{340 + 7.07}{340 - 14.14} \right) = 280 \left( \frac{347.07}{325.86} \right) \approx 298.2\; Hz$.
Thus,the observed frequency is approximately $298\; Hz$.

(B) Let the speed of sound be $v = 330 \ m/s$. The speed of the listener $v_L = 30 \ m/s$ and the speed of whistle $B$ is $v_B = 30 \ m/s$.
$1$. For whistle $B$: Since the listener and whistle $B$ are moving in the same direction with the same speed,their relative velocity is zero. Therefore,the frequency heard by the listener from $B$ remains $f_B' = 596 \ Hz$.
$2$. For whistle $A$: The listener is moving away from the stationary whistle $A$ with speed $v_L = 30 \ m/s$. The observed frequency $f_A'$ is given by the Doppler effect formula: $f_A' = f_A \left( \frac{v - v_L}{v} \right)$.
Substituting the values: $f_A' = 660 \left( \frac{330 - 30}{330} \right) = 660 \left( \frac{300}{330} \right) = 660 \times \frac{10}{11} = 600 \ Hz$.
$3$. The number of beats heard is the difference between the two observed frequencies: $\text{Beats} = |f_A' - f_B'| = |600 - 596| = 4 \ Hz$.

(A) The original wavelength of the sound emitted by the source is given by $\lambda = \frac{v}{f} = \frac{340}{170} = 2 \, m$.
When the source approaches a stationary observer with velocity $v_s = 17 \, ms^{-1}$,the apparent wavelength $\lambda'$ is given by the formula $\lambda' = \frac{v - v_s}{f} = \frac{340 - 17}{170} = \frac{323}{170} = 1.9 \, m$.
The apparent change in the wavelength is $\Delta \lambda = \lambda - \lambda' = 2 - 1.9 = 0.1 \, m$.

(B) Let $v$ be the speed of the motorcyclist and $v_s = 330 \ m/s$ be the speed of sound.
$1$. The frequency $n_1$ of the police car horn heard by the motorcyclist (who is moving away from the source) is given by the Doppler effect formula:
$n_1 = n_0 \left( \frac{v_s - v}{v_s - v_{police}} \right) = 176 \left( \frac{330 - v}{330 - 22} \right) = 176 \left( \frac{330 - v}{308} \right)$
$2$. The frequency $n_2$ of the stationary siren heard by the motorcyclist (who is moving towards the source) is:
$n_2 = n_s \left( \frac{v_s + v}{v_s} \right) = 165 \left( \frac{330 + v}{330} \right)$
$3$. Since the motorcyclist observes no beats,the frequencies must be equal $(n_1 = n_2)$:
$176 \left( \frac{330 - v}{308} \right) = 165 \left( \frac{330 + v}{330} \right)$
$4$. Simplifying the equation:
$\frac{176}{308} (330 - v) = \frac{165}{330} (330 + v)$
$\frac{4}{7} (330 - v) = \frac{1}{2} (330 + v)$
$8(330 - v) = 7(330 + v)$
$2640 - 8v = 2310 + 7v$
$15v = 330$
$v = 22 \ m/s$.

(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:
$f' = f \left( \frac{v + v_{car}}{v - v_{car}} \right)$
Given that the reflected frequency $f' = 2f$,we substitute this into the equation:
$2f = f \left( \frac{v + v_{car}}{v - v_{car}} \right)$
$2 = \frac{v + v_{car}}{v - v_{car}}$
$2(v - v_{car}) = v + v_{car}$
$2v - 2v_{car} = v + v_{car}$
$v = 3v_{car}$
$v_{car} = v/3$
Therefore,the velocity of the car is $v/3$.

(A) According to the Doppler effect,when a source of sound moves with a constant velocity $v_s$ relative to a stationary observer,the observed frequency $f$ is given by:
$1$. When the engine is approaching the observer: $f' = f_0 \left( \frac{v}{v - v_s} \right)$,where $v$ is the speed of sound and $f_0$ is the source frequency. This frequency is constant and higher than $f_0$.
$2$. When the engine is moving away from the observer: $f'' = f_0 \left( \frac{v}{v + v_s} \right)$. This frequency is also constant and lower than $f_0$.
Since the engine passes the observer at time $t_0$,the frequency heard by the observer will be a constant higher value for $t < t_0$ and will abruptly drop to a constant lower value for $t > t_0$. This behavior is represented by the graph in option $A$.

(B) For an observer approaching a stationary source,the apparent frequency $n'$ is given by the Doppler effect formula:
$n' = n \left( \frac{v + v_0}{v} \right)$
where $v$ is the speed of sound and $v_0$ is the velocity of the observer.
Since the observer starts from rest and moves with uniform acceleration $a$,their velocity at time $t$ is $v_0 = at$.
Substituting this into the frequency formula:
$n' = n \left( \frac{v + at}{v} \right) = n \left( 1 + \frac{at}{v} \right) = \left( \frac{na}{v} \right) t + n$
This is an equation of the form $y = mx + c$,which represents a straight line.
Here,the intercept on the $n'$ axis is $n$ (at $t = 0$),and the slope of the line is $\frac{na}{v}$,which is positive.
Therefore,the graph is a straight line with a positive slope starting from $n$.

(C) According to the Doppler effect, when a source moves towards a stationary observer, the observed frequency $n'$ is given by:
$n' = \left( \frac{V}{V - v_s} \right) n$
Given:
Frequency of source $n = 90 \, \text{vibrations/sec}$
Velocity of source $v_s = \frac{1}{10} V$
Substituting the values:
$n' = \left( \frac{V}{V - \frac{V}{10}} \right) \times 90$
$n' = \left( \frac{V}{\frac{9V}{10}} \right) \times 90$
$n' = \frac{10}{9} \times 90$
$n' = 100 \, \text{vibrations/sec}$

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

(C) Given: Velocity of car $v_s = 72 \,km/hr = 72 \times \frac{5}{18} = 20 \,ms^{-1}$.
Frequency of source $n = 124 \,vib/sec$.
Speed of sound $v = 330 \,ms^{-1}$.
The sound reflects from the wall. The wall acts as a stationary source for the reflected sound,and the car acts as a moving observer.
The frequency heard by the driver is given by the Doppler effect formula for a moving observer and stationary source:
$n' = n \left( \frac{v + v_o}{v - v_s} \right)$
Since the car is moving towards the wall,the wall receives sound at a higher frequency $n_w = n \left( \frac{v}{v - v_s} \right)$.
Then,the driver (moving towards the wall) receives the reflected sound at frequency $n' = n_w \left( \frac{v + v_o}{v} \right)$.
Substituting $v_o = v_s = 20 \,ms^{-1}$:
$n' = n \left( \frac{v + v_s}{v - v_s} \right) = 124 \left( \frac{330 + 20}{330 - 20} \right) = 124 \left( \frac{350}{310} \right) = 124 \times 1.129 \approx 140 \,vib/sec$.

(A) The sound reflects off the wall and returns to the driver. The wall acts as a stationary source of sound.
First,the frequency received by the wall is $f_w = f \left( \frac{v}{v - v_s} \right)$,where $v = 330 \, m/s$,$v_s = 30 \, m/s$,and $f = 600 \, Hz$.
$f_w = 600 \left( \frac{330}{330 - 30} \right) = 600 \left( \frac{330}{300} \right) = 600 \times 1.1 = 660 \, Hz$.
Now,the driver acts as an observer moving towards this reflected sound source (the wall) with velocity $v_o = 30 \, m/s$.
The frequency heard by the driver is $f' = f_w \left( \frac{v + v_o}{v} \right)$.
$f' = 660 \left( \frac{330 + 30}{330} \right) = 660 \left( \frac{360}{330} \right) = 2 \times 360 = 720 \, Hz$.

(B) The Doppler effect formula for the apparent frequency $f'$ is given by:
$f' = f \left( \frac{v + v_o}{v - v_s} \right)$
Here,the velocity of sound is $v = V$.
The observer is moving towards the source,so $v_o = V/10$.
The source is moving towards the observer,so $v_s = V/10$.
Substituting these values into the formula:
$f' = f \left( \frac{V + V/10}{V - V/10} \right)$
$f' = f \left( \frac{11V/10}{9V/10} \right)$
$f' = f \left( \frac{11}{9} \right)$
$f' \approx 1.22 f$
Therefore,the frequency heard by the observer is $1.22 f$.

