QBManager

PHXI15:WAVES

354571 A simple harmonic progressive wave is represented as $y = 0.03 \sin π (2 t - 0.01 x) m$ . At a given instant of time, the phase difference between two particles $25 m$ apart is

1

π r a d

2

\frac{π}{2} r a d

3

\frac{π}{4} r a d

4

\frac{π}{8} r a d

Explanation:

The given equation of $S H M$ wave is
$\begin{aligned} y = 0.03 \sin π (2 t - 0.01 x) m \\ y = 0.03 \sin (2 π t - 0.01 π x) m \end{aligned}$
Comparing it with general equation, we get
$y = a \sin (ω t - k x)$
where, $k = \frac{2 π}{λ} \Rightarrow λ = 200 m$
The phase difference between two particles is given by
$Δ ϕ = k x = \frac{2 π}{λ} \times x (1)$
here, $x = 25 m$
Substituting the values of $x$ and $λ$ in eq.(1), we get
$Δ ϕ = \frac{2 π}{200} \times 25 = \frac{π}{4} r a d$

PHXI15:WAVES

354572 The plane progressive wave is described by the equation $y = 3 \cos (\frac{x}{4} - 10 t - \frac{π}{2})$ , where $x$ and $y$ are in meters and $t$ in seconds. The maximum velocity of the particles of the medium due to this wave is

1

30 m / s

2

(\frac{3 π}{2}) m / s

3

3 / 4 m / s

4

40 m / s

Explanation:

$y = 3 \cos (\frac{x}{4} - 10 t - \frac{π}{2})$
maximum particle velocity
$= A ω = 3 \times 10 = 30 m / s$

PHXI15:WAVES

354573 The tansverse wave represented by the equation $y = \sin [3 x - 15 t]$

1 Wavelength

= 4 \frac{π}{3}

2 Amplitude

= 4

3 Period

= \frac{π}{15}

4 Speed of propagation

= 5

Explanation:

Compare the given equation with standard form
$\begin{aligned} y = A \sin [\frac{2 π x}{λ} - \frac{2 π t}{T}] \\ A = 1 \\ \frac{2 π}{λ} = 3, λ = \frac{2 π}{3} and \frac{2 π}{T} = 15 \\ T = \frac{2 π}{15} \Rightarrow v = \frac{λ}{T} = 5 \end{aligned}$

PHXI15:WAVES

354574 A transverse wave is represented by $y = 2 \sin (ω t - k x)$ The wavelength for which wave-velocity is equal to the maximum particle velocity, is

1

9.62 m

2

12.56 m

3

15.23 m

4

18.42 m

Explanation:

Wave velocity, $v = \frac{ω}{k}$
Particle velocity, $v_{p} = \frac{d y}{d t} = A ω \cos (ω t - k x)$
$∴ {(v_{p})}_{max} = A ω$
$A ω = \frac{ω}{k} (given)$
$\Rightarrow A = \frac{λ}{2 π}$
$\Rightarrow λ = 2 π A = 4 π = 12.56 m$

PHXI15:WAVES

354571 A simple harmonic progressive wave is represented as $y = 0.03 \sin π (2 t - 0.01 x) m$ . At a given instant of time, the phase difference between two particles $25 m$ apart is

1

π r a d

2

\frac{π}{2} r a d

3

\frac{π}{4} r a d

4

\frac{π}{8} r a d

Explanation:

The given equation of $S H M$ wave is
$\begin{aligned} y = 0.03 \sin π (2 t - 0.01 x) m \\ y = 0.03 \sin (2 π t - 0.01 π x) m \end{aligned}$
Comparing it with general equation, we get
$y = a \sin (ω t - k x)$
where, $k = \frac{2 π}{λ} \Rightarrow λ = 200 m$
The phase difference between two particles is given by
$Δ ϕ = k x = \frac{2 π}{λ} \times x (1)$
here, $x = 25 m$
Substituting the values of $x$ and $λ$ in eq.(1), we get
$Δ ϕ = \frac{2 π}{200} \times 25 = \frac{π}{4} r a d$

