QBManager

WAVES

172356 If two wave of intensities I and 4 I superpose, the ratio between maximum and minimum intensities is

1

9 : 1

2

5 : 2

3

4 : 3

4

3 : 1

5

6 : 1

Explanation:

A Given,
Intensities $(I_{1}) = I$ and $(I_{2}) = 4 I$
We know that,
Maximum intensity,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} \Rightarrow (\sqrt{I} + \sqrt{4 I})^{2}$
$I_{max} = I + 4 I + 2 \times 2 I$
$I_{max} = 9 I$
And, minimum intensity $(I_{min}) = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$(I_{min}) = (\sqrt{I} - \sqrt{4 I})^{2}$
$(I_{min}) = I + 4 I - 2 \times 2 I$
$(I_{min}) = I$
Hence, ratio between maximum and minimum intensities is-
$I_{max} : I_{min} = 9 : 1$

Kerala CEE 2020

WAVES

172357 Two progressive waves $Y_{1} = \sin 2 π (\frac{t}{0.4} - \frac{x}{4})$ and $Y_{2} = \sin 2 π (\frac{t}{0.4} + \frac{x}{4})$ superpose to form a standing wave. $x, Y_{1}$ and $Y_{2}$ are in SI system. Amplitude of the particle at $x = 0.5 m$ is $[\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}]$

1

2 \sqrt{2} m

2

2 m

3

\frac{1}{\sqrt{2}} m

4

\sqrt{2} m

Explanation:

D Given waves equation
$Y_{1} = \sin 2 π (\frac{t}{0.4} - \frac{x}{4}), Y_{2} = \sin 2 π (\frac{t}{0.4} + \frac{x}{4})$
We know, standard equation of wave $Y = A \sin (ω t - kx)$
Where, $ω = 2 π n$
$K = \frac{2 π}{λ}$
Then,
$Y = A \sin (2 π nt - \frac{2 π x}{λ})$
And $n = \frac{1}{T}$
So, $Y = A \sin 2 π (\frac{t}{T} - \frac{x}{λ})$
By comparing given waves equation-
$∴ λ = 4 m, T = 0.4 s$
We know that, stationary wave is
$Y = 2 \cos \frac{2 π x}{λ} \cdot \sin \frac{2 π t}{T}$
Amplitude of the particle $A = 2 \cos \frac{2 π x}{λ}$
At $x = 0.5 m$
$A = 2 \cos \frac{2 π \times 0.5}{4}$
Hence, $A = 2 \cos \frac{π}{4} = 2 \cos 45^{\circ} = \frac{2 \times 1}{\sqrt{2}} = \sqrt{2} m$

MHT-CET 2020

WAVES

172358 Two waves given as $y_{1} = 10 \sin ω t cm$ and $y_{2} = 10 \sin (ω t + \frac{π}{3}) cm$ are superimposed. What is the amplitude of the resultant wave? $[\cos \frac{π}{3} = \frac{1}{2}]$

1

10 \sqrt{2} cm

2

5 \sqrt{3} cm

3

10 \sqrt{3} cm

4

10 cm

Explanation:

C Given,
Two waves equations
$y_{1} = 10 \sin ω t, y_{2} = 10 \sin (ω t + \frac{π}{3})$
Where, $A_{1} = A_{2} = 10 cm$
$ϕ_{1} = ω t, ϕ_{2} = ω t + \frac{π}{3}$
Then, phase difference $Δ ϕ = ϕ_{2} - ϕ_{1} = \frac{π}{3}$
So, amplitude of resultant wave $= A$
$A = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos Δ ϕ}$
$A = \sqrt{(10)^{2} + (10)^{2} + 2 \times 10 \times 10 \cos \frac{π}{3}}$
$A = \sqrt{200 + 200 \times \frac{1}{2}} = \sqrt{300} = 10 \sqrt{3} cm$
$A = 10 \sqrt{3} cm$

MHT-CET 2020

WAVES

172359 The two waves are represented by $Y_{1} = 10^{- 2} \sin [50 t + \frac{x}{25} + 0.3] m$ and $Y_{2} = 10^{- 2} \cos [50 t + \frac{x}{25}] m$ where $x$ is in metre and time in second. The phase difference between the two waves is nearly

