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PHXI15:WAVES

354528 The equation of longitudinal wave represented as $y = 20 \cos π (50 t - x) c m,$ then its wavelength is

1

120 c m

2

50 c m

3

2 c m

4

5 c m

Explanation:

The standard equation of a wave of amplitude $a$ is given by
$\begin{matrix} (1) & y = a \cos (ω t - k x) \end{matrix}$
where, $ω$ is angular velocity, $k$ is the wave number and $t$ is the time.
Comparing with given equation, we get $k = π$ .
Wavelength, $λ = \frac{2 π}{k} = \frac{2 π}{π} = 2 c m$

PHXI15:WAVES

354529 The position of a wave (of wavelength $λ$ ) $y (x, t) = A \sin (k x - 23562 t)$ is shown at $t = 0$ , find $x$ -coordinate of point $P$ in metres, if the wave speed is $300 m / \sec [∴ y, x$ are measured in meters, $t$ is in seconds].
supporting img

1

\frac{27 λ}{12}

2

\frac{29 λ}{12}

3

\frac{25 λ}{12}

4

\frac{31 λ}{12}

Explanation:

At $t = 0, y = A \sin k x = A \sin \frac{2 π}{λ} x$
From the figure, $k x = \frac{2 π}{λ} x = 4 π + \frac{π}{6} \Rightarrow x = \frac{25 λ}{12}$

PHXI15:WAVES

354530 A wave equation is given by $y = 4 \sin [π (\frac{t}{5} - \frac{x}{9} + \frac{1}{6})]$ , where $x$ is in centimetre and $t$ is in second. Which of the following is true?

1

λ = 18 c m

2

v = 4 m s^{- 1}

3

a = 0.4 m

4

f = 50 H z

Explanation:

The given equation will be written as
$\begin{matrix} (1) & y = 4 \sin [π (\frac{t}{5} - \frac{x}{9} + \frac{1}{6})] \end{matrix}$
The standard wave equation can be written as
$\begin{aligned} y & = a \sin (ω t - k x + ϕ) \\ (2) & \Rightarrow y & = a \sin (\frac{2 π}{T} t - \frac{2 π}{λ} \cdot x + ϕ) \end{aligned}$
Comparing Eqs.(1) and (2),we get
Amplitude, $a = 4 c m = 0.04 m$
Frequency, $f = \frac{1}{T} = \frac{1}{10} H z = 0.1 H z$
Wavelength, $λ = 2 \times 9 = 18 c m$
Velocity, $v = f λ$
$= 0.1 \times 18 = 1.8 c m s^{- 1}$

PHXI15:WAVES

354531 A transverse wave is described by the equation $y = y_{0} \sin 2 π [n t - \frac{x}{λ}]$ . If the maximum particle velocity is four times the wave velocity, then $λ$ is equal to

1

y_{0}

2

\frac{y_{0}}{2}

3

π y_{0}

4

\frac{π y_{0}}{2}

Explanation:

The given equation is
$\begin{aligned} y = y_{o} \sin 2 π [n t - \frac{x}{λ}] \\ ω = 2 π n, k = \frac{2 π}{λ} \end{aligned}$
The velocity of the wave is $v = \frac{ω}{k} = n λ$
Given that $v_{p} = 4 v$
$y_{o} (2 π n) = 4 n λ \Rightarrow λ = \frac{π y_{o}}{2}$

PHXI15:WAVES

354528 The equation of longitudinal wave represented as $y = 20 \cos π (50 t - x) c m,$ then its wavelength is

1

120 c m

2

50 c m

3

2 c m

4

5 c m

Explanation:

The standard equation of a wave of amplitude $a$ is given by
$\begin{matrix} (1) & y = a \cos (ω t - k x) \end{matrix}$
where, $ω$ is angular velocity, $k$ is the wave number and $t$ is the time.
Comparing with given equation, we get $k = π$ .
Wavelength, $λ = \frac{2 π}{k} = \frac{2 π}{π} = 2 c m$

PHXI15:WAVES

354529 The position of a wave (of wavelength $λ$ ) $y (x, t) = A \sin (k x - 23562 t)$ is shown at $t = 0$ , find $x$ -coordinate of point $P$ in metres, if the wave speed is $300 m / \sec [∴ y, x$ are measured in meters, $t$ is in seconds].
supporting img

1

\frac{27 λ}{12}

2

\frac{29 λ}{12}

3

\frac{25 λ}{12}

4

\frac{31 λ}{12}

Explanation:

At $t = 0, y = A \sin k x = A \sin \frac{2 π}{λ} x$
From the figure, $k x = \frac{2 π}{λ} x = 4 π + \frac{π}{6} \Rightarrow x = \frac{25 λ}{12}$

PHXI15:WAVES

354530 A wave equation is given by $y = 4 \sin [π (\frac{t}{5} - \frac{x}{9} + \frac{1}{6})]$ , where $x$ is in centimetre and $t$ is in second. Which of the following is true?

