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PHXI15:WAVES

355130 Standing waves can be produced

1 On a string clamped at both ends

2 On a string clamped at one end and free at the other

3 When incident wave gets reflected from a wall

4 All the above

PHXI15:WAVES

355131 Equations of a stationary and travelling waves are as follows
$y_{1} = a \sin k x \cos ω t and y_{2} = a \sin (ω t - k x)$
The phase difference between two points $x_{1} = \frac{π}{3 k}$ and $x_{2} = \frac{3 π}{2 k}$ is $ϕ_{1}$ , in the standing wave $y_{1}$ and is $ϕ_{2}$ in travelling wave $(y_{2})$ then $\frac{ϕ_{1}}{ϕ_{2}}$ is

1

1

2

\frac{3}{4}

3

\frac{6}{7}

4

\frac{5}{6}

Explanation:

The positions of the points
$\begin{aligned} are x_{1} = \frac{π}{3 k} = \frac{λ}{6} \\ x_{2} = \frac{3 π}{2 k} = \frac{3 λ}{4} \\ Δ x = x_{2} - x_{1} = \frac{7 λ}{12} \end{aligned}$
The points $x_{1}$ and $x_{2}$ are shown the figures.
supporting img

$Δ ϕ_{1} = π$
$\frac{Δ ϕ_{1}}{Δ ϕ_{2}} = Δ x \times \frac{2 π}{λ} = \frac{6}{7} π$
If two points are present in the same loop of a standing wave then $Δ ϕ = 0$ and if they are present in adjacent loops then $Δ ϕ = π$
$\frac{Δ ϕ_{1}}{Δ ϕ_{2}} = \frac{6}{7}$

PHXI15:WAVES

355132 The equation of a stationary wave is $y = 10 \sin \frac{π x}{4} \cos 20 π t$ . The distance between two consecutive nodes (in $m$ ) is

1 4

2 2

3 5

4 8

Explanation:

$k = \frac{2 π}{λ} = \frac{π}{4} \Rightarrow λ = 8$
Now, distance between nodes $= λ / 2 = 4$

PHXI15:WAVES

355133 When a stationary wave is formed then its frequency is

1 Half that of the individual waves

2 Twice that of the individual waves

3 Same as that of the individual waves

4 None of these

PHXI15:WAVES

355134 The vibrations of a string of length $60 c m$ fixed at both the ends are represented by the equation $y = 2 \sin (\frac{4 π x}{15}) \cos (96 π t)$ where $x$ and $y$ are in $c m$ . The maximum number of loops that can be formed in it is

1

16

2

4

3

15

4

5

Explanation:

Let the string vibrates in $p$ loops, wavelength of the $p^{th}$ mode of vibration is given by
$λ_{p} = \frac{2 l}{p} (1)$
Given, $y = 2 \sin (\frac{4 π x}{15}) \cos (96 π t)$
$k = \frac{4 π}{15} = \frac{2 π}{λ}$
$λ = \frac{15}{2} (2)$
From (1) & (2) we get
$\Rightarrow p = 16 .$

PHXI15:WAVES

355130 Standing waves can be produced

1 On a string clamped at both ends

2 On a string clamped at one end and free at the other

3 When incident wave gets reflected from a wall

4 All the above

PHXI15:WAVES

355131 Equations of a stationary and travelling waves are as follows
$y_{1} = a \sin k x \cos ω t and y_{2} = a \sin (ω t - k x)$
The phase difference between two points $x_{1} = \frac{π}{3 k}$ and $x_{2} = \frac{3 π}{2 k}$ is $ϕ_{1}$ , in the standing wave $y_{1}$ and is $ϕ_{2}$ in travelling wave $(y_{2})$ then $\frac{ϕ_{1}}{ϕ_{2}}$ is

1

1

2

\frac{3}{4}

3

\frac{6}{7}

4

\frac{5}{6}

Explanation:

The positions of the points
$\begin{aligned} are x_{1} = \frac{π}{3 k} = \frac{λ}{6} \\ x_{2} = \frac{3 π}{2 k} = \frac{3 λ}{4} \\ Δ x = x_{2} - x_{1} = \frac{7 λ}{12} \end{aligned}$
The points $x_{1}$ and $x_{2}$ are shown the figures.
supporting img

$Δ ϕ_{1} = π$
$\frac{Δ ϕ_{1}}{Δ ϕ_{2}} = Δ x \times \frac{2 π}{λ} = \frac{6}{7} π$
If two points are present in the same loop of a standing wave then $Δ ϕ = 0$ and if they are present in adjacent loops then $Δ ϕ = π$
$\frac{Δ ϕ_{1}}{Δ ϕ_{2}} = \frac{6}{7}$

PHXI15:WAVES

355132 The equation of a stationary wave is $y = 10 \sin \frac{π x}{4} \cos 20 π t$ . The distance between two consecutive nodes (in $m$ ) is

1 4

2 2

3 5

4 8

Explanation:

$k = \frac{2 π}{λ} = \frac{π}{4} \Rightarrow λ = 8$
Now, distance between nodes $= λ / 2 = 4$

