physics-class-11-unit-15-chapter-05-problem-solving-doppler-effect-l-5-5-3dqhdomy6io

### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
### <primary style="font-size:24px"> Problems Based on Doppler Effect </primary>
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- Frequency = $\frac {velocity }{wavelength}$

- $\nu^\prime=\frac{\text {velocity of sound as measured by observer}}{\text {wavelength of sound}}$

- Wavelength = The distance between successive crest.

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<footer style = "font-size: 18px">Problem Solving Session Doppler Effect $\rightarrow$ Problems Based on Doppler Effect $\rightarrow$ <span style = "color:blue"> Problems Based on Doppler Effect </span> $\rightarrow$ Problems Based on Doppler Effect $\rightarrow$ Problems Based on Doppler Effect </footer>

### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
### <primary style="font-size:24px"> Problems Based on Doppler Effect </primary>
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- The source is moving away from the observer

- u is the speed of sound

- $L =u T_0-U T_0 = (u-U) T_0$

- $ \nu^{\prime}=\frac{u}{(u-U) T_0} $

- $ \frac{1}{T_0}=\nu $

- $ \nu^{\prime}=\nu \frac{u}{u-U}$

- $ \nu^{\prime}=\nu \frac{u}{u \pm U_s}$

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### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
### <primary style="font-size:24px"> Problems Based on Doppler Effect </primary>
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- Source is moving towards the observer.

- $L=u T_0$

- $\nu^{\prime}=\frac{u-(-V)}{u T_0}=\frac{u+V}{u T_0} $

- $\nu^{\prime}=\nu (\frac{u+V}{u})$

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### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
### <primary style="font-size:24px"> Problems Based on Doppler Effect </primary>
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- Both source and observer are moving towards each other.

- $\nu^\prime=\frac{\text {velocity of sound as measured by observer}}{\text {wavelength of sound}}$

- $=\frac{u-(-V)}{(u-U) T_0} $

- $\nu^{\prime}=\nu \frac{u+V}{u-U}$

- If both source and observer moving away from each other:

- $\nu^{\prime}=\nu \frac{u-V}{u+U}$

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-11-unit-15-chapter-05-problem-solving-doppler-effect-l-5_5-3dqhdomy6io-09.jpg)
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-11-unit-15-chapter-05-problem-solving-doppler-effect-l-5_5-3dqhdomy6io-10.jpg)

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### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- A police car with a siren of frequency $8 {kHz}$ is moving with uniform velocpress Esc to exit tul screen ards a tall building which reflect the sound waves. The speed of sound in air is $320 {m} /{s}$. The frequency of the siren heard by the car driver is        (2011)

- (A) $8.50{kHz}$

- (B) $8.25 \mathrm{kHz}$

- (C) $7.75 \mathrm{kHz}$

- (D) $7.50 \mathrm{kHz}$

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<footer style = "font-size: 18px">Problems Based on Doppler Effect $\rightarrow$ Problems Based on Doppler Effect $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- $V = 36 km/hr = 10 m/s$

- $\nu_1$ frequency measured by an observer (at rest) in the building

- $\nu_1  =\nu_s \frac{u}{(u-V)}=8 \frac{320}{310} {kHz}$

- $\nu^{\prime}  =\nu_1 \frac{u+v}{u} $

- $ =8 \times \frac{320}{310} \times \frac{330}{320}=8 \times \frac{33}{31}$

- $\approx 8.5 {KH}_2$

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### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
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- A police car moving at $22{m} / {s}$ sounds a horn at $176 {Hz}$ while chasing a motor-cyclist, Both, police car and motor-cyclist, are moving towards a stationary siren of frequency $165{Hz}$. What is the speexl of the motorcycle if it is given that the motorecyclist does not obrerve beats? (Speed of sound $=330 {m} / {s}$.)
- $(2003)$

- (A) $33 {m} /{s}$

-  (B) $22 {m} / {s}$

- (C) zero

-  (D) $11 {m} / {s}$
 
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### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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 - $\nu_1$ frequency measured by Motor cyclist when $s_1$ is taken as the source
 
- $ \nu_1=\nu_p \frac{u-v}{(u-v)}=176 \times \frac{330-v}{330-22} \rightarrow(i)$
 
- $ \nu_2=\nu_s \frac{u+v}{u}=165 \times \frac{330+v}{{330}}+(i i)$

- $ \nu_1-\nu_2=0 $

- $ \Rightarrow \frac{1764}{3087} \times(330-v)=\frac{1}{2} \times(330+v) $

- $ \Rightarrow \quad v=\frac{330}{15}=22 \{m} / {s}$
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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-11-unit-15-chapter-05-problem-solving-doppler-effect-l-5_5-3dqhdomy6io-14.jpg)
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-11-unit-15-chapter-05-problem-solving-doppler-effect-l-5_5-3dqhdomy6io-15.jpg)

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### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- A siren placed at a railway platform is emitting sound of frequency $5 \mathrm{kHz}$. A passenger sitting in a moving train $A$ records a frequency of $5.5 \mathrm{kHz}$, while the train approaches the siren. During his return journey in a different train $B$ he records a frequency of $6.0 \mathrm{kHz}$ while approaching the same siren. The ratio of the velocity of $\operatorname{train} B$ to that of $\operatorname{train} A$ is
- (2002)

- (A) $242 / 252$

- (B) 2

- (C) $5 / 6$

- (D) $11 / 6$

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### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\nu_A  ={\nu} \frac{u-v_1}{u}$