1

λ = 18 c m

2

v = 4 m s^{- 1}

3

a = 0.4 m

4

f = 50 H z

Explanation:

The given equation will be written as
$\begin{matrix} (1) & y = 4 \sin [π (\frac{t}{5} - \frac{x}{9} + \frac{1}{6})] \end{matrix}$
The standard wave equation can be written as
$\begin{aligned} y & = a \sin (ω t - k x + ϕ) \\ (2) & \Rightarrow y & = a \sin (\frac{2 π}{T} t - \frac{2 π}{λ} \cdot x + ϕ) \end{aligned}$
Comparing Eqs.(1) and (2),we get
Amplitude, $a = 4 c m = 0.04 m$
Frequency, $f = \frac{1}{T} = \frac{1}{10} H z = 0.1 H z$
Wavelength, $λ = 2 \times 9 = 18 c m$
Velocity, $v = f λ$
$= 0.1 \times 18 = 1.8 c m s^{- 1}$

PHXI15:WAVES

354531 A transverse wave is described by the equation $y = y_{0} \sin 2 π [n t - \frac{x}{λ}]$ . If the maximum particle velocity is four times the wave velocity, then $λ$ is equal to

1

y_{0}

2

\frac{y_{0}}{2}

3

π y_{0}

4

\frac{π y_{0}}{2}

Explanation:

The given equation is
$\begin{aligned} y = y_{o} \sin 2 π [n t - \frac{x}{λ}] \\ ω = 2 π n, k = \frac{2 π}{λ} \end{aligned}$
The velocity of the wave is $v = \frac{ω}{k} = n λ$
Given that $v_{p} = 4 v$
$y_{o} (2 π n) = 4 n λ \Rightarrow λ = \frac{π y_{o}}{2}$

PHXI15:WAVES

354528 The equation of longitudinal wave represented as $y = 20 \cos π (50 t - x) c m,$ then its wavelength is

1

120 c m

2

50 c m

3

2 c m

4

5 c m

Explanation:

The standard equation of a wave of amplitude $a$ is given by
$\begin{matrix} (1) & y = a \cos (ω t - k x) \end{matrix}$
where, $ω$ is angular velocity, $k$ is the wave number and $t$ is the time.
Comparing with given equation, we get $k = π$ .
Wavelength, $λ = \frac{2 π}{k} = \frac{2 π}{π} = 2 c m$

PHXI15:WAVES

354529 The position of a wave (of wavelength $λ$ ) $y (x, t) = A \sin (k x - 23562 t)$ is shown at $t = 0$ , find $x$ -coordinate of point $P$ in metres, if the wave speed is $300 m / \sec [∴ y, x$ are measured in meters, $t$ is in seconds].
supporting img

1

\frac{27 λ}{12}

2

\frac{29 λ}{12}

3

\frac{25 λ}{12}

4

\frac{31 λ}{12}

Explanation:

At $t = 0, y = A \sin k x = A \sin \frac{2 π}{λ} x$
From the figure, $k x = \frac{2 π}{λ} x = 4 π + \frac{π}{6} \Rightarrow x = \frac{25 λ}{12}$

PHXI15:WAVES

354530 A wave equation is given by $y = 4 \sin [π (\frac{t}{5} - \frac{x}{9} + \frac{1}{6})]$ , where $x$ is in centimetre and $t$ is in second. Which of the following is true?