PHXI15:WAVES

355133 When a stationary wave is formed then its frequency is

1 Half that of the individual waves

2 Twice that of the individual waves

3 Same as that of the individual waves

4 None of these

PHXI15:WAVES

355134 The vibrations of a string of length $60 c m$ fixed at both the ends are represented by the equation $y = 2 \sin (\frac{4 π x}{15}) \cos (96 π t)$ where $x$ and $y$ are in $c m$ . The maximum number of loops that can be formed in it is

1

16

2

4

3

15

4

5

Explanation:

Let the string vibrates in $p$ loops, wavelength of the $p^{th}$ mode of vibration is given by
$λ_{p} = \frac{2 l}{p} (1)$
Given, $y = 2 \sin (\frac{4 π x}{15}) \cos (96 π t)$
$k = \frac{4 π}{15} = \frac{2 π}{λ}$
$λ = \frac{15}{2} (2)$
From (1) & (2) we get
$\Rightarrow p = 16 .$

PHXI15:WAVES

355130 Standing waves can be produced

1 On a string clamped at both ends

2 On a string clamped at one end and free at the other

3 When incident wave gets reflected from a wall

4 All the above

PHXI15:WAVES

355131 Equations of a stationary and travelling waves are as follows
$y_{1} = a \sin k x \cos ω t and y_{2} = a \sin (ω t - k x)$
The phase difference between two points $x_{1} = \frac{π}{3 k}$ and $x_{2} = \frac{3 π}{2 k}$ is $ϕ_{1}$ , in the standing wave $y_{1}$ and is $ϕ_{2}$ in travelling wave $(y_{2})$ then $\frac{ϕ_{1}}{ϕ_{2}}$ is

1

1

2

\frac{3}{4}

3

\frac{6}{7}

4

\frac{5}{6}

Explanation:

The positions of the points
$\begin{aligned} are x_{1} = \frac{π}{3 k} = \frac{λ}{6} \\ x_{2} = \frac{3 π}{2 k} = \frac{3 λ}{4} \\ Δ x = x_{2} - x_{1} = \frac{7 λ}{12} \end{aligned}$
The points $x_{1}$ and $x_{2}$ are shown the figures.
supporting img

$Δ ϕ_{1} = π$
$\frac{Δ ϕ_{1}}{Δ ϕ_{2}} = Δ x \times \frac{2 π}{λ} = \frac{6}{7} π$
If two points are present in the same loop of a standing wave then $Δ ϕ = 0$ and if they are present in adjacent loops then $Δ ϕ = π$
$\frac{Δ ϕ_{1}}{Δ ϕ_{2}} = \frac{6}{7}$

PHXI15:WAVES

355132 The equation of a stationary wave is $y = 10 \sin \frac{π x}{4} \cos 20 π t$ . The distance between two consecutive nodes (in $m$ ) is

1 4

2 2

3 5

4 8

Explanation:

$k = \frac{2 π}{λ} = \frac{π}{4} \Rightarrow λ = 8$
Now, distance between nodes $= λ / 2 = 4$

PHXI15:WAVES

355133 When a stationary wave is formed then its frequency is

1 Half that of the individual waves

2 Twice that of the individual waves

3 Same as that of the individual waves

4 None of these

PHXI15:WAVES

355134 The vibrations of a string of length $60 c m$ fixed at both the ends are represented by the equation $y = 2 \sin (\frac{4 π x}{15}) \cos (96 π t)$ where $x$ and $y$ are in $c m$ . The maximum number of loops that can be formed in it is

1

16

2

4

3

15

4

5

Explanation:

Let the string vibrates in $p$ loops, wavelength of the $p^{th}$ mode of vibration is given by
$λ_{p} = \frac{2 l}{p} (1)$
Given, $y = 2 \sin (\frac{4 π x}{15}) \cos (96 π t)$
$k = \frac{4 π}{15} = \frac{2 π}{λ}$
$λ = \frac{15}{2} (2)$
From (1) & (2) we get
$\Rightarrow p = 16 .$

PHXI15:WAVES

355130 Standing waves can be produced

1 On a string clamped at both ends

2 On a string clamped at one end and free at the other

3 When incident wave gets reflected from a wall

4 All the above

PHXI15:WAVES

355131 Equations of a stationary and travelling waves are as follows
$y_{1} = a \sin k x \cos ω t and y_{2} = a \sin (ω t - k x)$
The phase difference between two points $x_{1} = \frac{π}{3 k}$ and $x_{2} = \frac{3 π}{2 k}$ is $ϕ_{1}$ , in the standing wave $y_{1}$ and is $ϕ_{2}$ in travelling wave $(y_{2})$ then $\frac{ϕ_{1}}{ϕ_{2}}$ is

1

1

2

\frac{3}{4}

3

\frac{6}{7}

4

\frac{5}{6}

Explanation:

The positions of the points
$\begin{aligned} are x_{1} = \frac{π}{3 k} = \frac{λ}{6} \\ x_{2} = \frac{3 π}{2 k} = \frac{3 λ}{4} \\ Δ x = x_{2} - x_{1} = \frac{7 λ}{12} \end{aligned}$
The points $x_{1}$ and $x_{2}$ are shown the figures.
supporting img

$Δ ϕ_{1} = π$
$\frac{Δ ϕ_{1}}{Δ ϕ_{2}} = Δ x \times \frac{2 π}{λ} = \frac{6}{7} π$
If two points are present in the same loop of a standing wave then $Δ ϕ = 0$ and if they are present in adjacent loops then $Δ ϕ = π$
$\frac{Δ ϕ_{1}}{Δ ϕ_{2}} = \frac{6}{7}$

PHXI15:WAVES

355132 The equation of a stationary wave is $y = 10 \sin \frac{π x}{4} \cos 20 π t$ . The distance between two consecutive nodes (in $m$ ) is

1 4

2 2

3 5

4 8

Explanation:

$k = \frac{2 π}{λ} = \frac{π}{4} \Rightarrow λ = 8$
Now, distance between nodes $= λ / 2 = 4$

PHXI15:WAVES

355133 When a stationary wave is formed then its frequency is

1 Half that of the individual waves

2 Twice that of the individual waves

3 Same as that of the individual waves

4 None of these

PHXI15:WAVES

355134 The vibrations of a string of length $60 c m$ fixed at both the ends are represented by the equation $y = 2 \sin (\frac{4 π x}{15}) \cos (96 π t)$ where $x$ and $y$ are in $c m$ . The maximum number of loops that can be formed in it is

1

16

2

4

3

15

4

5

Explanation:

Let the string vibrates in $p$ loops, wavelength of the $p^{th}$ mode of vibration is given by
$λ_{p} = \frac{2 l}{p} (1)$
Given, $y = 2 \sin (\frac{4 π x}{15}) \cos (96 π t)$
$k = \frac{4 π}{15} = \frac{2 π}{λ}$
$λ = \frac{15}{2} (2)$
From (1) & (2) we get
$\Rightarrow p = 16 .$

PHXI15:WAVES

355130 Standing waves can be produced

1 On a string clamped at both ends

2 On a string clamped at one end and free at the other

3 When incident wave gets reflected from a wall

4 All the above

PHXI15:WAVES

355131 Equations of a stationary and travelling waves are as follows
$y_{1} = a \sin k x \cos ω t and y_{2} = a \sin (ω t - k x)$
The phase difference between two points $x_{1} = \frac{π}{3 k}$ and $x_{2} = \frac{3 π}{2 k}$ is $ϕ_{1}$ , in the standing wave $y_{1}$ and is $ϕ_{2}$ in travelling wave $(y_{2})$ then $\frac{ϕ_{1}}{ϕ_{2}}$ is

1

1

2

\frac{3}{4}

3

\frac{6}{7}

4

\frac{5}{6}

Explanation:

The positions of the points
$\begin{aligned} are x_{1} = \frac{π}{3 k} = \frac{λ}{6} \\ x_{2} = \frac{3 π}{2 k} = \frac{3 λ}{4} \\ Δ x = x_{2} - x_{1} = \frac{7 λ}{12} \end{aligned}$
The points $x_{1}$ and $x_{2}$ are shown the figures.
supporting img

$Δ ϕ_{1} = π$
$\frac{Δ ϕ_{1}}{Δ ϕ_{2}} = Δ x \times \frac{2 π}{λ} = \frac{6}{7} π$
If two points are present in the same loop of a standing wave then $Δ ϕ = 0$ and if they are present in adjacent loops then $Δ ϕ = π$
$\frac{Δ ϕ_{1}}{Δ ϕ_{2}} = \frac{6}{7}$

PHXI15:WAVES

355132 The equation of a stationary wave is $y = 10 \sin \frac{π x}{4} \cos 20 π t$ . The distance between two consecutive nodes (in $m$ ) is

1 4

2 2

3 5

4 8

Explanation:

$k = \frac{2 π}{λ} = \frac{π}{4} \Rightarrow λ = 8$
Now, distance between nodes $= λ / 2 = 4$

PHXI15:WAVES

355133 When a stationary wave is formed then its frequency is

1 Half that of the individual waves

2 Twice that of the individual waves

3 Same as that of the individual waves

4 None of these

PHXI15:WAVES

355134 The vibrations of a string of length $60 c m$ fixed at both the ends are represented by the equation $y = 2 \sin (\frac{4 π x}{15}) \cos (96 π t)$ where $x$ and $y$ are in $c m$ . The maximum number of loops that can be formed in it is

1

16

2

4

3

15

4

5

Explanation:

Let the string vibrates in $p$ loops, wavelength of the $p^{th}$ mode of vibration is given by
$λ_{p} = \frac{2 l}{p} (1)$
Given, $y = 2 \sin (\frac{4 π x}{15}) \cos (96 π t)$
$k = \frac{4 π}{15} = \frac{2 π}{λ}$
$λ = \frac{15}{2} (2)$
From (1) & (2) we get
$\Rightarrow p = 16 .$

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