- $\Rightarrow  5 \cdot 5=5 \frac{u-v_1}{u}=5-5 \frac{v_1}{u} \rightarrow(i) $

- $\Rightarrow \quad \frac{1}{2}  =-5 \frac{v_1}{u} \rightarrow(i i) $

- $\nu_B  =\nu \frac{u-v_2}{u} $

- $\Rightarrow  6=5 \frac{u-v_2}{u}=5-5 \frac{v_2}{u}$

- $\Rightarrow  1=-5 \frac{v_2}{u} \rightarrow \text { (iii) }$

- $ \text { (iii) } / \text { (ii) }$

- $ \frac{v_2}{v_1}=2$
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
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- In hydrogen spectrum the wavelength of $\mathrm{H}_\alpha$ line is $656 \mathrm{~nm}$; whereas in the spectrum of a distant galaxy $\mathrm{H}_\alpha$ line wavelength is $706 {nm}$. Estimated speed of galaxy with respect to earth is         (1999)

- (A) $2 \times 10^8 {m} / {s}$

- (B) $2 \times 10^7 {m} / {s}$

- (C) $2 \times 10^6 {m} / {s}$

- (D) $2 \times 10^5 {m} / {s}$

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### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\nu^{\prime}=\frac {\text{velocity of light as measured by each observer}}{\text{wavelength of light measured by each observer}}$

- $\nu = \nu \frac{c}{c+v}$

- $\frac{1}{\lambda^{\prime}}=\frac{1}{\lambda} \frac{c}{c+v}$

- $\Rightarrow \quad \frac{656}{706}=\frac{c}{c+v} $
 
- $ \Rightarrow \quad 0.93 c+0.93 v=c $

- $ \Rightarrow \quad v=\frac{0.07}{0.93} c=2.3 \times 10^7 {m} / {s} $

- $ \Rightarrow \quad c=3 \times 10^8 {m} /{s}$

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### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
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- Two loudspeakers $M$ and $N$ are locatod $20 {m}$apart and emit sound at frequencies $118{Hz}$ and $121{Hz}_4$ respectively. A car is initially at a point $P, 1800 {m}$ away from the midpoint $Q$ of the line $M N$ and moves tomards $Q$ constantly at $60{km} / {hr}$ along the perpendicular bisector of MN. 
- It croses $Q$ and eventually reaches a point $R, 1800{m}$ away from $Q$. 
- Let $v(t)$ represent the beat frequency measured by a person sitting in the car at time $t$. Let $v_P, v_Q$ and $v_R$ be the beat frequencies measured at locations $P, Q$ and $R$, respectively. The speed of sound in air is $330{m} /{s}$. Which of the following statement $({s})$ is (are) true regarding the sound beard by the person?(2016)

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### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
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- (A) $v_P+v_R=2 v_0$.

- (B) The rate of chango in beat frequency is maximum when the car passes through Q.

- (C) The plot below represents schematically the waria. tion of beat frequency with time.

- (D) The plot below represemts shematically the variation of beat frequency with time.

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### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\nu_p=\nu_M^p-\nu_N^p $

- $\nu_M^P=\nu_M \frac{u+U \cos \theta}{u}$

- $ \nu_N^P=\nu_N \frac{u+U \cos \theta}{u}$

- ${\nu_P=\frac{u+U cos^\theta}{u}\left(\nu_M-\nu_N\right)}{\rightarrow(i)}$

- $\nu_Q=\nu_M-\nu_N \rightarrow(i i)$

- $\nu_R=$

- $\frac{u-U \cos \theta}{u}(\nu_M-\nu_N) \rightarrow($ iii $)$

- (i)+(iii)

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### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\nu_P+\nu_R$

- $=(\nu_M-\nu_N)[\frac{u+v \cos \theta}{u}+\frac{u-v \cos \theta}{u}] $

- $=2(\nu_M-\nu_N\) =2 \nu_Q $
 
- $\nu(t)=\frac{u+U \cos \theta(t)}{u}(\nu_M-\nu_N)$

- $ \cos \theta(t)=\frac{D-U t}{\sqrt{d^2+(D-U t)^2}}$

- $\frac{d \nu(t)}{d t}=\frac{d \nu}{d \theta} \frac{d \theta}{d t}$

- $=-\frac{U}{v} \sin \theta \dot{\theta}(\nu_M-\nu_N)$

- $ =-\frac{v^2}{u}(\nu_M-\nu_N) \frac{d^2}{\{d^2+(D-v t)\}^{3 / 2}}$

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### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
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- Two vehicles, each moving with speed $u$ on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity $w$. One of these vehicles blows a whistle of frequency $f_1$. An observer in the other vehicle hears the frequency of the whistle to be $f_2$. The speed of sound in still air is $v$. The correct statement(s) is (are)          (2013)

- (A) If the wind blows from the observer to the source, $f_2>f_1$.

- (B) If the wind blows from the source to the observer, $f_2>f_1$.

- (C) If the wind blows from the observer to the source, $f_2<f_1$.

- (D) If the wind blows from the source to the observer, $f_2<f_1$.

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### <Success style="font-size:25px"> Problem Solving Doppler Effect L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $f_2 = \frac{\text{velocity of sound as measured by B}}{\text{wavelength of sound as measure by B}}$

- V : Velocity of sound

- $=\frac{v+u}{(v-u) T_0}$

- $ =f_1 \frac{v+u}{v-u}$

- $f_2  =f_1 \frac{(v+w)+u}{(v+w)-u}$

- $f_2  =f_1 \frac{(v-w)+u}{(v-w)-u}$

- Wind is blowing from source to observer

- Wind is blowing from observer to source

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