QBManager

WAVES

172969 Two trains are moving towards each other with speeds of $20 m / s$ and $15 m / s$ relative to the ground. The first train sounds a whistle of frequency $600 Hz$ . The frequency of the whistle heard by a passenger in the second train before the train meets, is (the speed of sound in air is $340 m / s$ )

1

600 Hz

2

585 Hz

3

645 Hz

4

666 Hz

Explanation:

D Given that, $f = 600 Hz, v_{o} = 15 m / s$
$v_{s} = 20 m / s, v = 340 m / s$
We know that,
Apparent frequency
$f^{'} = f (\frac{v + v_{o}}{v - v_{s}})$
$f^{'} = 600 \times [\frac{340 + 15}{340 - 20}]$
$f^{'} = \frac{600 \times 355}{320} = \frac{21300}{32} = 665.625 ◻ 666 Hz$

BITSAT-2019

WAVES

172970 A police car moving at $22 {ms}^{- 1}$ chases a motor cyclist. The police man sounds horn at $176 Hz$ , while both of them move towards a stationary siren of frequency $165 Hz$ . If the number of beats heard by the motor cyclist per second is zero, then the speed of motorcycle is (Speed of sound in air $= 330 {ms}^{- 1}$ )

1

33 {ms}^{- 1}

2

22 {ms}^{- 1}

3

44 {ms}^{- 1}

4

11 {ms}^{- 1}

Explanation:

B Given, speed of a police car $(v_{s}) = 22 m / s$ , Frequency of the sound horn $(f_{horn}) = 176 Hz$
Frequency of siren $(f_{siren}) = 165 Hz, v = 330 m / s$

We know that,
$f^{'} = f_{horn} (\frac{v - v_{m}}{v - v_{s}})$
$f^{''} = f_{siren} (\frac{v + v_{m}}{v})$
According to question -
Motorcyclist does not observe any beats and this is possible only when the difference in the frequencies heard by the motorcyclist is zero.
$Thus, f^{'} - f^{''} = 0$
$f_{horn} (\frac{v - v_{m}}{v - v_{s}}) = f_{siren} (\frac{v + v_{m}}{v})$
$176 (\frac{v - v_{m}}{330 - 22}) = 165 (\frac{v + v_{m}}{330})$
$\frac{v - v_{m}}{v + v_{m}} = \frac{165}{176} \times \frac{308}{330} = \frac{7}{8}$
$8 v - 8 v_{m} = 7 v + 7 v_{m}$
$15 v_{m} = v$
$v_{m} = \frac{v}{15} = \frac{330}{15} = 22 m / s$

AP EAMCET (23.04.2019) Shift-I

WAVES

172971 A source producing sound of frequency $720 Hz$ is falling freely from the top of a tower of height $20 m$ . The frequency of sound heard by an observer on the top of the tower when the source just reaches the ground is (Acceleration due to gravity, $g = 10 {ms}^{- 2}$ and speed of sound in air $= 340 {m s}^{- 1}$ )

1

660 Hz

2

680 Hz

3

740 Hz

4

760 Hz

Explanation:

B Given,
$f = 720 Hz, v = 340 {ms}^{- 1}$
$h = 20 m, g = 10 m / s^{2}$
We know that the third equation of motion is
$v^{2} = u^{2} + 2 g h$
Here, final velocity of source
$v_{s}^{2} = u_{s}^{2} + 2 gh (u_{s} = 0)$
$v_{s} = \sqrt{2 gh}$
$v_{s} = \sqrt{2 \times 10 \times 20} = 20 m / s$
$f^{'} = f (\frac{v}{v + v_{s}})$
$f^{'} = 720 (\frac{340}{340 + 20}) = 680 Hz$

AP EAMCET (22.04.2019) Shift-II

WAVES

172972 A train approaching a railway crossing at a speed of $120 km / h$ sounds a whistle of frequency $576 Hz$ , when it is $288 m$ away from the crossing. The frequency heard by the observer standing on the road perpendicular to the track from the crossing at a distance of $384 m$ is (Speed of sound in air $= 340 {ms}^{- 1}$ )