1

1.22 rad

2

1.05 rad

3

1.15 rad

4

1.27 rad

Explanation:

D Given,
Two waves are-
$Y_{1} = 10^{- 2} \sin [50 t + \frac{x}{25} + 0.3]$
$Y_{2} = 10^{- 2} \cos [50 t + \frac{x}{25}]$
Then, $Y_{2}$ can be written as
$Y_{2} = 10^{- 2} \sin [50 t + \frac{x}{25} + \frac{π}{2}]$
So, $ϕ_{1} = 50 t + \frac{x}{25} + 0.3, ϕ_{2} = 50 t + \frac{x}{25} + \frac{π}{2}$
Phase difference $ϕ = ϕ_{2} - ϕ_{1}$
$ϕ = \frac{π}{2} - 0.3$
$ϕ = \frac{3.14}{2} - 0.3 = 1.57 - 0.3$
Hence, $ϕ = 1.27 rad$

MHT-CET 2020

WAVES

172356 If two wave of intensities I and 4 I superpose, the ratio between maximum and minimum intensities is

1

9 : 1

2

5 : 2

3

4 : 3

4

3 : 1

5

6 : 1

Explanation:

A Given,
Intensities $(I_{1}) = I$ and $(I_{2}) = 4 I$
We know that,
Maximum intensity,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} \Rightarrow (\sqrt{I} + \sqrt{4 I})^{2}$
$I_{max} = I + 4 I + 2 \times 2 I$
$I_{max} = 9 I$
And, minimum intensity $(I_{min}) = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$(I_{min}) = (\sqrt{I} - \sqrt{4 I})^{2}$
$(I_{min}) = I + 4 I - 2 \times 2 I$
$(I_{min}) = I$
Hence, ratio between maximum and minimum intensities is-
$I_{max} : I_{min} = 9 : 1$

Kerala CEE 2020

WAVES

172357 Two progressive waves $Y_{1} = \sin 2 π (\frac{t}{0.4} - \frac{x}{4})$ and $Y_{2} = \sin 2 π (\frac{t}{0.4} + \frac{x}{4})$ superpose to form a standing wave. $x, Y_{1}$ and $Y_{2}$ are in SI system. Amplitude of the particle at $x = 0.5 m$ is $[\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}]$

1

2 \sqrt{2} m

2

2 m

3

\frac{1}{\sqrt{2}} m

4

\sqrt{2} m

Explanation:

D Given waves equation
$Y_{1} = \sin 2 π (\frac{t}{0.4} - \frac{x}{4}), Y_{2} = \sin 2 π (\frac{t}{0.4} + \frac{x}{4})$
We know, standard equation of wave $Y = A \sin (ω t - kx)$
Where, $ω = 2 π n$
$K = \frac{2 π}{λ}$
Then,
$Y = A \sin (2 π nt - \frac{2 π x}{λ})$
And $n = \frac{1}{T}$
So, $Y = A \sin 2 π (\frac{t}{T} - \frac{x}{λ})$
By comparing given waves equation-
$∴ λ = 4 m, T = 0.4 s$
We know that, stationary wave is
$Y = 2 \cos \frac{2 π x}{λ} \cdot \sin \frac{2 π t}{T}$
Amplitude of the particle $A = 2 \cos \frac{2 π x}{λ}$
At $x = 0.5 m$
$A = 2 \cos \frac{2 π \times 0.5}{4}$
Hence, $A = 2 \cos \frac{π}{4} = 2 \cos 45^{\circ} = \frac{2 \times 1}{\sqrt{2}} = \sqrt{2} m$

MHT-CET 2020

WAVES

172358 Two waves given as $y_{1} = 10 \sin ω t cm$ and $y_{2} = 10 \sin (ω t + \frac{π}{3}) cm$ are superimposed. What is the amplitude of the resultant wave? $[\cos \frac{π}{3} = \frac{1}{2}]$

1

10 \sqrt{2} cm

2

5 \sqrt{3} cm

3

10 \sqrt{3} cm

4

10 cm

Explanation:

