QBManager

WAVES

172171 A wave is represented by $x = 0.4$ $\cos (8 t - \frac{y}{2})$ where $x$ and $y$ are in meters and $t$ in secs. The speed of the wave is

1

0.5 m \cdot s^{- 1}

2

8 m \cdot s^{- 1}

3

16 m \cdot s^{- 1}

4

0.1 m \cdot s^{- 1}

Explanation:

C Equation of the wave given by
$x = 0.4 \cos (8 t - \frac{y}{2})$
Standard equation of wave
$x = A \cos (ω t - ky)$
On comparing
$ω = 8, k = \frac{1}{2}$
Speed $(v) = \frac{ω}{k} = \frac{8}{1 / 2}$
$v = 16 m / \sec$
The speed wave is $16 m / s$

AP EAMCET-07.10.2020

WAVES

172172 The equation of the displacement of waves is $y$ (in $cm) = 10 (\sqrt{3} \sin 2 π t + \cos 2 π t)$
The amplitude of the wave is.

1

10 cm

2

17.3 cm

3

20 cm

4

40 cm

Explanation:

C Given that,
$Y = 10 (\sqrt{3} \sin 2 π t + \cos 2 π t)$
$Y = (10 \sqrt{3} \sin 2 π t + 10 \cos 2 π t)$
The amplitude of the wave-
$A_{resultant} = \sqrt{A_{1}^{2} + A_{2}^{2}}$
$= \sqrt{(10 \sqrt{3})^{2} + (10)^{2}}$
$= \sqrt{300 + 100}$
$A_{resultant} = 20 cm$

UPSEE 2020

WAVES

172173 A travelling wave in a medium is given by the equation $y = a \sin (ω t - k x)$ . The maximum acceleration of the particle in the medium is

1

a ω

2

a ω^{2}

3

\frac{ω}{k}

4

\frac{x}{t}

5

k ω

Explanation:

B Given, $y = a \sin (ω t - kx)$
Velocity, $\frac{dy}{dt} = a \frac{d}{dt} (\sin ω t - kx)$
$v = a ω \cos (ω t - kx)$
Acceleration, $\frac{dv}{dt} = a ω \times \frac{d}{dt} \cos (ω t - kx)$
Acceleration $= - a ω^{2} \sin (ω t - k x)$
Maximum acceleration $= a ω^{2}$

Kerala CEE 2020

WAVES

172175 A progressive wave of frequency $50 Hz$ is travelling with velocity $350 m / s$ through a medium. The change in phase at a given time interval of $0.01 s$ is

1

\frac{3 π}{2} rad

2

\frac{π}{4} rad

3

π rad

4

\frac{π}{2} rad

Explanation:

C Given that,
$Frequency, f = 50 Hz$
$t = 0.01 s$
$∵ T = \frac{1}{f} = \frac{1}{50} = 0.02 s$
So, phase change $ϕ = \frac{2 π t}{T} = \frac{2 π \times 0.01}{0.02}$
$ϕ = π rad$

MHT-CET 2020

WAVES

172171 A wave is represented by $x = 0.4$ $\cos (8 t - \frac{y}{2})$ where $x$ and $y$ are in meters and $t$ in secs. The speed of the wave is

1

0.5 m \cdot s^{- 1}

2

8 m \cdot s^{- 1}

3

16 m \cdot s^{- 1}

4

0.1 m \cdot s^{- 1}

Explanation:

C Equation of the wave given by
$x = 0.4 \cos (8 t - \frac{y}{2})$
Standard equation of wave
$x = A \cos (ω t - ky)$
On comparing
$ω = 8, k = \frac{1}{2}$
Speed $(v) = \frac{ω}{k} = \frac{8}{1 / 2}$
$v = 16 m / \sec$
The speed wave is $16 m / s$

AP EAMCET-07.10.2020

WAVES

172172 The equation of the displacement of waves is $y$ (in $cm) = 10 (\sqrt{3} \sin 2 π t + \cos 2 π t)$
The amplitude of the wave is.

