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PHXI15:WAVES

354715 A source of sound and an observer are approaching each other with the same speed, which is equal to $\frac{1}{10}$ times the speed of sound. The apparent relative change in the frequency of the source is

1

22.2 %

increase

2

18.2 %

increase

3

22.2 %

decrease

4

18.2 %

decrease

Explanation:

Speed of sound $= v$ (say)
Speed of source $=$ speed of observer $= \frac{v}{10}$
Let natural frequency of sound $= n$
Apparent frequency of sound,
$n^{'} = n (\frac{v + v_{o}}{v - v_{s}})$
Since source is moving in the direction of sound and observer moving towards source, then
$v_{o} = \frac{v}{10} and v_{s} = \frac{v}{10}$
$∴ n^{'} = n \frac{(v + \frac{v}{10})}{(v - \frac{1}{10})} = (\frac{\frac{11 v}{10}}{\frac{9 v}{10}})$
$= \frac{11}{9} n \Rightarrow \frac{n^{'} - n}{n} = \frac{2}{9}$
$∴ \frac{n^{'} - n}{n} \times 100 % = \frac{2}{9} \times 100 % = \frac{200}{9} % = 22.22$

PHXI15:WAVES

354716 A source of sound emits sound waves at frequency $f_{0}$ . It is moving towards an observer with fixed speed $v_{s} (v_{s} < v$ , where $v$ is the speed of sound in air). If the observer were to move towards the source with speed $v_{0}$ , one of the following two graphs ( $A$ and $B$ ) will given the correct variation of the frequency $f$ heard by the observer as $v_{0}$ is changed.
supporting img

1 Graph

A

with slope

= \frac{f_{0}}{(v + v_{s})}

2 Graph

B

with slope

= \frac{f_{0}}{(v - v_{s})}

3 Graph

A

with slope

= \frac{f_{0}}{(v - v_{s})}

4 Graph

B

with slope

= \frac{f_{0}}{(v + v_{s})}

Explanation:

According to Doppler's effect,
Apparent, frequency $f = (\frac{v - v_{0}}{v - v_{s}}) f_{0}$
Now, $f = (\frac{- f_{0}}{v - v_{s}}) v_{0} + \frac{v f_{0}}{v - v_{s}}$
So, slope $= \frac{- f_{0}}{v - v_{s}}$

JEE - 2015

PHXI15:WAVES

354717 In Doppler effect, when a source moves towards a stationary observer, the apparent increase in frequency is due to

1 Increase in wavelength of sound received by observer

2 Decrease in wavelength of sound received by observer

3 Increase in velocity of waves received by observer

4 Decrease in number of waves received by observer in one second

PHXI15:WAVES

354718 A source of sound is moving with constant velocity of $20 m / s$ emitting a note of frequency $1000 H z$ . The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be

1

8 : 9

2

9 : 8

3

9 : 10

4

1 : 1

Explanation:

When source is approaching the observer, the frequency heard $n_{a} = (\frac{v}{v - v_{s}}) \times n = (\frac{340}{340 - 20}) \times 1000 = 1063 H z$
When source is receding, the frequency heard
$\begin{aligned} n_{r} = (\frac{v}{v + v_{s}}) \times n = \frac{340}{340 + 20} \times 1000 = 944 \\ \Rightarrow n_{a} : n_{r} = 9 : 8 \end{aligned}$

PHXI15:WAVES

354715 A source of sound and an observer are approaching each other with the same speed, which is equal to $\frac{1}{10}$ times the speed of sound. The apparent relative change in the frequency of the source is

1

22.2 %

increase

2

18.2 %

increase

3

22.2 %

decrease

4

18.2 %

decrease

Explanation:

Speed of sound $= v$ (say)
Speed of source $=$ speed of observer $= \frac{v}{10}$
Let natural frequency of sound $= n$
Apparent frequency of sound,
$n^{'} = n (\frac{v + v_{o}}{v - v_{s}})$
Since source is moving in the direction of sound and observer moving towards source, then
$v_{o} = \frac{v}{10} and v_{s} = \frac{v}{10}$
$∴ n^{'} = n \frac{(v + \frac{v}{10})}{(v - \frac{1}{10})} = (\frac{\frac{11 v}{10}}{\frac{9 v}{10}})$
$= \frac{11}{9} n \Rightarrow \frac{n^{'} - n}{n} = \frac{2}{9}$
$∴ \frac{n^{'} - n}{n} \times 100 % = \frac{2}{9} \times 100 % = \frac{200}{9} % = 22.22$