(A) Let the speed of sound be $v = 320 \, m/s$,the speed of the source be $v_s = 4 \, m/s$,and the source frequency be $n = 240 \, Hz$.
For the train moving towards the man,the apparent frequency is $n_1 = \frac{v}{v - v_s} \cdot n = \frac{320}{320 - 4} \cdot 240 = \frac{320}{316} \cdot 240 \approx 243.038 \, Hz$.
For the train moving away from the man,the apparent frequency is $n_2 = \frac{v}{v + v_s} \cdot n = \frac{320}{320 + 4} \cdot 240 = \frac{320}{324} \cdot 240 \approx 237.037 \, Hz$.
The number of beats heard per second is the difference between the apparent frequencies: $n_{beat} = n_1 - n_2$.
$n_{beat} = 320 \cdot 240 \cdot \left( \frac{1}{316} - \frac{1}{324} \right) = 76800 \cdot \left( \frac{324 - 316}{316 \cdot 324} \right) = 76800 \cdot \frac{8}{102384} \approx 6 \, Hz$.

(D) According to the Doppler effect,the apparent frequency $n'$ heard by a stationary observer when the source moves with velocity $v_s$ towards the observer is given by:
$n' = n \left( \frac{v}{v - v_s} \right)$
Given that the observed frequency is half the original frequency,$n' = \frac{n}{2}$.
Substituting this into the formula:
$\frac{n}{2} = n \left( \frac{v}{v - v_s} \right)$
$\frac{1}{2} = \frac{v}{v - v_s}$
$v - v_s = 2v$
$v_s = -v$
The negative sign indicates that the source must move away from the observer at a speed of $v$ to hear half the frequency.

(B) The car acts as the source and the hill acts as the observer.
First,we calculate the frequency $f_1$ heard by the hill:
$f_1 = f_0 \times \frac{v}{v - v_s} = 600 \times \frac{330}{330 - 30} = 600 \times \frac{330}{300} = 660 \, Hz$.
Now,the hill reflects this sound,acting as the source,and the driver acts as the observer moving towards the source.
The frequency $f_2$ heard by the driver is:
$f_2 = f_1 \times \frac{v + v_o}{v} = 660 \times \frac{330 + 30}{330} = 660 \times \frac{360}{330} = 2 \times 360 = 720 \, Hz$.

(C) The speed of the train (source and observer) is $v_T = 220\, m s^{-1}$.
The speed of sound in air is $v = 330\, m s^{-1}$.
The source (train) emits a frequency $f_0 = 1000\, Hz$.
First,the sound reaches the stationary object. The frequency received by the object is $f_1 = f_0 \left( \frac{v}{v - v_T} \right)$.
Then,the object reflects this sound back to the moving train. The train acts as an observer moving towards the reflected sound source. The frequency detected by the driver is $f' = f_1 \left( \frac{v + v_T}{v} \right)$.
Substituting $f_1$ into the equation for $f'$,we get $f' = f_0 \left( \frac{v}{v - v_T} \right) \left( \frac{v + v_T}{v} \right) = f_0 \left( \frac{v + v_T}{v - v_T} \right)$.
Calculating the value: $f' = 1000 \left( \frac{330 + 220}{330 - 220} \right) = 1000 \left( \frac{550}{110} \right) = 1000 \times 5 = 5000\, Hz$.

(C) The speed of the motorcyclist (observer) is $v_o = 36\, km\, h^{-1} = 36 \times \frac{5}{18} = 10\, m s^{-1}$.
Since the motorcyclist is moving towards the source (car),the observer's velocity is taken as positive.
The speed of the car (source) is $v_s = 18\, km\, h^{-1} = 18 \times \frac{5}{18} = 5\, m s^{-1}$.
Since the source is moving away from the observer,the source's velocity is taken as positive.
The speed of sound is $v = 343\, m s^{-1}$.
The frequency of the source is $f_0 = 1392\, Hz$.
Using the Doppler effect formula for frequency heard by the observer,$f' = f_0 \left( \frac{v + v_o}{v + v_s} \right)$,
$f' = 1392 \left( \frac{343 + 10}{343 + 5} \right)$,
$f' = 1392 \left( \frac{353}{348} \right)$,
$f' = 4 \times 353 = 1412\, Hz$.