1

632 Hz

2

612 Hz

3

512 Hz

4

472 Hz

Explanation:

B Given,
Speed of source $(v_{s}) = 120 km / h = \frac{100}{3} {ms}^{- 1}$
Original frequency of the source $(f_{o}) = 576 Hz$
Speed of sound in air $= 340 m / s$

From diagram -
$\cos θ = \frac{288}{AB} = \frac{288}{AB}$
$AC = \sqrt{(288)^{2} + (384)^{2}} = 480$
$∴ \cos θ = \frac{288}{480} = \frac{3}{5}$
$Therefore, v_{s} \cos θ = \frac{100}{3} \times \frac{3}{5} = 20 m / s$
$The frequency heard by the observer is given by$
$f = f 。 (\frac{v}{v - v_{s} \cos θ})$
$f = 576 (\frac{340}{340 - 20})$
$f = 576 \times \frac{340}{320}$
$f = 612 Hz$

AP EAMCET (21.04.2019) Shift-II

WAVES

172969 Two trains are moving towards each other with speeds of $20 m / s$ and $15 m / s$ relative to the ground. The first train sounds a whistle of frequency $600 Hz$ . The frequency of the whistle heard by a passenger in the second train before the train meets, is (the speed of sound in air is $340 m / s$ )

1

600 Hz

2

585 Hz

3

645 Hz

4

666 Hz

Explanation:

D Given that, $f = 600 Hz, v_{o} = 15 m / s$
$v_{s} = 20 m / s, v = 340 m / s$
We know that,
Apparent frequency
$f^{'} = f (\frac{v + v_{o}}{v - v_{s}})$
$f^{'} = 600 \times [\frac{340 + 15}{340 - 20}]$
$f^{'} = \frac{600 \times 355}{320} = \frac{21300}{32} = 665.625 ◻ 666 Hz$

BITSAT-2019

WAVES

172970 A police car moving at $22 {ms}^{- 1}$ chases a motor cyclist. The police man sounds horn at $176 Hz$ , while both of them move towards a stationary siren of frequency $165 Hz$ . If the number of beats heard by the motor cyclist per second is zero, then the speed of motorcycle is (Speed of sound in air $= 330 {ms}^{- 1}$ )

1

33 {ms}^{- 1}

2

22 {ms}^{- 1}

3

44 {ms}^{- 1}

4

11 {ms}^{- 1}

Explanation:

B Given, speed of a police car $(v_{s}) = 22 m / s$ , Frequency of the sound horn $(f_{horn}) = 176 Hz$
Frequency of siren $(f_{siren}) = 165 Hz, v = 330 m / s$

We know that,
$f^{'} = f_{horn} (\frac{v - v_{m}}{v - v_{s}})$
$f^{''} = f_{siren} (\frac{v + v_{m}}{v})$
According to question -
Motorcyclist does not observe any beats and this is possible only when the difference in the frequencies heard by the motorcyclist is zero.
$Thus, f^{'} - f^{''} = 0$
$f_{horn} (\frac{v - v_{m}}{v - v_{s}}) = f_{siren} (\frac{v + v_{m}}{v})$
$176 (\frac{v - v_{m}}{330 - 22}) = 165 (\frac{v + v_{m}}{330})$
$\frac{v - v_{m}}{v + v_{m}} = \frac{165}{176} \times \frac{308}{330} = \frac{7}{8}$
$8 v - 8 v_{m} = 7 v + 7 v_{m}$
$15 v_{m} = v$
$v_{m} = \frac{v}{15} = \frac{330}{15} = 22 m / s$

AP EAMCET (23.04.2019) Shift-I

WAVES

172971 A source producing sound of frequency $720 Hz$ is falling freely from the top of a tower of height $20 m$ . The frequency of sound heard by an observer on the top of the tower when the source just reaches the ground is (Acceleration due to gravity, $g = 10 {ms}^{- 2}$ and speed of sound in air $= 340 {m s}^{- 1}$ )

1

660 Hz

2

680 Hz

3

740 Hz

4

760 Hz

Explanation:

B Given,
$f = 720 Hz, v = 340 {ms}^{- 1}$
$h = 20 m, g = 10 m / s^{2}$
We know that the third equation of motion is
$v^{2} = u^{2} + 2 g h$
Here, final velocity of source
$v_{s}^{2} = u_{s}^{2} + 2 gh (u_{s} = 0)$
$v_{s} = \sqrt{2 gh}$
$v_{s} = \sqrt{2 \times 10 \times 20} = 20 m / s$
$f^{'} = f (\frac{v}{v + v_{s}})$
$f^{'} = 720 (\frac{340}{340 + 20}) = 680 Hz$

AP EAMCET (22.04.2019) Shift-II

WAVES

172972 A train approaching a railway crossing at a speed of $120 km / h$ sounds a whistle of frequency $576 Hz$ , when it is $288 m$ away from the crossing. The frequency heard by the observer standing on the road perpendicular to the track from the crossing at a distance of $384 m$ is (Speed of sound in air $= 340 {ms}^{- 1}$ )

1

632 Hz

2

612 Hz

3

512 Hz

4

472 Hz

Explanation:

B Given,
Speed of source $(v_{s}) = 120 km / h = \frac{100}{3} {ms}^{- 1}$
Original frequency of the source $(f_{o}) = 576 Hz$
Speed of sound in air $= 340 m / s$

From diagram -
$\cos θ = \frac{288}{AB} = \frac{288}{AB}$
$AC = \sqrt{(288)^{2} + (384)^{2}} = 480$
$∴ \cos θ = \frac{288}{480} = \frac{3}{5}$
$Therefore, v_{s} \cos θ = \frac{100}{3} \times \frac{3}{5} = 20 m / s$
$The frequency heard by the observer is given by$
$f = f 。 (\frac{v}{v - v_{s} \cos θ})$
$f = 576 (\frac{340}{340 - 20})$
$f = 576 \times \frac{340}{320}$
$f = 612 Hz$

AP EAMCET (21.04.2019) Shift-II

WAVES

172969 Two trains are moving towards each other with speeds of $20 m / s$ and $15 m / s$ relative to the ground. The first train sounds a whistle of frequency $600 Hz$ . The frequency of the whistle heard by a passenger in the second train before the train meets, is (the speed of sound in air is $340 m / s$ )

1

600 Hz

2

585 Hz

3

645 Hz

4

666 Hz

Explanation:

D Given that, $f = 600 Hz, v_{o} = 15 m / s$
$v_{s} = 20 m / s, v = 340 m / s$
We know that,
Apparent frequency
$f^{'} = f (\frac{v + v_{o}}{v - v_{s}})$
$f^{'} = 600 \times [\frac{340 + 15}{340 - 20}]$
$f^{'} = \frac{600 \times 355}{320} = \frac{21300}{32} = 665.625 ◻ 666 Hz$

BITSAT-2019

WAVES

172970 A police car moving at $22 {ms}^{- 1}$ chases a motor cyclist. The police man sounds horn at $176 Hz$ , while both of them move towards a stationary siren of frequency $165 Hz$ . If the number of beats heard by the motor cyclist per second is zero, then the speed of motorcycle is (Speed of sound in air $= 330 {ms}^{- 1}$ )

1

33 {ms}^{- 1}

2

22 {ms}^{- 1}

3

44 {ms}^{- 1}

4

11 {ms}^{- 1}

Explanation:

B Given, speed of a police car $(v_{s}) = 22 m / s$ , Frequency of the sound horn $(f_{horn}) = 176 Hz$
Frequency of siren $(f_{siren}) = 165 Hz, v = 330 m / s$