PHXI15:WAVES

354716 A source of sound emits sound waves at frequency $f_{0}$ . It is moving towards an observer with fixed speed $v_{s} (v_{s} < v$ , where $v$ is the speed of sound in air). If the observer were to move towards the source with speed $v_{0}$ , one of the following two graphs ( $A$ and $B$ ) will given the correct variation of the frequency $f$ heard by the observer as $v_{0}$ is changed.
supporting img

1 Graph

A

with slope

= \frac{f_{0}}{(v + v_{s})}

2 Graph

B

with slope

= \frac{f_{0}}{(v - v_{s})}

3 Graph

A

with slope

= \frac{f_{0}}{(v - v_{s})}

4 Graph

B

with slope

= \frac{f_{0}}{(v + v_{s})}

Explanation:

According to Doppler's effect,
Apparent, frequency $f = (\frac{v - v_{0}}{v - v_{s}}) f_{0}$
Now, $f = (\frac{- f_{0}}{v - v_{s}}) v_{0} + \frac{v f_{0}}{v - v_{s}}$
So, slope $= \frac{- f_{0}}{v - v_{s}}$

JEE - 2015

PHXI15:WAVES

354717 In Doppler effect, when a source moves towards a stationary observer, the apparent increase in frequency is due to

1 Increase in wavelength of sound received by observer

2 Decrease in wavelength of sound received by observer

3 Increase in velocity of waves received by observer

4 Decrease in number of waves received by observer in one second

PHXI15:WAVES

354718 A source of sound is moving with constant velocity of $20 m / s$ emitting a note of frequency $1000 H z$ . The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be

1

8 : 9

2

9 : 8

3

9 : 10

4

1 : 1

Explanation:

When source is approaching the observer, the frequency heard $n_{a} = (\frac{v}{v - v_{s}}) \times n = (\frac{340}{340 - 20}) \times 1000 = 1063 H z$
When source is receding, the frequency heard
$\begin{aligned} n_{r} = (\frac{v}{v + v_{s}}) \times n = \frac{340}{340 + 20} \times 1000 = 944 \\ \Rightarrow n_{a} : n_{r} = 9 : 8 \end{aligned}$

PHXI15:WAVES

354715 A source of sound and an observer are approaching each other with the same speed, which is equal to $\frac{1}{10}$ times the speed of sound. The apparent relative change in the frequency of the source is

1

22.2 %

increase

2

18.2 %

increase

3

22.2 %

decrease

4

18.2 %

decrease

Explanation:

Speed of sound $= v$ (say)
Speed of source $=$ speed of observer $= \frac{v}{10}$
Let natural frequency of sound $= n$
Apparent frequency of sound,
$n^{'} = n (\frac{v + v_{o}}{v - v_{s}})$
Since source is moving in the direction of sound and observer moving towards source, then
$v_{o} = \frac{v}{10} and v_{s} = \frac{v}{10}$
$∴ n^{'} = n \frac{(v + \frac{v}{10})}{(v - \frac{1}{10})} = (\frac{\frac{11 v}{10}}{\frac{9 v}{10}})$
$= \frac{11}{9} n \Rightarrow \frac{n^{'} - n}{n} = \frac{2}{9}$
$∴ \frac{n^{'} - n}{n} \times 100 % = \frac{2}{9} \times 100 % = \frac{200}{9} % = 22.22$

PHXI15:WAVES

354716 A source of sound emits sound waves at frequency $f_{0}$ . It is moving towards an observer with fixed speed $v_{s} (v_{s} < v$ , where $v$ is the speed of sound in air). If the observer were to move towards the source with speed $v_{0}$ , one of the following two graphs ( $A$ and $B$ ) will given the correct variation of the frequency $f$ heard by the observer as $v_{0}$ is changed.
supporting img

1 Graph

A

with slope

= \frac{f_{0}}{(v + v_{s})}

2 Graph

B

with slope

= \frac{f_{0}}{(v - v_{s})}

3 Graph

A

with slope

= \frac{f_{0}}{(v - v_{s})}

4 Graph

B

with slope

= \frac{f_{0}}{(v + v_{s})}

Explanation:

According to Doppler's effect,
Apparent, frequency $f = (\frac{v - v_{0}}{v - v_{s}}) f_{0}$
Now, $f = (\frac{- f_{0}}{v - v_{s}}) v_{0} + \frac{v f_{0}}{v - v_{s}}$
So, slope $= \frac{- f_{0}}{v - v_{s}}$