(A) Given:
Frequency of source,$f_{0} = 100 \, Hz$
Velocity of source,$v_{s} = 19.4 \, m s^{-1}$
Velocity of sound in air,$v = 330 \, m s^{-1}$
The component of the source velocity along the line joining the source and the observer is $v_{s} \cos 60^{\circ}$. Since the source is moving away from the observer,the apparent frequency $f'$ is given by the Doppler effect formula:
$f' = f_{0} \left( \frac{v}{v + v_{s} \cos 60^{\circ}} \right)$
Substituting the values:
$f' = 100 \left( \frac{330}{330 + 19.4 \times \cos 60^{\circ}} \right)$
$f' = 100 \left( \frac{330}{330 + 19.4 \times 0.5} \right)$
$f' = 100 \left( \frac{330}{330 + 9.7} \right)$
$f' = 100 \left( \frac{330}{339.7} \right) \approx 97.14 \, Hz$
Rounding to the nearest integer,the apparent frequency is $97 \, Hz$.

(B) The frequency of the sound emitted by the siren is $f_0 = 800 \, Hz$.
The speed of the source (siren) is $v_s = 15 \, m s^{-1}$.
The speed of sound in air is $v = 330 \, m s^{-1}$.
Since the siren is moving towards the cliff,the cliff acts as an observer receiving the sound. The frequency $f'$ received by the cliff is given by the Doppler effect formula for a moving source and stationary observer:
$f' = f_0 \left( \frac{v}{v - v_s} \right) = 800 \left( \frac{330}{330 - 15} \right) = 800 \left( \frac{330}{315} \right) \approx 838.09 \, Hz$.
The cliff reflects this sound back to the original observer. Since the cliff is stationary,it acts as a stationary source emitting sound at frequency $f'$. The original observer is stationary relative to the cliff,so they hear the reflected sound at the same frequency $f'$.
Therefore,the frequency heard by the observer is approximately $838 \, Hz$.

(B) The apparent frequency $v^{\prime}$ heard by the observer is given by the Doppler effect formula for sound:
$v^{\prime} = v \left( \frac{v + v_{o}}{v - v_{s}} \right)$
Here,the velocity of sound $v = 340 \, m s^{-1}$.
The source is the first car,so $v_{s} = 22 \, m s^{-1}$.
The observer is the second car,so $v_{o} = 16.5 \, m s^{-1}$.
The source frequency $v = 400 \, Hz$.
Substituting the values into the formula:
$v^{\prime} = 400 \left( \frac{340 + 16.5}{340 - 22} \right)$
$v^{\prime} = 400 \left( \frac{356.5}{318} \right)$
$v^{\prime} = 400 \times 1.121069...$
$v^{\prime} \approx 448.42 \, Hz$
Rounding to the nearest integer,the frequency heard is $448 \, Hz$.

(A) The Doppler effect depends on the relative velocity between the source and the observer along the line joining them.
In this scenario,the engine (source) is moving along a circular path,and the observer is at the center of the circle.
The velocity vector of the engine is always tangential to the circular path,while the line joining the engine and the observer is along the radius.
Since the tangent to a circle is always perpendicular to the radius at the point of contact,the velocity of the source along the line joining the source and the observer is always zero.
Because there is no component of velocity along the line of sight (radial direction),there is no relative motion between the source and the observer along the line joining them.
Therefore,the apparent frequency heard by the observer remains equal to the original frequency of the whistle,which is $500 \ Hz$.

(A) $1$. First,the enemy submarine acts as a receiver moving towards the source (Indian submarine). The frequency $f'$ received by the enemy submarine is: $f' = f \left( \frac{v + v_E}{v} \right) = 1000 \left( \frac{5500 + 70}{5500} \right) \approx 1012.73 \ Hz$.
$2$. The enemy submarine then acts as a source reflecting this frequency back to the Indian submarine. The Indian submarine is moving towards this source. The frequency $f''$ detected by the Indian submarine is: $f'' = f' \left( \frac{v + v_I}{v} \right) = 1012.73 \left( \frac{5500 + 50}{5500} \right) \approx 1012.73 \times 1.00909 \approx 1021.94 \ Hz$.
$3$. Converting to $kHz$,we get $f'' \approx 1.02 \ kHz$.