C Given,
Two waves equations
$y_{1} = 10 \sin ω t, y_{2} = 10 \sin (ω t + \frac{π}{3})$
Where, $A_{1} = A_{2} = 10 cm$
$ϕ_{1} = ω t, ϕ_{2} = ω t + \frac{π}{3}$
Then, phase difference $Δ ϕ = ϕ_{2} - ϕ_{1} = \frac{π}{3}$
So, amplitude of resultant wave $= A$
$A = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos Δ ϕ}$
$A = \sqrt{(10)^{2} + (10)^{2} + 2 \times 10 \times 10 \cos \frac{π}{3}}$
$A = \sqrt{200 + 200 \times \frac{1}{2}} = \sqrt{300} = 10 \sqrt{3} cm$
$A = 10 \sqrt{3} cm$

MHT-CET 2020

WAVES

172359 The two waves are represented by $Y_{1} = 10^{- 2} \sin [50 t + \frac{x}{25} + 0.3] m$ and $Y_{2} = 10^{- 2} \cos [50 t + \frac{x}{25}] m$ where $x$ is in metre and time in second. The phase difference between the two waves is nearly

1

1.22 rad

2

1.05 rad

3

1.15 rad

4

1.27 rad

Explanation:

D Given,
Two waves are-
$Y_{1} = 10^{- 2} \sin [50 t + \frac{x}{25} + 0.3]$
$Y_{2} = 10^{- 2} \cos [50 t + \frac{x}{25}]$
Then, $Y_{2}$ can be written as
$Y_{2} = 10^{- 2} \sin [50 t + \frac{x}{25} + \frac{π}{2}]$
So, $ϕ_{1} = 50 t + \frac{x}{25} + 0.3, ϕ_{2} = 50 t + \frac{x}{25} + \frac{π}{2}$
Phase difference $ϕ = ϕ_{2} - ϕ_{1}$
$ϕ = \frac{π}{2} - 0.3$
$ϕ = \frac{3.14}{2} - 0.3 = 1.57 - 0.3$
Hence, $ϕ = 1.27 rad$

MHT-CET 2020

WAVES

172356 If two wave of intensities I and 4 I superpose, the ratio between maximum and minimum intensities is

1

9 : 1

2

5 : 2

3

4 : 3

4

3 : 1

5

6 : 1

Explanation:

A Given,
Intensities $(I_{1}) = I$ and $(I_{2}) = 4 I$
We know that,
Maximum intensity,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} \Rightarrow (\sqrt{I} + \sqrt{4 I})^{2}$
$I_{max} = I + 4 I + 2 \times 2 I$
$I_{max} = 9 I$
And, minimum intensity $(I_{min}) = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$(I_{min}) = (\sqrt{I} - \sqrt{4 I})^{2}$
$(I_{min}) = I + 4 I - 2 \times 2 I$
$(I_{min}) = I$
Hence, ratio between maximum and minimum intensities is-
$I_{max} : I_{min} = 9 : 1$

Kerala CEE 2020

WAVES

172357 Two progressive waves $Y_{1} = \sin 2 π (\frac{t}{0.4} - \frac{x}{4})$ and $Y_{2} = \sin 2 π (\frac{t}{0.4} + \frac{x}{4})$ superpose to form a standing wave. $x, Y_{1}$ and $Y_{2}$ are in SI system. Amplitude of the particle at $x = 0.5 m$ is $[\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}]$

1

2 \sqrt{2} m

2

2 m

3

\frac{1}{\sqrt{2}} m

4

\sqrt{2} m

Explanation:

D Given waves equation
$Y_{1} = \sin 2 π (\frac{t}{0.4} - \frac{x}{4}), Y_{2} = \sin 2 π (\frac{t}{0.4} + \frac{x}{4})$
We know, standard equation of wave $Y = A \sin (ω t - kx)$
Where, $ω = 2 π n$
$K = \frac{2 π}{λ}$
Then,
$Y = A \sin (2 π nt - \frac{2 π x}{λ})$
And $n = \frac{1}{T}$
So, $Y = A \sin 2 π (\frac{t}{T} - \frac{x}{λ})$
By comparing given waves equation-
$∴ λ = 4 m, T = 0.4 s$
We know that, stationary wave is
$Y = 2 \cos \frac{2 π x}{λ} \cdot \sin \frac{2 π t}{T}$
Amplitude of the particle $A = 2 \cos \frac{2 π x}{λ}$
At $x = 0.5 m$
$A = 2 \cos \frac{2 π \times 0.5}{4}$
Hence, $A = 2 \cos \frac{π}{4} = 2 \cos 45^{\circ} = \frac{2 \times 1}{\sqrt{2}} = \sqrt{2} m$