1

10 cm

2

17.3 cm

3

20 cm

4

40 cm

Explanation:

C Given that,
$Y = 10 (\sqrt{3} \sin 2 π t + \cos 2 π t)$
$Y = (10 \sqrt{3} \sin 2 π t + 10 \cos 2 π t)$
The amplitude of the wave-
$A_{resultant} = \sqrt{A_{1}^{2} + A_{2}^{2}}$
$= \sqrt{(10 \sqrt{3})^{2} + (10)^{2}}$
$= \sqrt{300 + 100}$
$A_{resultant} = 20 cm$

UPSEE 2020

WAVES

172173 A travelling wave in a medium is given by the equation $y = a \sin (ω t - k x)$ . The maximum acceleration of the particle in the medium is

1

a ω

2

a ω^{2}

3

\frac{ω}{k}

4

\frac{x}{t}

5

k ω

Explanation:

B Given, $y = a \sin (ω t - kx)$
Velocity, $\frac{dy}{dt} = a \frac{d}{dt} (\sin ω t - kx)$
$v = a ω \cos (ω t - kx)$
Acceleration, $\frac{dv}{dt} = a ω \times \frac{d}{dt} \cos (ω t - kx)$
Acceleration $= - a ω^{2} \sin (ω t - k x)$
Maximum acceleration $= a ω^{2}$

Kerala CEE 2020

WAVES

172175 A progressive wave of frequency $50 Hz$ is travelling with velocity $350 m / s$ through a medium. The change in phase at a given time interval of $0.01 s$ is

1

\frac{3 π}{2} rad

2

\frac{π}{4} rad

3

π rad

4

\frac{π}{2} rad

Explanation:

C Given that,
$Frequency, f = 50 Hz$
$t = 0.01 s$
$∵ T = \frac{1}{f} = \frac{1}{50} = 0.02 s$
So, phase change $ϕ = \frac{2 π t}{T} = \frac{2 π \times 0.01}{0.02}$
$ϕ = π rad$

MHT-CET 2020

WAVES

172171 A wave is represented by $x = 0.4$ $\cos (8 t - \frac{y}{2})$ where $x$ and $y$ are in meters and $t$ in secs. The speed of the wave is

1

0.5 m \cdot s^{- 1}

2

8 m \cdot s^{- 1}

3

16 m \cdot s^{- 1}

4

0.1 m \cdot s^{- 1}

Explanation:

C Equation of the wave given by
$x = 0.4 \cos (8 t - \frac{y}{2})$
Standard equation of wave
$x = A \cos (ω t - ky)$
On comparing
$ω = 8, k = \frac{1}{2}$
Speed $(v) = \frac{ω}{k} = \frac{8}{1 / 2}$
$v = 16 m / \sec$
The speed wave is $16 m / s$

AP EAMCET-07.10.2020

WAVES

172172 The equation of the displacement of waves is $y$ (in $cm) = 10 (\sqrt{3} \sin 2 π t + \cos 2 π t)$
The amplitude of the wave is.

1

10 cm

2

17.3 cm

3

20 cm

4

40 cm

Explanation:

C Given that,
$Y = 10 (\sqrt{3} \sin 2 π t + \cos 2 π t)$
$Y = (10 \sqrt{3} \sin 2 π t + 10 \cos 2 π t)$
The amplitude of the wave-
$A_{resultant} = \sqrt{A_{1}^{2} + A_{2}^{2}}$
$= \sqrt{(10 \sqrt{3})^{2} + (10)^{2}}$
$= \sqrt{300 + 100}$
$A_{resultant} = 20 cm$

UPSEE 2020

WAVES

172173 A travelling wave in a medium is given by the equation $y = a \sin (ω t - k x)$ . The maximum acceleration of the particle in the medium is

1

a ω

2

a ω^{2}

3

\frac{ω}{k}

4

\frac{x}{t}

5

k ω

Explanation:

B Given, $y = a \sin (ω t - kx)$
Velocity, $\frac{dy}{dt} = a \frac{d}{dt} (\sin ω t - kx)$
$v = a ω \cos (ω t - kx)$
Acceleration, $\frac{dv}{dt} = a ω \times \frac{d}{dt} \cos (ω t - kx)$
Acceleration $= - a ω^{2} \sin (ω t - k x)$
Maximum acceleration $= a ω^{2}$

Kerala CEE 2020

WAVES

172175 A progressive wave of frequency $50 Hz$ is travelling with velocity $350 m / s$ through a medium. The change in phase at a given time interval of $0.01 s$ is

1

\frac{3 π}{2} rad

2

\frac{π}{4} rad

3

π rad

4

\frac{π}{2} rad

Explanation:

C Given that,
$Frequency, f = 50 Hz$
$t = 0.01 s$
$∵ T = \frac{1}{f} = \frac{1}{50} = 0.02 s$
So, phase change $ϕ = \frac{2 π t}{T} = \frac{2 π \times 0.01}{0.02}$
$ϕ = π rad$

MHT-CET 2020

WAVES

172171 A wave is represented by $x = 0.4$ $\cos (8 t - \frac{y}{2})$ where $x$ and $y$ are in meters and $t$ in secs. The speed of the wave is

1

0.5 m \cdot s^{- 1}

2

8 m \cdot s^{- 1}

3

16 m \cdot s^{- 1}

4

0.1 m \cdot s^{- 1}

Explanation:

C Equation of the wave given by
$x = 0.4 \cos (8 t - \frac{y}{2})$
Standard equation of wave
$x = A \cos (ω t - ky)$
On comparing
$ω = 8, k = \frac{1}{2}$
Speed $(v) = \frac{ω}{k} = \frac{8}{1 / 2}$
$v = 16 m / \sec$
The speed wave is $16 m / s$

AP EAMCET-07.10.2020

WAVES

172172 The equation of the displacement of waves is $y$ (in $cm) = 10 (\sqrt{3} \sin 2 π t + \cos 2 π t)$
The amplitude of the wave is.

1

10 cm

2

17.3 cm

3

20 cm

4

40 cm

Explanation:

C Given that,
$Y = 10 (\sqrt{3} \sin 2 π t + \cos 2 π t)$
$Y = (10 \sqrt{3} \sin 2 π t + 10 \cos 2 π t)$
The amplitude of the wave-
$A_{resultant} = \sqrt{A_{1}^{2} + A_{2}^{2}}$
$= \sqrt{(10 \sqrt{3})^{2} + (10)^{2}}$
$= \sqrt{300 + 100}$
$A_{resultant} = 20 cm$

UPSEE 2020

WAVES

172173 A travelling wave in a medium is given by the equation $y = a \sin (ω t - k x)$ . The maximum acceleration of the particle in the medium is

1

a ω

2

a ω^{2}

3

\frac{ω}{k}

4

\frac{x}{t}

5

k ω

Explanation:

B Given, $y = a \sin (ω t - kx)$
Velocity, $\frac{dy}{dt} = a \frac{d}{dt} (\sin ω t - kx)$
$v = a ω \cos (ω t - kx)$
Acceleration, $\frac{dv}{dt} = a ω \times \frac{d}{dt} \cos (ω t - kx)$
Acceleration $= - a ω^{2} \sin (ω t - k x)$
Maximum acceleration $= a ω^{2}$

Kerala CEE 2020

WAVES

172175 A progressive wave of frequency $50 Hz$ is travelling with velocity $350 m / s$ through a medium. The change in phase at a given time interval of $0.01 s$ is

1

\frac{3 π}{2} rad

2

\frac{π}{4} rad

3

π rad

4

\frac{π}{2} rad

Explanation:

C Given that,
$Frequency, f = 50 Hz$
$t = 0.01 s$
$∵ T = \frac{1}{f} = \frac{1}{50} = 0.02 s$
So, phase change $ϕ = \frac{2 π t}{T} = \frac{2 π \times 0.01}{0.02}$
$ϕ = π rad$

MHT-CET 2020