(C) The apparent frequency $n_1$ of the train coming towards the observer is given by $n_1 = n \left( \frac{v}{v - u} \right)$.
The apparent frequency $n_2$ of the train going away from the observer is given by $n_2 = n \left( \frac{v}{v + u} \right)$.
The number of beats produced is the difference between the two apparent frequencies: $\Delta n = n_1 - n_2$.
$\Delta n = n \left( \frac{v}{v - u} - \frac{v}{v + u} \right) = n \left( \frac{v(v + u) - v(v - u)}{v^2 - u^2} \right) = n \left( \frac{2uv}{v^2 - u^2} \right)$.
Substituting the given values: $n = 300 \; Hz$,$v = 332 \; m/s$,$u = 4 \; m/s$.
$\Delta n = 300 \times \left( \frac{2 \times 4 \times 332}{332^2 - 4^2} \right) = 300 \times \left( \frac{2656}{110224 - 16} \right) = 300 \times \left( \frac{2656}{110208} \right) \approx 300 \times 0.0241 = 7.23$.
Rounding to the nearest integer,the number of beats is $7$.

(D) When the source approaches the stationary observer,the apparent frequency $\nu'$ is given by $\nu' = \nu \left( \frac{v}{v - v_s} \right)$.
Given that the frequency increases by $10\%$,we have $\frac{\nu'}{\nu} = 1.1 = \frac{11}{10}$.
So,$\frac{v}{v - v_s} = \frac{11}{10} \implies 10v = 11v - 11v_s \implies 11v_s = v \implies v_s = \frac{v}{11}$.
When the source recedes from the observer,the apparent frequency $\nu''$ is given by $\nu'' = \nu \left( \frac{v}{v + v_s} \right)$.
Substituting $v_s = \frac{v}{11}$,we get $\frac{\nu''}{\nu} = \frac{v}{v + v/11} = \frac{v}{12v/11} = \frac{11}{12} \approx 0.9167$.
The change in frequency is $\nu - \nu'' = \nu - 0.9167\nu = 0.0833\nu$.
The percentage change is $0.0833 \times 100 \% \approx 8.33 \%$,which is approximately $8.5 \%$.

(C) According to the Doppler effect,the apparent frequency $f'$ heard by an observer is given by $f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right)$,where $f$ is the source frequency,$v$ is the speed of sound,$v_o$ is the velocity of the observer,and $v_s$ is the velocity of the source.
In this problem,the source $S_1$,the observer $O$,and the source $S_2$ are all moving with the same velocity $v_s = v_o = 30\,m/s$ in the same direction.
For source $S_1$: The observer is moving away from $S_1$ and $S_1$ is moving towards the observer. The apparent frequency $f_1$ is $f_1 = f \left( \frac{v - v_o}{v - v_s} \right)$. Since $v_o = v_s$,$f_1 = f \left( \frac{v - v_s}{v - v_s} \right) = f = 100\,Hz$.
For source $S_2$: The observer is moving towards $S_2$ and $S_2$ is moving away from the observer. The apparent frequency $f_2$ is $f_2 = f \left( \frac{v + v_o}{v + v_s} \right)$. Since $v_o = v_s$,$f_2 = f \left( \frac{v + v_s}{v + v_s} \right) = f = 100\,Hz$.
The beat frequency is the difference between the two apparent frequencies: $f_{beat} = |f_1 - f_2| = |100 - 100| = 0\,Hz$.