MHT-CET 2020

WAVES

172358 Two waves given as $y_{1} = 10 \sin ω t cm$ and $y_{2} = 10 \sin (ω t + \frac{π}{3}) cm$ are superimposed. What is the amplitude of the resultant wave? $[\cos \frac{π}{3} = \frac{1}{2}]$

1

10 \sqrt{2} cm

2

5 \sqrt{3} cm

3

10 \sqrt{3} cm

4

10 cm

Explanation:

C Given,
Two waves equations
$y_{1} = 10 \sin ω t, y_{2} = 10 \sin (ω t + \frac{π}{3})$
Where, $A_{1} = A_{2} = 10 cm$
$ϕ_{1} = ω t, ϕ_{2} = ω t + \frac{π}{3}$
Then, phase difference $Δ ϕ = ϕ_{2} - ϕ_{1} = \frac{π}{3}$
So, amplitude of resultant wave $= A$
$A = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos Δ ϕ}$
$A = \sqrt{(10)^{2} + (10)^{2} + 2 \times 10 \times 10 \cos \frac{π}{3}}$
$A = \sqrt{200 + 200 \times \frac{1}{2}} = \sqrt{300} = 10 \sqrt{3} cm$
$A = 10 \sqrt{3} cm$

MHT-CET 2020

WAVES

172359 The two waves are represented by $Y_{1} = 10^{- 2} \sin [50 t + \frac{x}{25} + 0.3] m$ and $Y_{2} = 10^{- 2} \cos [50 t + \frac{x}{25}] m$ where $x$ is in metre and time in second. The phase difference between the two waves is nearly

1

1.22 rad

2

1.05 rad

3

1.15 rad

4

1.27 rad

Explanation:

D Given,
Two waves are-
$Y_{1} = 10^{- 2} \sin [50 t + \frac{x}{25} + 0.3]$
$Y_{2} = 10^{- 2} \cos [50 t + \frac{x}{25}]$
Then, $Y_{2}$ can be written as
$Y_{2} = 10^{- 2} \sin [50 t + \frac{x}{25} + \frac{π}{2}]$
So, $ϕ_{1} = 50 t + \frac{x}{25} + 0.3, ϕ_{2} = 50 t + \frac{x}{25} + \frac{π}{2}$
Phase difference $ϕ = ϕ_{2} - ϕ_{1}$
$ϕ = \frac{π}{2} - 0.3$
$ϕ = \frac{3.14}{2} - 0.3 = 1.57 - 0.3$
Hence, $ϕ = 1.27 rad$

MHT-CET 2020

WAVES

172356 If two wave of intensities I and 4 I superpose, the ratio between maximum and minimum intensities is

1

9 : 1

2

5 : 2

3

4 : 3

4

3 : 1

5

6 : 1

Explanation:

A Given,
Intensities $(I_{1}) = I$ and $(I_{2}) = 4 I$
We know that,
Maximum intensity,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} \Rightarrow (\sqrt{I} + \sqrt{4 I})^{2}$
$I_{max} = I + 4 I + 2 \times 2 I$
$I_{max} = 9 I$
And, minimum intensity $(I_{min}) = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$(I_{min}) = (\sqrt{I} - \sqrt{4 I})^{2}$
$(I_{min}) = I + 4 I - 2 \times 2 I$
$(I_{min}) = I$
Hence, ratio between maximum and minimum intensities is-
$I_{max} : I_{min} = 9 : 1$