We know that,
$f^{'} = f_{horn} (\frac{v - v_{m}}{v - v_{s}})$
$f^{''} = f_{siren} (\frac{v + v_{m}}{v})$
According to question -
Motorcyclist does not observe any beats and this is possible only when the difference in the frequencies heard by the motorcyclist is zero.
$Thus, f^{'} - f^{''} = 0$
$f_{horn} (\frac{v - v_{m}}{v - v_{s}}) = f_{siren} (\frac{v + v_{m}}{v})$
$176 (\frac{v - v_{m}}{330 - 22}) = 165 (\frac{v + v_{m}}{330})$
$\frac{v - v_{m}}{v + v_{m}} = \frac{165}{176} \times \frac{308}{330} = \frac{7}{8}$
$8 v - 8 v_{m} = 7 v + 7 v_{m}$
$15 v_{m} = v$
$v_{m} = \frac{v}{15} = \frac{330}{15} = 22 m / s$

AP EAMCET (23.04.2019) Shift-I

WAVES

172971 A source producing sound of frequency $720 Hz$ is falling freely from the top of a tower of height $20 m$ . The frequency of sound heard by an observer on the top of the tower when the source just reaches the ground is (Acceleration due to gravity, $g = 10 {ms}^{- 2}$ and speed of sound in air $= 340 {m s}^{- 1}$ )

1

660 Hz

2

680 Hz

3

740 Hz

4

760 Hz

Explanation:

B Given,
$f = 720 Hz, v = 340 {ms}^{- 1}$
$h = 20 m, g = 10 m / s^{2}$
We know that the third equation of motion is
$v^{2} = u^{2} + 2 g h$
Here, final velocity of source
$v_{s}^{2} = u_{s}^{2} + 2 gh (u_{s} = 0)$
$v_{s} = \sqrt{2 gh}$
$v_{s} = \sqrt{2 \times 10 \times 20} = 20 m / s$
$f^{'} = f (\frac{v}{v + v_{s}})$
$f^{'} = 720 (\frac{340}{340 + 20}) = 680 Hz$

AP EAMCET (22.04.2019) Shift-II

WAVES

172972 A train approaching a railway crossing at a speed of $120 km / h$ sounds a whistle of frequency $576 Hz$ , when it is $288 m$ away from the crossing. The frequency heard by the observer standing on the road perpendicular to the track from the crossing at a distance of $384 m$ is (Speed of sound in air $= 340 {ms}^{- 1}$ )

1

632 Hz

2

612 Hz

3

512 Hz

4

472 Hz

Explanation:

B Given,
Speed of source $(v_{s}) = 120 km / h = \frac{100}{3} {ms}^{- 1}$
Original frequency of the source $(f_{o}) = 576 Hz$
Speed of sound in air $= 340 m / s$

From diagram -
$\cos θ = \frac{288}{AB} = \frac{288}{AB}$
$AC = \sqrt{(288)^{2} + (384)^{2}} = 480$
$∴ \cos θ = \frac{288}{480} = \frac{3}{5}$
$Therefore, v_{s} \cos θ = \frac{100}{3} \times \frac{3}{5} = 20 m / s$
$The frequency heard by the observer is given by$
$f = f 。 (\frac{v}{v - v_{s} \cos θ})$
$f = 576 (\frac{340}{340 - 20})$
$f = 576 \times \frac{340}{320}$
$f = 612 Hz$

AP EAMCET (21.04.2019) Shift-II

WAVES

172969 Two trains are moving towards each other with speeds of $20 m / s$ and $15 m / s$ relative to the ground. The first train sounds a whistle of frequency $600 Hz$ . The frequency of the whistle heard by a passenger in the second train before the train meets, is (the speed of sound in air is $340 m / s$ )

1

600 Hz

2

585 Hz

3

645 Hz

4

666 Hz

Explanation:

D Given that, $f = 600 Hz, v_{o} = 15 m / s$
$v_{s} = 20 m / s, v = 340 m / s$
We know that,
Apparent frequency
$f^{'} = f (\frac{v + v_{o}}{v - v_{s}})$
$f^{'} = 600 \times [\frac{340 + 15}{340 - 20}]$
$f^{'} = \frac{600 \times 355}{320} = \frac{21300}{32} = 665.625 ◻ 666 Hz$