(B) Let $n_1$ be the frequency observed by the observer before the source crosses the observer,and $n_2$ be the frequency after the source crosses the observer. We are given that $n_2 = f n_1$ ... $(1)$.
The natural frequency of the source is $n_0$. Using the Doppler effect formula for a moving source and stationary observer:
$n_1 = \frac{V}{V - V_s} n_0$
$n_2 = \frac{V}{V + V_s} n_0$
Substituting these into equation $(1)$:
$\frac{V}{V + V_s} n_0 = f \left( \frac{V}{V - V_s} n_0 \right)$
$\frac{1}{V + V_s} = \frac{f}{V - V_s} \implies V - V_s = fV + fV_s$
$V(1 - f) = V_s(1 + f) \implies \frac{V_s}{V} = \frac{1 - f}{1 + f}$ ... $(2)$.
The difference in frequencies is $n_1 - n_2 = n_0 \left( \frac{V}{V - V_s} - \frac{V}{V + V_s} \right) = n_0 \left( \frac{1}{1 - V_s/V} - \frac{1}{1 + V_s/V} \right)$.
Substituting $V_s/V = \frac{1 - f}{1 + f}$:
$n_1 - n_2 = n_0 \left( \frac{1}{1 - \frac{1-f}{1+f}} - \frac{1}{1 + \frac{1-f}{1+f}} \right) = n_0 \left( \frac{1+f}{2f} - \frac{1+f}{2} \right) = n_0 \left( \frac{1+f - f(1+f)}{2f} \right) = n_0 \left( \frac{1+f - f - f^2}{2f} \right) = \frac{1}{2} n_0 \left( \frac{1 - f^2}{f} \right)$.

(B) Let the speed of the observer be $v_o = v$ and the speed of the source be $v_s = 2v$.
At time $t$,the position of the observer is $(0, vt)$ and the position of the source is $(2vt, 0)$.
The velocity vector of the observer is $\vec{v}_o = v \hat{j}$ and the velocity vector of the source is $\vec{v}_s = 2v \hat{i}$.
The unit vector from the source to the observer is $\hat{r} = \frac{-2vt \hat{i} + vt \hat{j}}{\sqrt{(2vt)^2 + (vt)^2}} = \frac{-2 \hat{i} + \hat{j}}{\sqrt{5}}$.
The component of the source velocity along the line of sight is $v_{s,r} = \vec{v}_s \cdot \hat{r} = (2v \hat{i}) \cdot \left( \frac{-2 \hat{i} + \hat{j}}{\sqrt{5}} \right) = -\frac{4v}{\sqrt{5}}$.
The component of the observer velocity along the line of sight is $v_{o,r} = \vec{v}_o \cdot \hat{r} = (v \hat{j}) \cdot \left( \frac{-2 \hat{i} + \hat{j}}{\sqrt{5}} \right) = \frac{v}{\sqrt{5}}$.
The apparent frequency is given by $f = f_0 \left( \frac{c - v_{o,r}}{c - v_{s,r}} \right)$,where $c$ is the speed of sound.
Substituting the values,$f = f_0 \left( \frac{c - v/\sqrt{5}}{c + 4v/\sqrt{5}} \right)$.
Since $v_{s,r}$ and $v_{o,r}$ are constant,the apparent frequency $f$ is constant over time.
Therefore,the graph of $f$ versus $t$ is a horizontal straight line. Comparing this with the given options,Graph $B$ represents a constant frequency $f < f_0$.

(D) The problem involves two steps of the Doppler effect.
Step $1$: The reflecting surface acts as an observer moving away from the source $S$. The frequency $f'$ received by the surface is given by $f' = f \left( \frac{V - u}{V} \right)$,where $f = 334 \, Hz$,$V = 330 \, m/s$,and $u = 2 \, m/s$.
$f' = 334 \left( \frac{330 - 2}{330} \right) = 334 \left( \frac{328}{330} \right) \approx 331.97 \, Hz$.
Step $2$: The reflecting surface now acts as a moving source reflecting the sound back to the stationary observer $O$. The frequency $f''$ heard by the observer is $f'' = f' \left( \frac{V}{V + u} \right)$.
$f'' = 334 \left( \frac{328}{330} \right) \times \left( \frac{330}{330 + 2} \right) = 334 \left( \frac{328}{332} \right) = 334 \times 0.98795 \approx 330 \, Hz$.

(C) The frequency observed by a moving detector approaching a stationary source is given by the Doppler effect formula: $f = f_0 \left( \frac{v + v_0}{v} \right)$,where $v$ is the speed of sound and $v_0$ is the speed of the detector.
Since the detector is released from rest and falls under gravity,its speed at time $t$ is $v_0 = gt = 10t$.
Substituting this into the formula: $f = 10^3 \left( \frac{v + 10t}{v} \right) = 10^3 + \left( \frac{10^4}{v} \right)t$.
This equation represents a straight line with a slope $m = \frac{10^4}{v}$.
From the graph,the frequency changes from $1000 \, Hz$ to $2000 \, Hz$ in $30 \, s$.
Therefore,the slope is $m = \frac{2000 - 1000}{30} = \frac{1000}{30} = \frac{100}{3}$.
Equating the slopes: $\frac{10^4}{v} = \frac{100}{3}$.
Solving for $v$: $v = \frac{10^4 \times 3}{100} = 100 \times 3 = 300 \, m/s$.