Kerala CEE 2020

WAVES

172357 Two progressive waves $Y_{1} = \sin 2 π (\frac{t}{0.4} - \frac{x}{4})$ and $Y_{2} = \sin 2 π (\frac{t}{0.4} + \frac{x}{4})$ superpose to form a standing wave. $x, Y_{1}$ and $Y_{2}$ are in SI system. Amplitude of the particle at $x = 0.5 m$ is $[\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}]$

1

2 \sqrt{2} m

2

2 m

3

\frac{1}{\sqrt{2}} m

4

\sqrt{2} m

Explanation:

D Given waves equation
$Y_{1} = \sin 2 π (\frac{t}{0.4} - \frac{x}{4}), Y_{2} = \sin 2 π (\frac{t}{0.4} + \frac{x}{4})$
We know, standard equation of wave $Y = A \sin (ω t - kx)$
Where, $ω = 2 π n$
$K = \frac{2 π}{λ}$
Then,
$Y = A \sin (2 π nt - \frac{2 π x}{λ})$
And $n = \frac{1}{T}$
So, $Y = A \sin 2 π (\frac{t}{T} - \frac{x}{λ})$
By comparing given waves equation-
$∴ λ = 4 m, T = 0.4 s$
We know that, stationary wave is
$Y = 2 \cos \frac{2 π x}{λ} \cdot \sin \frac{2 π t}{T}$
Amplitude of the particle $A = 2 \cos \frac{2 π x}{λ}$
At $x = 0.5 m$
$A = 2 \cos \frac{2 π \times 0.5}{4}$
Hence, $A = 2 \cos \frac{π}{4} = 2 \cos 45^{\circ} = \frac{2 \times 1}{\sqrt{2}} = \sqrt{2} m$

MHT-CET 2020

WAVES

172358 Two waves given as $y_{1} = 10 \sin ω t cm$ and $y_{2} = 10 \sin (ω t + \frac{π}{3}) cm$ are superimposed. What is the amplitude of the resultant wave? $[\cos \frac{π}{3} = \frac{1}{2}]$

1

10 \sqrt{2} cm

2

5 \sqrt{3} cm

3

10 \sqrt{3} cm

4

10 cm

Explanation:

C Given,
Two waves equations
$y_{1} = 10 \sin ω t, y_{2} = 10 \sin (ω t + \frac{π}{3})$
Where, $A_{1} = A_{2} = 10 cm$
$ϕ_{1} = ω t, ϕ_{2} = ω t + \frac{π}{3}$
Then, phase difference $Δ ϕ = ϕ_{2} - ϕ_{1} = \frac{π}{3}$
So, amplitude of resultant wave $= A$
$A = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos Δ ϕ}$
$A = \sqrt{(10)^{2} + (10)^{2} + 2 \times 10 \times 10 \cos \frac{π}{3}}$
$A = \sqrt{200 + 200 \times \frac{1}{2}} = \sqrt{300} = 10 \sqrt{3} cm$
$A = 10 \sqrt{3} cm$

MHT-CET 2020

WAVES

172359 The two waves are represented by $Y_{1} = 10^{- 2} \sin [50 t + \frac{x}{25} + 0.3] m$ and $Y_{2} = 10^{- 2} \cos [50 t + \frac{x}{25}] m$ where $x$ is in metre and time in second. The phase difference between the two waves is nearly

1

1.22 rad

2

1.05 rad

3

1.15 rad

4

1.27 rad

Explanation:

D Given,
Two waves are-
$Y_{1} = 10^{- 2} \sin [50 t + \frac{x}{25} + 0.3]$
$Y_{2} = 10^{- 2} \cos [50 t + \frac{x}{25}]$
Then, $Y_{2}$ can be written as
$Y_{2} = 10^{- 2} \sin [50 t + \frac{x}{25} + \frac{π}{2}]$
So, $ϕ_{1} = 50 t + \frac{x}{25} + 0.3, ϕ_{2} = 50 t + \frac{x}{25} + \frac{π}{2}$
Phase difference $ϕ = ϕ_{2} - ϕ_{1}$
$ϕ = \frac{π}{2} - 0.3$
$ϕ = \frac{3.14}{2} - 0.3 = 1.57 - 0.3$
Hence, $ϕ = 1.27 rad$

MHT-CET 2020