1

λ = 18 c m

2

v = 4 m s^{- 1}

3

a = 0.4 m

4

f = 50 H z

Explanation:

The given equation will be written as
$\begin{matrix} (1) & y = 4 \sin [π (\frac{t}{5} - \frac{x}{9} + \frac{1}{6})] \end{matrix}$
The standard wave equation can be written as
$\begin{aligned} y & = a \sin (ω t - k x + ϕ) \\ (2) & \Rightarrow y & = a \sin (\frac{2 π}{T} t - \frac{2 π}{λ} \cdot x + ϕ) \end{aligned}$
Comparing Eqs.(1) and (2),we get
Amplitude, $a = 4 c m = 0.04 m$
Frequency, $f = \frac{1}{T} = \frac{1}{10} H z = 0.1 H z$
Wavelength, $λ = 2 \times 9 = 18 c m$
Velocity, $v = f λ$
$= 0.1 \times 18 = 1.8 c m s^{- 1}$

PHXI15:WAVES

354531 A transverse wave is described by the equation $y = y_{0} \sin 2 π [n t - \frac{x}{λ}]$ . If the maximum particle velocity is four times the wave velocity, then $λ$ is equal to

1

y_{0}

2

\frac{y_{0}}{2}

3

π y_{0}

4

\frac{π y_{0}}{2}

Explanation:

The given equation is
$\begin{aligned} y = y_{o} \sin 2 π [n t - \frac{x}{λ}] \\ ω = 2 π n, k = \frac{2 π}{λ} \end{aligned}$
The velocity of the wave is $v = \frac{ω}{k} = n λ$
Given that $v_{p} = 4 v$
$y_{o} (2 π n) = 4 n λ \Rightarrow λ = \frac{π y_{o}}{2}$

PHXI15:WAVES

354528 The equation of longitudinal wave represented as $y = 20 \cos π (50 t - x) c m,$ then its wavelength is

1

120 c m

2

50 c m

3

2 c m

4

5 c m

Explanation:

The standard equation of a wave of amplitude $a$ is given by
$\begin{matrix} (1) & y = a \cos (ω t - k x) \end{matrix}$
where, $ω$ is angular velocity, $k$ is the wave number and $t$ is the time.
Comparing with given equation, we get $k = π$ .
Wavelength, $λ = \frac{2 π}{k} = \frac{2 π}{π} = 2 c m$

PHXI15:WAVES

354529 The position of a wave (of wavelength $λ$ ) $y (x, t) = A \sin (k x - 23562 t)$ is shown at $t = 0$ , find $x$ -coordinate of point $P$ in metres, if the wave speed is $300 m / \sec [∴ y, x$ are measured in meters, $t$ is in seconds].
supporting img

1

\frac{27 λ}{12}

2

\frac{29 λ}{12}

3

\frac{25 λ}{12}

4

\frac{31 λ}{12}

Explanation:

At $t = 0, y = A \sin k x = A \sin \frac{2 π}{λ} x$
From the figure, $k x = \frac{2 π}{λ} x = 4 π + \frac{π}{6} \Rightarrow x = \frac{25 λ}{12}$

PHXI15:WAVES

354530 A wave equation is given by $y = 4 \sin [π (\frac{t}{5} - \frac{x}{9} + \frac{1}{6})]$ , where $x$ is in centimetre and $t$ is in second. Which of the following is true?

1

λ = 18 c m

2

v = 4 m s^{- 1}

3

a = 0.4 m

4

f = 50 H z

Explanation:

The given equation will be written as
$\begin{matrix} (1) & y = 4 \sin [π (\frac{t}{5} - \frac{x}{9} + \frac{1}{6})] \end{matrix}$
The standard wave equation can be written as
$\begin{aligned} y & = a \sin (ω t - k x + ϕ) \\ (2) & \Rightarrow y & = a \sin (\frac{2 π}{T} t - \frac{2 π}{λ} \cdot x + ϕ) \end{aligned}$
Comparing Eqs.(1) and (2),we get
Amplitude, $a = 4 c m = 0.04 m$
Frequency, $f = \frac{1}{T} = \frac{1}{10} H z = 0.1 H z$
Wavelength, $λ = 2 \times 9 = 18 c m$
Velocity, $v = f λ$
$= 0.1 \times 18 = 1.8 c m s^{- 1}$

PHXI15:WAVES

354531 A transverse wave is described by the equation $y = y_{0} \sin 2 π [n t - \frac{x}{λ}]$ . If the maximum particle velocity is four times the wave velocity, then $λ$ is equal to

1

y_{0}

2

\frac{y_{0}}{2}

3

π y_{0}

4

\frac{π y_{0}}{2}

Explanation:

The given equation is
$\begin{aligned} y = y_{o} \sin 2 π [n t - \frac{x}{λ}] \\ ω = 2 π n, k = \frac{2 π}{λ} \end{aligned}$
The velocity of the wave is $v = \frac{ω}{k} = n λ$
Given that $v_{p} = 4 v$
$y_{o} (2 π n) = 4 n λ \Rightarrow λ = \frac{π y_{o}}{2}$