(A) For an observer approaching a stationary source,the apparent frequency $f'$ is given by the Doppler effect formula:
$f' = f \left( \frac{v + v_0}{v} \right)$
where $v$ is the speed of sound and $v_0$ is the velocity of the observer.
Since the observer starts from rest and moves with uniform acceleration $a$,their velocity at time $t$ is $v_0 = at$.
Substituting this into the frequency formula:
$f' = f \left( \frac{v + at}{v} \right) = f \left( 1 + \frac{a}{v} t \right) = \left( \frac{fa}{v} \right) t + f$
This equation is of the form $y = mx + c$,which represents a straight line.
Here,the intercept on the $f'$ axis is $f$ (at $t = 0$) and the slope is $\frac{fa}{v}$,which is positive.
Therefore,the graph is a straight line with a positive intercept and a positive slope.

(A) After $t = 1\, s$ of free fall,the velocity of the source is $v_s = g \times t = 10 \times 1 = 10\, m/s$ (downward).
The observer in the balloon is moving upward with $v_o = 2\, m/s$.
Let $v = 300\, m/s$ be the speed of sound.
Before crossing,the source is moving towards the observer and the observer is moving towards the source. The observed frequency $n_1$ is:
$n_1 = f_0 \left( \frac{v + v_o}{v - v_s} \right) = 150 \left( \frac{300 + 2}{300 - 10} \right) = 150 \left( \frac{302}{290} \right) \approx 156.21\, Hz$.
After crossing,the source is moving away from the observer and the observer is moving away from the source. The observed frequency $n_2$ is:
$n_2 = f_0 \left( \frac{v - v_o}{v + v_s} \right) = 150 \left( \frac{300 - 2}{300 + 10} \right) = 150 \left( \frac{298}{310} \right) \approx 144.19\, Hz$.
The difference in frequency is $\Delta n = n_1 - n_2 = 156.21 - 144.19 = 12.02\, Hz \approx 12\, Hz$.

(D) The Doppler effect formula is given by $n' = n \left( \frac{v \pm v_o}{v \mp v_s} \right)$.
For whistle $A$ (source stationary,$v_s = 0$) and observer moving away at $25\,m/s$ $(v_o = 25\,m/s)$:
$n'_A = 500 \left( \frac{350 - 25}{350} \right) = 500 \times \frac{325}{350} \approx 464.28\,Hz$.
For whistle $B$ (source moving away at $50\,m/s$,$v_s = 50\,m/s$) and observer moving towards it at $25\,m/s$ $(v_o = 25\,m/s)$:
$n'_B = 500 \left( \frac{350 + 25}{350 + 50} \right) = 500 \times \frac{375}{400} = 468.75\,Hz$.
Difference in frequencies heard by the observer: $\Delta n = |n'_B - n'_A| = |468.75 - 464.28| = 4.47\,Hz \approx 4.5\,Hz$.
Thus,statements $(B)$ and $(C)$ are correct.

(C) Let $v$ be the speed of sound and $n$ be the frequency of the sources. Since $\lambda = v/n$,we have $n = v/\lambda$.
For the $1^{st}$ source,the observer is moving away with velocity $u$. The apparent frequency is $n_1 = n \left( \frac{v - u}{v} \right) = n \left( 1 - \frac{u}{v} \right)$.
For the $2^{nd}$ source,the observer is moving towards it with velocity $u$. The apparent frequency is $n_2 = n \left( \frac{v + u}{v} \right) = n \left( 1 + \frac{u}{v} \right)$.
The beat frequency is the difference between the two apparent frequencies:
$|n_2 - n_1| = n \left( 1 + \frac{u}{v} \right) - n \left( 1 - \frac{u}{v} \right) = n \left( \frac{2u}{v} \right)$.
Substituting $n = v/\lambda$,we get:
Beat frequency $= \left( \frac{v}{\lambda} \right) \left( \frac{2u}{v} \right) = \frac{2u}{\lambda}$.