BITSAT-2019

WAVES

172970 A police car moving at $22 {ms}^{- 1}$ chases a motor cyclist. The police man sounds horn at $176 Hz$ , while both of them move towards a stationary siren of frequency $165 Hz$ . If the number of beats heard by the motor cyclist per second is zero, then the speed of motorcycle is (Speed of sound in air $= 330 {ms}^{- 1}$ )

1

33 {ms}^{- 1}

2

22 {ms}^{- 1}

3

44 {ms}^{- 1}

4

11 {ms}^{- 1}

Explanation:

B Given, speed of a police car $(v_{s}) = 22 m / s$ , Frequency of the sound horn $(f_{horn}) = 176 Hz$
Frequency of siren $(f_{siren}) = 165 Hz, v = 330 m / s$

We know that,
$f^{'} = f_{horn} (\frac{v - v_{m}}{v - v_{s}})$
$f^{''} = f_{siren} (\frac{v + v_{m}}{v})$
According to question -
Motorcyclist does not observe any beats and this is possible only when the difference in the frequencies heard by the motorcyclist is zero.
$Thus, f^{'} - f^{''} = 0$
$f_{horn} (\frac{v - v_{m}}{v - v_{s}}) = f_{siren} (\frac{v + v_{m}}{v})$
$176 (\frac{v - v_{m}}{330 - 22}) = 165 (\frac{v + v_{m}}{330})$
$\frac{v - v_{m}}{v + v_{m}} = \frac{165}{176} \times \frac{308}{330} = \frac{7}{8}$
$8 v - 8 v_{m} = 7 v + 7 v_{m}$
$15 v_{m} = v$
$v_{m} = \frac{v}{15} = \frac{330}{15} = 22 m / s$

AP EAMCET (23.04.2019) Shift-I

WAVES

172971 A source producing sound of frequency $720 Hz$ is falling freely from the top of a tower of height $20 m$ . The frequency of sound heard by an observer on the top of the tower when the source just reaches the ground is (Acceleration due to gravity, $g = 10 {ms}^{- 2}$ and speed of sound in air $= 340 {m s}^{- 1}$ )

1

660 Hz

2

680 Hz

3

740 Hz

4

760 Hz

Explanation:

B Given,
$f = 720 Hz, v = 340 {ms}^{- 1}$
$h = 20 m, g = 10 m / s^{2}$
We know that the third equation of motion is
$v^{2} = u^{2} + 2 g h$
Here, final velocity of source
$v_{s}^{2} = u_{s}^{2} + 2 gh (u_{s} = 0)$
$v_{s} = \sqrt{2 gh}$
$v_{s} = \sqrt{2 \times 10 \times 20} = 20 m / s$
$f^{'} = f (\frac{v}{v + v_{s}})$
$f^{'} = 720 (\frac{340}{340 + 20}) = 680 Hz$

AP EAMCET (22.04.2019) Shift-II

WAVES

172972 A train approaching a railway crossing at a speed of $120 km / h$ sounds a whistle of frequency $576 Hz$ , when it is $288 m$ away from the crossing. The frequency heard by the observer standing on the road perpendicular to the track from the crossing at a distance of $384 m$ is (Speed of sound in air $= 340 {ms}^{- 1}$ )

1

632 Hz

2

612 Hz

3

512 Hz

4

472 Hz

Explanation:

B Given,
Speed of source $(v_{s}) = 120 km / h = \frac{100}{3} {ms}^{- 1}$
Original frequency of the source $(f_{o}) = 576 Hz$
Speed of sound in air $= 340 m / s$

From diagram -
$\cos θ = \frac{288}{AB} = \frac{288}{AB}$
$AC = \sqrt{(288)^{2} + (384)^{2}} = 480$
$∴ \cos θ = \frac{288}{480} = \frac{3}{5}$
$Therefore, v_{s} \cos θ = \frac{100}{3} \times \frac{3}{5} = 20 m / s$
$The frequency heard by the observer is given by$
$f = f 。 (\frac{v}{v - v_{s} \cos θ})$
$f = 576 (\frac{340}{340 - 20})$
$f = 576 \times \frac{340}{320}$
$f = 612 Hz$

AP EAMCET (21.04.2019) Shift-II