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PHXI15:WAVES

355050 $S_{1}$ and $S_{2}$ are two coherent sources of sound having no initial phase difference. The velocity of sound is $330 m / s$ . No minima will be formed on the line passing through $S_{2}$ and perpendicular to the line joining $S_{1}$ and $S_{2}$ , if the frequency of both the sources may be :
supporting img

1

50 H z

2

60 H z

3

70 H z

4

80 H z

Explanation:

For minimum,
$Δ x = (2 n - 1) \frac{λ}{2}$
The maximum possible path difference $=$ distance between the sources $= 3 m$ .
For no minimum
$\begin{aligned} \frac{λ}{2} > 3 \Rightarrow λ > 6 \\ ∴ f = \frac{v}{λ} < \frac{330}{6} = 55 . \end{aligned}$
If $∴ f < 55 H z$ . No minimum will occur.

PHXI15:WAVES

355051 Two sound waves, each of amplitude $A$ and frequency $ω$ superpose at a point with phase difference of $\frac{π}{2}$ . The amplitude and frequency of the resultant wave are respectively

1

\frac{A}{\sqrt{2}}, \frac{ω}{2}

2

\frac{A}{\sqrt{2}}, ω

3

\sqrt{2} A, \frac{ω}{2}

4

\sqrt{2} A, ω

Explanation:

Let $y_{1} = A \sin (ω t)$ and
$y_{2} = A \sin (ω t + \frac{π}{2}) (1)$
Resultant amplitude
$R = A^{2} + A^{2} + 2 A^{2} \cos (\frac{π}{2})$
$\Rightarrow R = \sqrt{2 A^{2} + 2 A^{2} \times 0} \Rightarrow R = \sqrt{2 A^{2}}$
$\Rightarrow R = \sqrt{2} A$
However, both will have the same frequency on superimposing.

PHXI15:WAVES

355052 The phenomena arising due to the superposition of waves is/are

1 beats

2 stationary waves

3 Lissajous figure

4 All of these

PHXI15:WAVES

355053 Sound signal is sent through a composite tube as shown in Fig. The radius of the semicircle is $r$ . Speed of sound in air is $v$ . The source of sound is capable to generate frequencies in the range $f_{1}$ to $f_{2} (f_{2} > f_{1})$ . If $n$ is an integer then frequency for maximum intensity at ' $p$ ' is given by
supporting img

1

\frac{n v}{r (π - 2)}

2

\frac{n v}{r}

3

\frac{n v}{(r - 2) π}

4

\frac{n v}{π r}

Explanation:

Path difference $π r - 2 r = n λ$
$r (π - 2) = \frac{n v}{f} thus f = \frac{n v}{r (π - 2)}$

PHXI15:WAVES

355054 The principle of superposition of waves can not be used to explain wave phenomena like

1 Interference

2 Diffraction

3 Stationary Waves and beats

4 Dispersion

PHXI15:WAVES

355050 $S_{1}$ and $S_{2}$ are two coherent sources of sound having no initial phase difference. The velocity of sound is $330 m / s$ . No minima will be formed on the line passing through $S_{2}$ and perpendicular to the line joining $S_{1}$ and $S_{2}$ , if the frequency of both the sources may be :
supporting img

1

50 H z

2

60 H z

3

70 H z

4

80 H z

Explanation:

For minimum,
$Δ x = (2 n - 1) \frac{λ}{2}$
The maximum possible path difference $=$ distance between the sources $= 3 m$ .
For no minimum
$\begin{aligned} \frac{λ}{2} > 3 \Rightarrow λ > 6 \\ ∴ f = \frac{v}{λ} < \frac{330}{6} = 55 . \end{aligned}$
If $∴ f < 55 H z$ . No minimum will occur.

PHXI15:WAVES

355051 Two sound waves, each of amplitude $A$ and frequency $ω$ superpose at a point with phase difference of $\frac{π}{2}$ . The amplitude and frequency of the resultant wave are respectively

1

\frac{A}{\sqrt{2}}, \frac{ω}{2}

2

\frac{A}{\sqrt{2}}, ω

3

\sqrt{2} A, \frac{ω}{2}

4

\sqrt{2} A, ω

Explanation:

Let $y_{1} = A \sin (ω t)$ and
$y_{2} = A \sin (ω t + \frac{π}{2}) (1)$
Resultant amplitude
$R = A^{2} + A^{2} + 2 A^{2} \cos (\frac{π}{2})$
$\Rightarrow R = \sqrt{2 A^{2} + 2 A^{2} \times 0} \Rightarrow R = \sqrt{2 A^{2}}$
$\Rightarrow R = \sqrt{2} A$
However, both will have the same frequency on superimposing.

PHXI15:WAVES

355052 The phenomena arising due to the superposition of waves is/are

1 beats

2 stationary waves

3 Lissajous figure

4 All of these

PHXI15:WAVES

355053 Sound signal is sent through a composite tube as shown in Fig. The radius of the semicircle is $r$ . Speed of sound in air is $v$ . The source of sound is capable to generate frequencies in the range $f_{1}$ to $f_{2} (f_{2} > f_{1})$ . If $n$ is an integer then frequency for maximum intensity at ' $p$ ' is given by
supporting img

1

\frac{n v}{r (π - 2)}

2

\frac{n v}{r}

3

\frac{n v}{(r - 2) π}

4

\frac{n v}{π r}

Explanation:

Path difference $π r - 2 r = n λ$
$r (π - 2) = \frac{n v}{f} thus f = \frac{n v}{r (π - 2)}$

PHXI15:WAVES

355054 The principle of superposition of waves can not be used to explain wave phenomena like

1 Interference

2 Diffraction

3 Stationary Waves and beats

4 Dispersion

PHXI15:WAVES

355050 $S_{1}$ and $S_{2}$ are two coherent sources of sound having no initial phase difference. The velocity of sound is $330 m / s$ . No minima will be formed on the line passing through $S_{2}$ and perpendicular to the line joining $S_{1}$ and $S_{2}$ , if the frequency of both the sources may be :
supporting img

1

50 H z

2

60 H z

3

70 H z

4

80 H z

Explanation:

For minimum,
$Δ x = (2 n - 1) \frac{λ}{2}$
The maximum possible path difference $=$ distance between the sources $= 3 m$ .
For no minimum
$\begin{aligned} \frac{λ}{2} > 3 \Rightarrow λ > 6 \\ ∴ f = \frac{v}{λ} < \frac{330}{6} = 55 . \end{aligned}$
If $∴ f < 55 H z$ . No minimum will occur.

PHXI15:WAVES

355051 Two sound waves, each of amplitude $A$ and frequency $ω$ superpose at a point with phase difference of $\frac{π}{2}$ . The amplitude and frequency of the resultant wave are respectively

1

\frac{A}{\sqrt{2}}, \frac{ω}{2}

2

\frac{A}{\sqrt{2}}, ω

3

\sqrt{2} A, \frac{ω}{2}

4

\sqrt{2} A, ω

Explanation:

Let $y_{1} = A \sin (ω t)$ and
$y_{2} = A \sin (ω t + \frac{π}{2}) (1)$
Resultant amplitude
$R = A^{2} + A^{2} + 2 A^{2} \cos (\frac{π}{2})$
$\Rightarrow R = \sqrt{2 A^{2} + 2 A^{2} \times 0} \Rightarrow R = \sqrt{2 A^{2}}$
$\Rightarrow R = \sqrt{2} A$
However, both will have the same frequency on superimposing.

PHXI15:WAVES

355052 The phenomena arising due to the superposition of waves is/are

1 beats

2 stationary waves

3 Lissajous figure

4 All of these

PHXI15:WAVES

355053 Sound signal is sent through a composite tube as shown in Fig. The radius of the semicircle is $r$ . Speed of sound in air is $v$ . The source of sound is capable to generate frequencies in the range $f_{1}$ to $f_{2} (f_{2} > f_{1})$ . If $n$ is an integer then frequency for maximum intensity at ' $p$ ' is given by
supporting img

1

\frac{n v}{r (π - 2)}

2

\frac{n v}{r}

3

\frac{n v}{(r - 2) π}

4

\frac{n v}{π r}

Explanation:

Path difference $π r - 2 r = n λ$
$r (π - 2) = \frac{n v}{f} thus f = \frac{n v}{r (π - 2)}$

PHXI15:WAVES

355054 The principle of superposition of waves can not be used to explain wave phenomena like

1 Interference

2 Diffraction

3 Stationary Waves and beats

4 Dispersion

PHXI15:WAVES

355050 $S_{1}$ and $S_{2}$ are two coherent sources of sound having no initial phase difference. The velocity of sound is $330 m / s$ . No minima will be formed on the line passing through $S_{2}$ and perpendicular to the line joining $S_{1}$ and $S_{2}$ , if the frequency of both the sources may be :
supporting img

1

50 H z

2

60 H z

3

70 H z

4

80 H z

Explanation:

For minimum,
$Δ x = (2 n - 1) \frac{λ}{2}$
The maximum possible path difference $=$ distance between the sources $= 3 m$ .
For no minimum
$\begin{aligned} \frac{λ}{2} > 3 \Rightarrow λ > 6 \\ ∴ f = \frac{v}{λ} < \frac{330}{6} = 55 . \end{aligned}$
If $∴ f < 55 H z$ . No minimum will occur.

PHXI15:WAVES

355051 Two sound waves, each of amplitude $A$ and frequency $ω$ superpose at a point with phase difference of $\frac{π}{2}$ . The amplitude and frequency of the resultant wave are respectively

1

\frac{A}{\sqrt{2}}, \frac{ω}{2}

2

\frac{A}{\sqrt{2}}, ω

3

\sqrt{2} A, \frac{ω}{2}

4

\sqrt{2} A, ω

Explanation:

Let $y_{1} = A \sin (ω t)$ and
$y_{2} = A \sin (ω t + \frac{π}{2}) (1)$
Resultant amplitude
$R = A^{2} + A^{2} + 2 A^{2} \cos (\frac{π}{2})$
$\Rightarrow R = \sqrt{2 A^{2} + 2 A^{2} \times 0} \Rightarrow R = \sqrt{2 A^{2}}$
$\Rightarrow R = \sqrt{2} A$
However, both will have the same frequency on superimposing.

PHXI15:WAVES

355052 The phenomena arising due to the superposition of waves is/are

1 beats

2 stationary waves

3 Lissajous figure

4 All of these

PHXI15:WAVES

355053 Sound signal is sent through a composite tube as shown in Fig. The radius of the semicircle is $r$ . Speed of sound in air is $v$ . The source of sound is capable to generate frequencies in the range $f_{1}$ to $f_{2} (f_{2} > f_{1})$ . If $n$ is an integer then frequency for maximum intensity at ' $p$ ' is given by
supporting img

1

\frac{n v}{r (π - 2)}

2

\frac{n v}{r}

3

\frac{n v}{(r - 2) π}

4

\frac{n v}{π r}

Explanation:

Path difference $π r - 2 r = n λ$
$r (π - 2) = \frac{n v}{f} thus f = \frac{n v}{r (π - 2)}$

PHXI15:WAVES

355054 The principle of superposition of waves can not be used to explain wave phenomena like

1 Interference

2 Diffraction

3 Stationary Waves and beats

4 Dispersion

PHXI15:WAVES

355050 $S_{1}$ and $S_{2}$ are two coherent sources of sound having no initial phase difference. The velocity of sound is $330 m / s$ . No minima will be formed on the line passing through $S_{2}$ and perpendicular to the line joining $S_{1}$ and $S_{2}$ , if the frequency of both the sources may be :
supporting img

1

50 H z

2

60 H z

3

70 H z

4

80 H z

Explanation:

For minimum,
$Δ x = (2 n - 1) \frac{λ}{2}$
The maximum possible path difference $=$ distance between the sources $= 3 m$ .
For no minimum
$\begin{aligned} \frac{λ}{2} > 3 \Rightarrow λ > 6 \\ ∴ f = \frac{v}{λ} < \frac{330}{6} = 55 . \end{aligned}$
If $∴ f < 55 H z$ . No minimum will occur.

PHXI15:WAVES

355051 Two sound waves, each of amplitude $A$ and frequency $ω$ superpose at a point with phase difference of $\frac{π}{2}$ . The amplitude and frequency of the resultant wave are respectively

1

\frac{A}{\sqrt{2}}, \frac{ω}{2}

2

\frac{A}{\sqrt{2}}, ω

3

\sqrt{2} A, \frac{ω}{2}

4

\sqrt{2} A, ω

Explanation:

Let $y_{1} = A \sin (ω t)$ and
$y_{2} = A \sin (ω t + \frac{π}{2}) (1)$
Resultant amplitude
$R = A^{2} + A^{2} + 2 A^{2} \cos (\frac{π}{2})$
$\Rightarrow R = \sqrt{2 A^{2} + 2 A^{2} \times 0} \Rightarrow R = \sqrt{2 A^{2}}$
$\Rightarrow R = \sqrt{2} A$
However, both will have the same frequency on superimposing.

PHXI15:WAVES

355052 The phenomena arising due to the superposition of waves is/are

1 beats

2 stationary waves

3 Lissajous figure

4 All of these

PHXI15:WAVES

355053 Sound signal is sent through a composite tube as shown in Fig. The radius of the semicircle is $r$ . Speed of sound in air is $v$ . The source of sound is capable to generate frequencies in the range $f_{1}$ to $f_{2} (f_{2} > f_{1})$ . If $n$ is an integer then frequency for maximum intensity at ' $p$ ' is given by
supporting img

1

\frac{n v}{r (π - 2)}

2

\frac{n v}{r}

3

\frac{n v}{(r - 2) π}

4

\frac{n v}{π r}

Explanation:

Path difference $π r - 2 r = n λ$
$r (π - 2) = \frac{n v}{f} thus f = \frac{n v}{r (π - 2)}$

PHXI15:WAVES

355054 The principle of superposition of waves can not be used to explain wave phenomena like

1 Interference

2 Diffraction

3 Stationary Waves and beats

4 Dispersion

Explanation:

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