(B) The car acts as a source moving towards a stationary hill with speed $v_c$. The wavelength of the sound waves emitted by the car and reaching the hill is given by $\lambda = \frac{v - v_c}{f}$.
The sound reflects off the hill and acts as a source moving away from the observer. The frequency of the reflected sound heard by the observer is $f' = f \left( \frac{v + v_o}{v - v_c} \right)$.
The frequency of the direct sound heard by the observer is $f'' = f \left( \frac{v + v_o}{v - v_c} \right)$.
Wait,the beat frequency is the difference between the frequency of the reflected sound and the direct sound. The reflected sound frequency $f_{ref} = f \left( \frac{v + v_o}{v - v_c} \right)$ and the direct sound frequency $f_{dir} = f \left( \frac{v - v_o}{v - v_c} \right)$.
The beat frequency $f_b = |f_{ref} - f_{dir}| = f \left( \frac{v + v_o}{v - v_c} - \frac{v - v_o}{v - v_c} \right) = f \left( \frac{2v_o}{v - v_c} \right)$.
Re-evaluating the provided options,option $(B)$ is correct as the wavelength is $\frac{v - v_c}{f}$. Therefore,$(B)$ is the correct choice.

(D) The apparent frequency $f'$ heard by a stationary observer when the source approaches with velocity $v_s$ is given by the Doppler effect formula: $f' = f \left( \frac{v}{v - v_s} \right)$.
Given: source frequency $f = 9500 \ Hz$,speed of sound $v = 300 \ ms^{-1}$,and maximum audible frequency $f' = 10000 \ Hz$.
Substituting the values: $10000 = 9500 \left( \frac{300}{300 - v} \right)$.
Simplifying the equation: $\frac{10000}{9500} = \frac{300}{300 - v} \Rightarrow \frac{20}{19} = \frac{300}{300 - v}$.
Cross-multiplying: $20(300 - v) = 19 \times 300 \Rightarrow 6000 - 20v = 5700$.
$20v = 300 \Rightarrow v = 15 \ ms^{-1}$.
Thus,the maximum value of $v$ is $15 \ ms^{-1}$.

(B) The observer (motorcycle) is in motion and the source (siren) is at rest. The Doppler effect formula is given by $n' = n \left( \frac{v - v_O}{v} \right)$,where $v$ is the speed of sound and $v_O$ is the speed of the observer.
Given $n' = 0.94n$,we have $0.94n = n \left( \frac{330 - v_O}{330} \right)$.
Simplifying this,$0.94 \times 330 = 330 - v_O$.
$v_O = 330 - 310.2 = 19.8 \; m/s$.
The motorcycle starts from rest $(u = 0)$ and accelerates at $a = 2 \; m/s^2$. Using the equation of motion $v_O^2 = u^2 + 2as$:
$(19.8)^2 = 0^2 + 2 \times 2 \times s$.
$392.04 = 4s$.
$s = 98.01 \; m \approx 98 \; m$.

(A) The frequency heard by the observer when the train is approaching is given by $f_1 = f \left( \frac{v}{v - v_s} \right)$, where $v = 320 \ ms^{-1}$ and $v_s = 20 \ ms^{-1}$.
$f_1 = 1000 \times \left( \frac{320}{320 - 20} \right) = 1000 \times \frac{320}{300} \ Hz$.
When the train is moving away, the frequency heard is $f_2 = f \left( \frac{v}{v + v_s} \right)$.
$f_2 = 1000 \times \left( \frac{320}{320 + 20} \right) = 1000 \times \frac{320}{340} \ Hz$.
The change in frequency is $\Delta f = f_1 - f_2$.
The percentage change relative to the approaching frequency is $\frac{f_1 - f_2}{f_1} \times 100 = \left( 1 - \frac{f_2}{f_1} \right) \times 100$.
Substituting the values: $\left( 1 - \frac{320/340}{320/300} \right) \times 100 = \left( 1 - \frac{300}{340} \right) \times 100 = \left( \frac{40}{340} \right) \times 100 \approx 11.76 \% \approx 12 \%$.