PHXI15:WAVES

354572 The plane progressive wave is described by the equation $y = 3 \cos (\frac{x}{4} - 10 t - \frac{π}{2})$ , where $x$ and $y$ are in meters and $t$ in seconds. The maximum velocity of the particles of the medium due to this wave is

1

30 m / s

2

(\frac{3 π}{2}) m / s

3

3 / 4 m / s

4

40 m / s

Explanation:

$y = 3 \cos (\frac{x}{4} - 10 t - \frac{π}{2})$
maximum particle velocity
$= A ω = 3 \times 10 = 30 m / s$

PHXI15:WAVES

354573 The tansverse wave represented by the equation $y = \sin [3 x - 15 t]$

1 Wavelength

= 4 \frac{π}{3}

2 Amplitude

= 4

3 Period

= \frac{π}{15}

4 Speed of propagation

= 5

Explanation:

Compare the given equation with standard form
$\begin{aligned} y = A \sin [\frac{2 π x}{λ} - \frac{2 π t}{T}] \\ A = 1 \\ \frac{2 π}{λ} = 3, λ = \frac{2 π}{3} and \frac{2 π}{T} = 15 \\ T = \frac{2 π}{15} \Rightarrow v = \frac{λ}{T} = 5 \end{aligned}$

PHXI15:WAVES

354574 A transverse wave is represented by $y = 2 \sin (ω t - k x)$ The wavelength for which wave-velocity is equal to the maximum particle velocity, is

1

9.62 m

2

12.56 m

3

15.23 m

4

18.42 m

Explanation:

Wave velocity, $v = \frac{ω}{k}$
Particle velocity, $v_{p} = \frac{d y}{d t} = A ω \cos (ω t - k x)$
$∴ {(v_{p})}_{max} = A ω$
$A ω = \frac{ω}{k} (given)$
$\Rightarrow A = \frac{λ}{2 π}$
$\Rightarrow λ = 2 π A = 4 π = 12.56 m$

PHXI15:WAVES

354571 A simple harmonic progressive wave is represented as $y = 0.03 \sin π (2 t - 0.01 x) m$ . At a given instant of time, the phase difference between two particles $25 m$ apart is

1

π r a d

2

\frac{π}{2} r a d

3

\frac{π}{4} r a d

4

\frac{π}{8} r a d

Explanation:

The given equation of $S H M$ wave is
$\begin{aligned} y = 0.03 \sin π (2 t - 0.01 x) m \\ y = 0.03 \sin (2 π t - 0.01 π x) m \end{aligned}$
Comparing it with general equation, we get
$y = a \sin (ω t - k x)$
where, $k = \frac{2 π}{λ} \Rightarrow λ = 200 m$
The phase difference between two particles is given by
$Δ ϕ = k x = \frac{2 π}{λ} \times x (1)$
here, $x = 25 m$
Substituting the values of $x$ and $λ$ in eq.(1), we get
$Δ ϕ = \frac{2 π}{200} \times 25 = \frac{π}{4} r a d$

PHXI15:WAVES

354572 The plane progressive wave is described by the equation $y = 3 \cos (\frac{x}{4} - 10 t - \frac{π}{2})$ , where $x$ and $y$ are in meters and $t$ in seconds. The maximum velocity of the particles of the medium due to this wave is

1

30 m / s

2

(\frac{3 π}{2}) m / s

3

3 / 4 m / s

4

40 m / s

Explanation:

$y = 3 \cos (\frac{x}{4} - 10 t - \frac{π}{2})$
maximum particle velocity
$= A ω = 3 \times 10 = 30 m / s$

PHXI15:WAVES

354573 The tansverse wave represented by the equation $y = \sin [3 x - 15 t]$

1 Wavelength

= 4 \frac{π}{3}

2 Amplitude

= 4

3 Period

= \frac{π}{15}

4 Speed of propagation

= 5

Explanation:

Compare the given equation with standard form
$\begin{aligned} y = A \sin [\frac{2 π x}{λ} - \frac{2 π t}{T}] \\ A = 1 \\ \frac{2 π}{λ} = 3, λ = \frac{2 π}{3} and \frac{2 π}{T} = 15 \\ T = \frac{2 π}{15} \Rightarrow v = \frac{λ}{T} = 5 \end{aligned}$

PHXI15:WAVES

354574 A transverse wave is represented by $y = 2 \sin (ω t - k x)$ The wavelength for which wave-velocity is equal to the maximum particle velocity, is

1

9.62 m

2

12.56 m

3

15.23 m

4

18.42 m

Explanation:

Wave velocity, $v = \frac{ω}{k}$
Particle velocity, $v_{p} = \frac{d y}{d t} = A ω \cos (ω t - k x)$
$∴ {(v_{p})}_{max} = A ω$
$A ω = \frac{ω}{k} (given)$
$\Rightarrow A = \frac{λ}{2 π}$
$\Rightarrow λ = 2 π A = 4 π = 12.56 m$

PHXI15:WAVES

354571 A simple harmonic progressive wave is represented as $y = 0.03 \sin π (2 t - 0.01 x) m$ . At a given instant of time, the phase difference between two particles $25 m$ apart is

1

π r a d

2

\frac{π}{2} r a d

3

\frac{π}{4} r a d

4

\frac{π}{8} r a d

Explanation:

The given equation of $S H M$ wave is
$\begin{aligned} y = 0.03 \sin π (2 t - 0.01 x) m \\ y = 0.03 \sin (2 π t - 0.01 π x) m \end{aligned}$
Comparing it with general equation, we get
$y = a \sin (ω t - k x)$
where, $k = \frac{2 π}{λ} \Rightarrow λ = 200 m$
The phase difference between two particles is given by
$Δ ϕ = k x = \frac{2 π}{λ} \times x (1)$
here, $x = 25 m$
Substituting the values of $x$ and $λ$ in eq.(1), we get
$Δ ϕ = \frac{2 π}{200} \times 25 = \frac{π}{4} r a d$

PHXI15:WAVES

354572 The plane progressive wave is described by the equation $y = 3 \cos (\frac{x}{4} - 10 t - \frac{π}{2})$ , where $x$ and $y$ are in meters and $t$ in seconds. The maximum velocity of the particles of the medium due to this wave is

1

30 m / s

2

(\frac{3 π}{2}) m / s

3

3 / 4 m / s

4

40 m / s

Explanation:

$y = 3 \cos (\frac{x}{4} - 10 t - \frac{π}{2})$
maximum particle velocity
$= A ω = 3 \times 10 = 30 m / s$

PHXI15:WAVES

354573 The tansverse wave represented by the equation $y = \sin [3 x - 15 t]$

1 Wavelength

= 4 \frac{π}{3}

2 Amplitude

= 4

3 Period

= \frac{π}{15}

4 Speed of propagation

= 5

Explanation:

Compare the given equation with standard form
$\begin{aligned} y = A \sin [\frac{2 π x}{λ} - \frac{2 π t}{T}] \\ A = 1 \\ \frac{2 π}{λ} = 3, λ = \frac{2 π}{3} and \frac{2 π}{T} = 15 \\ T = \frac{2 π}{15} \Rightarrow v = \frac{λ}{T} = 5 \end{aligned}$

PHXI15:WAVES

354574 A transverse wave is represented by $y = 2 \sin (ω t - k x)$ The wavelength for which wave-velocity is equal to the maximum particle velocity, is

1

9.62 m

2

12.56 m

3

15.23 m

4

18.42 m

Explanation:

Wave velocity, $v = \frac{ω}{k}$
Particle velocity, $v_{p} = \frac{d y}{d t} = A ω \cos (ω t - k x)$
$∴ {(v_{p})}_{max} = A ω$
$A ω = \frac{ω}{k} (given)$
$\Rightarrow A = \frac{λ}{2 π}$
$\Rightarrow λ = 2 π A = 4 π = 12.56 m$

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: