QBManager

WAVES

172364 Two harmonic travelling waves are described by the equations $y_{1} = a \sin (k x - ω t)$ and $y_{2} = a$ $\sin (- k x + ω t + ϕ)$ . The amplitude of the superimposed wave is

1

2 acos \frac{ϕ}{2}

2

2 asin ϕ

3

2 acos ϕ

4

2 asin \frac{ϕ}{2}

Explanation:

D Given,
$y_{1} = a \sin (kx - ω t)$ and $y_{2} = a \sin (- k x + ω t + ϕ)$
Let, $k x - ω t = α$
$y_{1} = a \sin α, y_{2} = a \sin (α - ϕ)$
Superimposed wave, $y = y_{1} + y_{2}$
$y = [a \sin α + a \sin ϕ \cos α - a \cos ϕ \sin α]$
$y = [a (1 - \cos ϕ) \sin α + (a \cos ϕ) \cos α]$
Let $R \cos θ = a (1 - \cos ϕ)$ and $R \sin θ = a \sin ϕ$ .
Then, magnitude of $R$ will be the magnitude of resultant wave,
$R^{2} = a^{2} [\sin^{2} ϕ + (1 - \cos ϕ)^{2}]$
$R^{2} = a^{2} [1 + 1 - 2 \cos ϕ]$
$R^{2} = 4 a^{2} \sin^{2} \frac{ϕ}{2}$
Hence, $R = 2 a \sin \frac{ϕ}{2}$

TS- EAMCET-04.05.2018

WAVES

172365 Two periodic waves of intensities $I_{1}$ and $I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

1

I_{1} + I_{2}

2

{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}

3

{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}

4

2 (I_{1} + I_{2})

Explanation:

D Resultant intensity of two periodic waves is given by
$I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2} \cos ϕ}$
Where $ϕ$ is the phase difference between two waves for maximum intensity
$ϕ = 2 π n (n = 0, 1, 2 \dots \dots \dots)$
For maxima
$\cos ϕ = 1$
$I_{max} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}}$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$I_{min} when \cos 180^{\circ} = - 1$
$I_{min} = I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}}$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
So, the sum of the maximum and minimum intensity
$I_{max} + I_{min} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} + {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$I_{max} + I_{min} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} + I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}}$
$I_{max} + I_{min} = 2 (I_{1} + I_{2})$

AIPMT-2008

WAVES

172366 Two waves represented by $y = a \sin (ω t - k x)$ and $y = a \cos (ω t - k x)$ are superimposed. The resultant wave will have an amplitude

1 a

2

\sqrt{2} a

3

2 a

4 zero

Explanation:

B Given,
$y_{1} = a \sin (ω t - k x)$
$y_{2} = a \cos (ω t - k x)$
Then, $y = y_{1} + y_{2}$
$y = a \sin (ω t - k x) + \cos (ω t - k x)$
$y = a [\sin (ω t - k x) + \sin {\frac{π}{2} - (ω t - k x)}]$
$= a [2 \sin {\frac{ω t - k x + \frac{π}{2} - (ω t - k x)}{2}} \cos {\frac{ω t - k x - \frac{π}{2} + (ω t - k x)}{2}}]$
$= a [2 \sin \frac{π}{4} \times \cos \frac{{2 (ω t - kx) - \frac{π}{2}}}{2}]$
$= 2 a \times \frac{1}{\sqrt{2}} \cos [\frac{2 (ω t - kx) - \frac{π}{2}}{2}]$
$y = \sqrt{2} a \cos [(ω t - k x) - \frac{π}{4}]$
Hence, the resultant wave will have an amplitude $= \sqrt{2} a$

UPSEE - 2008

WAVES

172367 Three sinusoidal waves of the same frequency travel along a string in the positive $x$ -direction. Their amplitudes are $y, y / 2$ and $y / 3$ and their phase constants are $0, π / 2$ and $π$ respectively. What is the amplitude of the resultant wave?

