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PHXI15:WAVES

354749 Consider two sound sources $S_{1}$ and $S_{2}$ having same frequency $100 H z$ and the observer $O$ located between them. All the three are moving with same velocity in same direction. The beat frequency as heard by the observer is
supporting img

1 Zero

2 2

3 3

4 4

PHXI15:WAVES

354750 The observer is moving with velocity ' $v_{0}$ ' towards the stationary source of sound and then after crossing moves away from the source with velocity $v_{0}$ . Assume that the medium through which the sound wave travel is at rest. If $v$ is velocity of sound and $n$ is the frequency emitted by the source, then the difference between apparent frequencies heard by the observer is

1

\frac{2 n v_{0}}{v}

2

\frac{n v_{0}}{v}

3

\frac{v}{2 n v_{0}}

4

\frac{v}{n v_{0}}

Explanation:

When the observer is moving towards the source and source is stationary, then the apparent frequency is given as,
$n^{'} = n (\frac{v + v_{0}}{v}) (1)$
When the observer is moving away from the source and the source is stationary, then the apparent frequency is given as
$n^{,,} = n (\frac{v - v_{0}}{v}) (2)$
From eq.(1) and eq.(2), we get
$\begin{matrix} n^{'} - n^{''} = \frac{n}{v} (v + v_{0} - v + v_{0}) \\ = \frac{2 n v_{0}}{v} \end{matrix}$

MHTCET - 2017

PHXI15:WAVES

354751 An observer is moving away from a source at rest. The pitch of the note heard by the observer is less because

1 The pitch of the source decreases

2 The velocity of sound in air increases

3 Wavelength of the wave becomes small

4 Wavelength of the wave remains unchanged but observer receives less number of waves

PHXI15:WAVES

354752 With what velocity should an observer approach a stationary sound source so that the apparent frequency of sound should appear double the actual frequency? ( $v$ is velocity of sound).

1

2 v

2

v

3

v / 2

4

3 v

Explanation:

When the actual frequency is $f$ , the apparent frequency of the sound to the observer is
$f^{'} = f \times \frac{(v + v_{0})}{v}$
Here, $v =$ speed of sound and $v_{0} =$ speed of observer.
When $f^{'} = 2 f, \frac{(v + v_{0})}{v} = 2 \Rightarrow v_{0} = v$
$∴$ When the observer approaches the stationary source at the speed of sound, the apparent frequency is double the actual frequency.

KCET - 2008

PHXI15:WAVES

354749 Consider two sound sources $S_{1}$ and $S_{2}$ having same frequency $100 H z$ and the observer $O$ located between them. All the three are moving with same velocity in same direction. The beat frequency as heard by the observer is
supporting img

1 Zero

2 2

3 3

4 4

PHXI15:WAVES

354750 The observer is moving with velocity ' $v_{0}$ ' towards the stationary source of sound and then after crossing moves away from the source with velocity $v_{0}$ . Assume that the medium through which the sound wave travel is at rest. If $v$ is velocity of sound and $n$ is the frequency emitted by the source, then the difference between apparent frequencies heard by the observer is

1

\frac{2 n v_{0}}{v}

2

\frac{n v_{0}}{v}

3

\frac{v}{2 n v_{0}}

4

\frac{v}{n v_{0}}

Explanation:

When the observer is moving towards the source and source is stationary, then the apparent frequency is given as,
$n^{'} = n (\frac{v + v_{0}}{v}) (1)$
When the observer is moving away from the source and the source is stationary, then the apparent frequency is given as
$n^{,,} = n (\frac{v - v_{0}}{v}) (2)$
From eq.(1) and eq.(2), we get
$\begin{matrix} n^{'} - n^{''} = \frac{n}{v} (v + v_{0} - v + v_{0}) \\ = \frac{2 n v_{0}}{v} \end{matrix}$

MHTCET - 2017

PHXI15:WAVES

354751 An observer is moving away from a source at rest. The pitch of the note heard by the observer is less because

1 The pitch of the source decreases

2 The velocity of sound in air increases

3 Wavelength of the wave becomes small

4 Wavelength of the wave remains unchanged but observer receives less number of waves

PHXI15:WAVES

354752 With what velocity should an observer approach a stationary sound source so that the apparent frequency of sound should appear double the actual frequency? ( $v$ is velocity of sound).

1

2 v

2

v

3

v / 2

4

3 v

Explanation:

When the actual frequency is $f$ , the apparent frequency of the sound to the observer is
$f^{'} = f \times \frac{(v + v_{0})}{v}$
Here, $v =$ speed of sound and $v_{0} =$ speed of observer.
When $f^{'} = 2 f, \frac{(v + v_{0})}{v} = 2 \Rightarrow v_{0} = v$
$∴$ When the observer approaches the stationary source at the speed of sound, the apparent frequency is double the actual frequency.