JEE - 2015

PHXI15:WAVES

354717 In Doppler effect, when a source moves towards a stationary observer, the apparent increase in frequency is due to

1 Increase in wavelength of sound received by observer

2 Decrease in wavelength of sound received by observer

3 Increase in velocity of waves received by observer

4 Decrease in number of waves received by observer in one second

PHXI15:WAVES

354718 A source of sound is moving with constant velocity of $20 m / s$ emitting a note of frequency $1000 H z$ . The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be

1

8 : 9

2

9 : 8

3

9 : 10

4

1 : 1

Explanation:

When source is approaching the observer, the frequency heard $n_{a} = (\frac{v}{v - v_{s}}) \times n = (\frac{340}{340 - 20}) \times 1000 = 1063 H z$
When source is receding, the frequency heard
$\begin{aligned} n_{r} = (\frac{v}{v + v_{s}}) \times n = \frac{340}{340 + 20} \times 1000 = 944 \\ \Rightarrow n_{a} : n_{r} = 9 : 8 \end{aligned}$

PHXI15:WAVES

354715 A source of sound and an observer are approaching each other with the same speed, which is equal to $\frac{1}{10}$ times the speed of sound. The apparent relative change in the frequency of the source is

1

22.2 %

increase

2

18.2 %

increase

3

22.2 %

decrease

4

18.2 %

decrease

Explanation:

Speed of sound $= v$ (say)
Speed of source $=$ speed of observer $= \frac{v}{10}$
Let natural frequency of sound $= n$
Apparent frequency of sound,
$n^{'} = n (\frac{v + v_{o}}{v - v_{s}})$
Since source is moving in the direction of sound and observer moving towards source, then
$v_{o} = \frac{v}{10} and v_{s} = \frac{v}{10}$
$∴ n^{'} = n \frac{(v + \frac{v}{10})}{(v - \frac{1}{10})} = (\frac{\frac{11 v}{10}}{\frac{9 v}{10}})$
$= \frac{11}{9} n \Rightarrow \frac{n^{'} - n}{n} = \frac{2}{9}$
$∴ \frac{n^{'} - n}{n} \times 100 % = \frac{2}{9} \times 100 % = \frac{200}{9} % = 22.22$

PHXI15:WAVES

354716 A source of sound emits sound waves at frequency $f_{0}$ . It is moving towards an observer with fixed speed $v_{s} (v_{s} < v$ , where $v$ is the speed of sound in air). If the observer were to move towards the source with speed $v_{0}$ , one of the following two graphs ( $A$ and $B$ ) will given the correct variation of the frequency $f$ heard by the observer as $v_{0}$ is changed.
supporting img

1 Graph

A

with slope

= \frac{f_{0}}{(v + v_{s})}

2 Graph

B

with slope

= \frac{f_{0}}{(v - v_{s})}

3 Graph

A

with slope

= \frac{f_{0}}{(v - v_{s})}

4 Graph

B

with slope

= \frac{f_{0}}{(v + v_{s})}

Explanation:

According to Doppler's effect,
Apparent, frequency $f = (\frac{v - v_{0}}{v - v_{s}}) f_{0}$
Now, $f = (\frac{- f_{0}}{v - v_{s}}) v_{0} + \frac{v f_{0}}{v - v_{s}}$
So, slope $= \frac{- f_{0}}{v - v_{s}}$

JEE - 2015

PHXI15:WAVES

354717 In Doppler effect, when a source moves towards a stationary observer, the apparent increase in frequency is due to

1 Increase in wavelength of sound received by observer

2 Decrease in wavelength of sound received by observer

3 Increase in velocity of waves received by observer

4 Decrease in number of waves received by observer in one second

PHXI15:WAVES

354718 A source of sound is moving with constant velocity of $20 m / s$ emitting a note of frequency $1000 H z$ . The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be

1

8 : 9

2

9 : 8

3

9 : 10

4

1 : 1

Explanation:

When source is approaching the observer, the frequency heard $n_{a} = (\frac{v}{v - v_{s}}) \times n = (\frac{340}{340 - 20}) \times 1000 = 1063 H z$
When source is receding, the frequency heard
$\begin{aligned} n_{r} = (\frac{v}{v + v_{s}}) \times n = \frac{340}{340 + 20} \times 1000 = 944 \\ \Rightarrow n_{a} : n_{r} = 9 : 8 \end{aligned}$