1

0.63 y

2

0.72 y

3

0.83 y

4

0.52 y

Explanation:

C Equation for sinusoidal waves
$Y = A \sin (ω t + ϕ)$
Given that, $A_{1} = y, A_{2} = y / 2 A_{3} = y / 3$
$and ϕ_{1} = 0, ϕ_{2} = π / 2, ϕ_{3} = π$
$ω = Constant$
Hence three sinusoidal waves are given by
$Y_{1} = y \sin (ω t + 0)$
$Y_{2} = y / 2 \sin (ω t + π / 2)$
$Y_{3} = y / 3 \sin (ω t + π)$
According to superposition principle
$Y = Y_{1} + Y_{2} + Y_{3}$
$= y \sin ω t + \frac{y}{2} \sin (ω t + \frac{π}{2}) + \frac{y}{3} \sin (ω t + π)$
$= y \sin ω t + \frac{y}{2} \cos ω t + \frac{y}{3} (- \sin ω t)$
$= y \sin ω t - \frac{y}{3} \sin ω t + \frac{y}{2} \cos ω t$
$= \frac{2 y}{3} \sin ω t + \frac{y}{2} \cos ω t$
$= y (\frac{2}{3} \sin ω t + \frac{1}{2} \cos ω t)$
$= \frac{y}{6} (4 \sin ω t + 3 \cos ω t)$
$y = \frac{4 y}{6} \sin ω t + \frac{3 y}{6} \cos ω t$
$A_{1} = \frac{4 y}{6}, A_{2} = \frac{3 y}{6}$
$∴ A_{1} = \frac{2 y}{3}, A_{2} = \frac{y}{2}$
Net amplitude $A = \sqrt{A_{1}^{2} + A_{2}^{2}}$
$= \sqrt{{(\frac{2 y}{3})}^{2} + {(\frac{y}{2})}^{2}}$
$= y \sqrt{\frac{4}{9} + \frac{1}{4}}$
$= y \sqrt{\frac{25}{36}}$
$= \frac{5 y}{6}$
$= 0.83 y$

AMU-2011

WAVES

172368 Two waves $y_{1} = 2 \sin ω t$ and $y_{2} = 4 \sin (ω t + δ)$ superimpose. The ratio of the maximum to the minimum intensity of the resultant wave is

1 9

2 3

3 infinity

4 zero

Explanation:

A Given that,
$y_{1} = 2 \sin ω t$
$y_{2} = 4 \sin (ω t + δ)$
Comparing the resultant wave equation,
$y_{1} = A_{1} \sin ω t$
$y_{2} = A_{2} \sin (ω t + ϕ)$
Then, $A_{1} = 2, A_{2} = 4$
So, $\frac{I_{max}}{I_{min}} = \frac{{(A_{1} + A_{2})}^{2}}{{(A_{1} - A_{2})}^{2}}$
$\frac{I_{max}}{I_{min}} = \frac{(2 + 4)^{2}}{(2 - 4)^{2}}$
$\frac{I_{max}}{I_{min}} = \frac{36}{4} = 9$

UPSEE - 2011

WAVES

172364 Two harmonic travelling waves are described by the equations $y_{1} = a \sin (k x - ω t)$ and $y_{2} = a$ $\sin (- k x + ω t + ϕ)$ . The amplitude of the superimposed wave is

1

2 acos \frac{ϕ}{2}

2

2 asin ϕ

3

2 acos ϕ

4

2 asin \frac{ϕ}{2}

Explanation:

D Given,
$y_{1} = a \sin (kx - ω t)$ and $y_{2} = a \sin (- k x + ω t + ϕ)$
Let, $k x - ω t = α$
$y_{1} = a \sin α, y_{2} = a \sin (α - ϕ)$
Superimposed wave, $y = y_{1} + y_{2}$
$y = [a \sin α + a \sin ϕ \cos α - a \cos ϕ \sin α]$
$y = [a (1 - \cos ϕ) \sin α + (a \cos ϕ) \cos α]$
Let $R \cos θ = a (1 - \cos ϕ)$ and $R \sin θ = a \sin ϕ$ .
Then, magnitude of $R$ will be the magnitude of resultant wave,
$R^{2} = a^{2} [\sin^{2} ϕ + (1 - \cos ϕ)^{2}]$
$R^{2} = a^{2} [1 + 1 - 2 \cos ϕ]$
$R^{2} = 4 a^{2} \sin^{2} \frac{ϕ}{2}$
Hence, $R = 2 a \sin \frac{ϕ}{2}$

TS- EAMCET-04.05.2018

WAVES

172365 Two periodic waves of intensities $I_{1}$ and $I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

1

I_{1} + I_{2}

2

{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}

3

{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}

4

2 (I_{1} + I_{2})

Explanation:

D Resultant intensity of two periodic waves is given by
$I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2} \cos ϕ}$
Where $ϕ$ is the phase difference between two waves for maximum intensity
$ϕ = 2 π n (n = 0, 1, 2 \dots \dots \dots)$
For maxima
$\cos ϕ = 1$
$I_{max} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}}$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$I_{min} when \cos 180^{\circ} = - 1$
$I_{min} = I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}}$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
So, the sum of the maximum and minimum intensity
$I_{max} + I_{min} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} + {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$I_{max} + I_{min} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} + I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}}$
$I_{max} + I_{min} = 2 (I_{1} + I_{2})$

AIPMT-2008

WAVES

172366 Two waves represented by $y = a \sin (ω t - k x)$ and $y = a \cos (ω t - k x)$ are superimposed. The resultant wave will have an amplitude

1 a

2

\sqrt{2} a

3

2 a

4 zero

Explanation:

B Given,
$y_{1} = a \sin (ω t - k x)$
$y_{2} = a \cos (ω t - k x)$
Then, $y = y_{1} + y_{2}$
$y = a \sin (ω t - k x) + \cos (ω t - k x)$
$y = a [\sin (ω t - k x) + \sin {\frac{π}{2} - (ω t - k x)}]$
$= a [2 \sin {\frac{ω t - k x + \frac{π}{2} - (ω t - k x)}{2}} \cos {\frac{ω t - k x - \frac{π}{2} + (ω t - k x)}{2}}]$
$= a [2 \sin \frac{π}{4} \times \cos \frac{{2 (ω t - kx) - \frac{π}{2}}}{2}]$
$= 2 a \times \frac{1}{\sqrt{2}} \cos [\frac{2 (ω t - kx) - \frac{π}{2}}{2}]$
$y = \sqrt{2} a \cos [(ω t - k x) - \frac{π}{4}]$
Hence, the resultant wave will have an amplitude $= \sqrt{2} a$

UPSEE - 2008

WAVES

172367 Three sinusoidal waves of the same frequency travel along a string in the positive $x$ -direction. Their amplitudes are $y, y / 2$ and $y / 3$ and their phase constants are $0, π / 2$ and $π$ respectively. What is the amplitude of the resultant wave?

1

0.63 y

2

0.72 y

3

0.83 y

4

0.52 y

Explanation:

C Equation for sinusoidal waves
$Y = A \sin (ω t + ϕ)$
Given that, $A_{1} = y, A_{2} = y / 2 A_{3} = y / 3$
$and ϕ_{1} = 0, ϕ_{2} = π / 2, ϕ_{3} = π$
$ω = Constant$
Hence three sinusoidal waves are given by
$Y_{1} = y \sin (ω t + 0)$
$Y_{2} = y / 2 \sin (ω t + π / 2)$
$Y_{3} = y / 3 \sin (ω t + π)$
According to superposition principle
$Y = Y_{1} + Y_{2} + Y_{3}$
$= y \sin ω t + \frac{y}{2} \sin (ω t + \frac{π}{2}) + \frac{y}{3} \sin (ω t + π)$
$= y \sin ω t + \frac{y}{2} \cos ω t + \frac{y}{3} (- \sin ω t)$
$= y \sin ω t - \frac{y}{3} \sin ω t + \frac{y}{2} \cos ω t$
$= \frac{2 y}{3} \sin ω t + \frac{y}{2} \cos ω t$
$= y (\frac{2}{3} \sin ω t + \frac{1}{2} \cos ω t)$
$= \frac{y}{6} (4 \sin ω t + 3 \cos ω t)$
$y = \frac{4 y}{6} \sin ω t + \frac{3 y}{6} \cos ω t$
$A_{1} = \frac{4 y}{6}, A_{2} = \frac{3 y}{6}$
$∴ A_{1} = \frac{2 y}{3}, A_{2} = \frac{y}{2}$
Net amplitude $A = \sqrt{A_{1}^{2} + A_{2}^{2}}$
$= \sqrt{{(\frac{2 y}{3})}^{2} + {(\frac{y}{2})}^{2}}$
$= y \sqrt{\frac{4}{9} + \frac{1}{4}}$
$= y \sqrt{\frac{25}{36}}$
$= \frac{5 y}{6}$
$= 0.83 y$

AMU-2011

WAVES

172368 Two waves $y_{1} = 2 \sin ω t$ and $y_{2} = 4 \sin (ω t + δ)$ superimpose. The ratio of the maximum to the minimum intensity of the resultant wave is

1 9

2 3

3 infinity

4 zero

Explanation:

A Given that,
$y_{1} = 2 \sin ω t$
$y_{2} = 4 \sin (ω t + δ)$
Comparing the resultant wave equation,
$y_{1} = A_{1} \sin ω t$
$y_{2} = A_{2} \sin (ω t + ϕ)$
Then, $A_{1} = 2, A_{2} = 4$
So, $\frac{I_{max}}{I_{min}} = \frac{{(A_{1} + A_{2})}^{2}}{{(A_{1} - A_{2})}^{2}}$
$\frac{I_{max}}{I_{min}} = \frac{(2 + 4)^{2}}{(2 - 4)^{2}}$
$\frac{I_{max}}{I_{min}} = \frac{36}{4} = 9$

UPSEE - 2011

WAVES

172364 Two harmonic travelling waves are described by the equations $y_{1} = a \sin (k x - ω t)$ and $y_{2} = a$ $\sin (- k x + ω t + ϕ)$ . The amplitude of the superimposed wave is

1

2 acos \frac{ϕ}{2}

2

2 asin ϕ

3

2 acos ϕ

4

2 asin \frac{ϕ}{2}

Explanation:

D Given,
$y_{1} = a \sin (kx - ω t)$ and $y_{2} = a \sin (- k x + ω t + ϕ)$
Let, $k x - ω t = α$
$y_{1} = a \sin α, y_{2} = a \sin (α - ϕ)$
Superimposed wave, $y = y_{1} + y_{2}$
$y = [a \sin α + a \sin ϕ \cos α - a \cos ϕ \sin α]$
$y = [a (1 - \cos ϕ) \sin α + (a \cos ϕ) \cos α]$
Let $R \cos θ = a (1 - \cos ϕ)$ and $R \sin θ = a \sin ϕ$ .
Then, magnitude of $R$ will be the magnitude of resultant wave,
$R^{2} = a^{2} [\sin^{2} ϕ + (1 - \cos ϕ)^{2}]$
$R^{2} = a^{2} [1 + 1 - 2 \cos ϕ]$
$R^{2} = 4 a^{2} \sin^{2} \frac{ϕ}{2}$
Hence, $R = 2 a \sin \frac{ϕ}{2}$

TS- EAMCET-04.05.2018

WAVES

172365 Two periodic waves of intensities $I_{1}$ and $I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

1

I_{1} + I_{2}

2

{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}

3

{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}

4

2 (I_{1} + I_{2})

Explanation:

D Resultant intensity of two periodic waves is given by
$I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2} \cos ϕ}$
Where $ϕ$ is the phase difference between two waves for maximum intensity
$ϕ = 2 π n (n = 0, 1, 2 \dots \dots \dots)$
For maxima
$\cos ϕ = 1$
$I_{max} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}}$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$I_{min} when \cos 180^{\circ} = - 1$
$I_{min} = I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}}$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
So, the sum of the maximum and minimum intensity
$I_{max} + I_{min} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} + {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$I_{max} + I_{min} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} + I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}}$
$I_{max} + I_{min} = 2 (I_{1} + I_{2})$

AIPMT-2008

WAVES

172366 Two waves represented by $y = a \sin (ω t - k x)$ and $y = a \cos (ω t - k x)$ are superimposed. The resultant wave will have an amplitude

1 a

2

\sqrt{2} a

3

2 a

4 zero

Explanation:

B Given,
$y_{1} = a \sin (ω t - k x)$
$y_{2} = a \cos (ω t - k x)$
Then, $y = y_{1} + y_{2}$
$y = a \sin (ω t - k x) + \cos (ω t - k x)$
$y = a [\sin (ω t - k x) + \sin {\frac{π}{2} - (ω t - k x)}]$
$= a [2 \sin {\frac{ω t - k x + \frac{π}{2} - (ω t - k x)}{2}} \cos {\frac{ω t - k x - \frac{π}{2} + (ω t - k x)}{2}}]$
$= a [2 \sin \frac{π}{4} \times \cos \frac{{2 (ω t - kx) - \frac{π}{2}}}{2}]$
$= 2 a \times \frac{1}{\sqrt{2}} \cos [\frac{2 (ω t - kx) - \frac{π}{2}}{2}]$
$y = \sqrt{2} a \cos [(ω t - k x) - \frac{π}{4}]$
Hence, the resultant wave will have an amplitude $= \sqrt{2} a$

UPSEE - 2008

WAVES

172367 Three sinusoidal waves of the same frequency travel along a string in the positive $x$ -direction. Their amplitudes are $y, y / 2$ and $y / 3$ and their phase constants are $0, π / 2$ and $π$ respectively. What is the amplitude of the resultant wave?