KCET - 2008

PHXI15:WAVES

354749 Consider two sound sources $S_{1}$ and $S_{2}$ having same frequency $100 H z$ and the observer $O$ located between them. All the three are moving with same velocity in same direction. The beat frequency as heard by the observer is
supporting img

1 Zero

2 2

3 3

4 4

PHXI15:WAVES

354750 The observer is moving with velocity ' $v_{0}$ ' towards the stationary source of sound and then after crossing moves away from the source with velocity $v_{0}$ . Assume that the medium through which the sound wave travel is at rest. If $v$ is velocity of sound and $n$ is the frequency emitted by the source, then the difference between apparent frequencies heard by the observer is

1

\frac{2 n v_{0}}{v}

2

\frac{n v_{0}}{v}

3

\frac{v}{2 n v_{0}}

4

\frac{v}{n v_{0}}

Explanation:

When the observer is moving towards the source and source is stationary, then the apparent frequency is given as,
$n^{'} = n (\frac{v + v_{0}}{v}) (1)$
When the observer is moving away from the source and the source is stationary, then the apparent frequency is given as
$n^{,,} = n (\frac{v - v_{0}}{v}) (2)$
From eq.(1) and eq.(2), we get
$\begin{matrix} n^{'} - n^{''} = \frac{n}{v} (v + v_{0} - v + v_{0}) \\ = \frac{2 n v_{0}}{v} \end{matrix}$

MHTCET - 2017

PHXI15:WAVES

354751 An observer is moving away from a source at rest. The pitch of the note heard by the observer is less because

1 The pitch of the source decreases

2 The velocity of sound in air increases

3 Wavelength of the wave becomes small

4 Wavelength of the wave remains unchanged but observer receives less number of waves

PHXI15:WAVES

354752 With what velocity should an observer approach a stationary sound source so that the apparent frequency of sound should appear double the actual frequency? ( $v$ is velocity of sound).

1

2 v

2

v

3

v / 2

4

3 v

Explanation:

When the actual frequency is $f$ , the apparent frequency of the sound to the observer is
$f^{'} = f \times \frac{(v + v_{0})}{v}$
Here, $v =$ speed of sound and $v_{0} =$ speed of observer.
When $f^{'} = 2 f, \frac{(v + v_{0})}{v} = 2 \Rightarrow v_{0} = v$
$∴$ When the observer approaches the stationary source at the speed of sound, the apparent frequency is double the actual frequency.

KCET - 2008

PHXI15:WAVES

354749 Consider two sound sources $S_{1}$ and $S_{2}$ having same frequency $100 H z$ and the observer $O$ located between them. All the three are moving with same velocity in same direction. The beat frequency as heard by the observer is
supporting img

1 Zero

2 2

3 3

4 4

PHXI15:WAVES

354750 The observer is moving with velocity ' $v_{0}$ ' towards the stationary source of sound and then after crossing moves away from the source with velocity $v_{0}$ . Assume that the medium through which the sound wave travel is at rest. If $v$ is velocity of sound and $n$ is the frequency emitted by the source, then the difference between apparent frequencies heard by the observer is

1

\frac{2 n v_{0}}{v}

2

\frac{n v_{0}}{v}

3

\frac{v}{2 n v_{0}}

4

\frac{v}{n v_{0}}

Explanation:

When the observer is moving towards the source and source is stationary, then the apparent frequency is given as,
$n^{'} = n (\frac{v + v_{0}}{v}) (1)$
When the observer is moving away from the source and the source is stationary, then the apparent frequency is given as
$n^{,,} = n (\frac{v - v_{0}}{v}) (2)$
From eq.(1) and eq.(2), we get
$\begin{matrix} n^{'} - n^{''} = \frac{n}{v} (v + v_{0} - v + v_{0}) \\ = \frac{2 n v_{0}}{v} \end{matrix}$

MHTCET - 2017

PHXI15:WAVES

354751 An observer is moving away from a source at rest. The pitch of the note heard by the observer is less because

1 The pitch of the source decreases

2 The velocity of sound in air increases

3 Wavelength of the wave becomes small

4 Wavelength of the wave remains unchanged but observer receives less number of waves

PHXI15:WAVES

354752 With what velocity should an observer approach a stationary sound source so that the apparent frequency of sound should appear double the actual frequency? ( $v$ is velocity of sound).

1

2 v

2

v

3

v / 2

4

3 v

Explanation:

When the actual frequency is $f$ , the apparent frequency of the sound to the observer is
$f^{'} = f \times \frac{(v + v_{0})}{v}$
Here, $v =$ speed of sound and $v_{0} =$ speed of observer.
When $f^{'} = 2 f, \frac{(v + v_{0})}{v} = 2 \Rightarrow v_{0} = v$
$∴$ When the observer approaches the stationary source at the speed of sound, the apparent frequency is double the actual frequency.

KCET - 2008

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