1

0.63 y

2

0.72 y

3

0.83 y

4

0.52 y

Explanation:

C Equation for sinusoidal waves
$Y = A \sin (ω t + ϕ)$
Given that, $A_{1} = y, A_{2} = y / 2 A_{3} = y / 3$
$and ϕ_{1} = 0, ϕ_{2} = π / 2, ϕ_{3} = π$
$ω = Constant$
Hence three sinusoidal waves are given by
$Y_{1} = y \sin (ω t + 0)$
$Y_{2} = y / 2 \sin (ω t + π / 2)$
$Y_{3} = y / 3 \sin (ω t + π)$
According to superposition principle
$Y = Y_{1} + Y_{2} + Y_{3}$
$= y \sin ω t + \frac{y}{2} \sin (ω t + \frac{π}{2}) + \frac{y}{3} \sin (ω t + π)$
$= y \sin ω t + \frac{y}{2} \cos ω t + \frac{y}{3} (- \sin ω t)$
$= y \sin ω t - \frac{y}{3} \sin ω t + \frac{y}{2} \cos ω t$
$= \frac{2 y}{3} \sin ω t + \frac{y}{2} \cos ω t$
$= y (\frac{2}{3} \sin ω t + \frac{1}{2} \cos ω t)$
$= \frac{y}{6} (4 \sin ω t + 3 \cos ω t)$
$y = \frac{4 y}{6} \sin ω t + \frac{3 y}{6} \cos ω t$
$A_{1} = \frac{4 y}{6}, A_{2} = \frac{3 y}{6}$
$∴ A_{1} = \frac{2 y}{3}, A_{2} = \frac{y}{2}$
Net amplitude $A = \sqrt{A_{1}^{2} + A_{2}^{2}}$
$= \sqrt{{(\frac{2 y}{3})}^{2} + {(\frac{y}{2})}^{2}}$
$= y \sqrt{\frac{4}{9} + \frac{1}{4}}$
$= y \sqrt{\frac{25}{36}}$
$= \frac{5 y}{6}$
$= 0.83 y$

AMU-2011

WAVES

172368 Two waves $y_{1} = 2 \sin ω t$ and $y_{2} = 4 \sin (ω t + δ)$ superimpose. The ratio of the maximum to the minimum intensity of the resultant wave is

1 9

2 3

3 infinity

4 zero

Explanation:

A Given that,
$y_{1} = 2 \sin ω t$
$y_{2} = 4 \sin (ω t + δ)$
Comparing the resultant wave equation,
$y_{1} = A_{1} \sin ω t$
$y_{2} = A_{2} \sin (ω t + ϕ)$
Then, $A_{1} = 2, A_{2} = 4$
So, $\frac{I_{max}}{I_{min}} = \frac{{(A_{1} + A_{2})}^{2}}{{(A_{1} - A_{2})}^{2}}$
$\frac{I_{max}}{I_{min}} = \frac{(2 + 4)^{2}}{(2 - 4)^{2}}$
$\frac{I_{max}}{I_{min}} = \frac{36}{4} = 9$

UPSEE - 2011

WAVES

172364 Two harmonic travelling waves are described by the equations $y_{1} = a \sin (k x - ω t)$ and $y_{2} = a$ $\sin (- k x + ω t + ϕ)$ . The amplitude of the superimposed wave is

1

2 acos \frac{ϕ}{2}

2

2 asin ϕ

3

2 acos ϕ

4

2 asin \frac{ϕ}{2}

Explanation:

D Given,
$y_{1} = a \sin (kx - ω t)$ and $y_{2} = a \sin (- k x + ω t + ϕ)$
Let, $k x - ω t = α$
$y_{1} = a \sin α, y_{2} = a \sin (α - ϕ)$
Superimposed wave, $y = y_{1} + y_{2}$
$y = [a \sin α + a \sin ϕ \cos α - a \cos ϕ \sin α]$
$y = [a (1 - \cos ϕ) \sin α + (a \cos ϕ) \cos α]$
Let $R \cos θ = a (1 - \cos ϕ)$ and $R \sin θ = a \sin ϕ$ .
Then, magnitude of $R$ will be the magnitude of resultant wave,
$R^{2} = a^{2} [\sin^{2} ϕ + (1 - \cos ϕ)^{2}]$
$R^{2} = a^{2} [1 + 1 - 2 \cos ϕ]$
$R^{2} = 4 a^{2} \sin^{2} \frac{ϕ}{2}$
Hence, $R = 2 a \sin \frac{ϕ}{2}$

TS- EAMCET-04.05.2018

WAVES

172365 Two periodic waves of intensities $I_{1}$ and $I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

1

I_{1} + I_{2}

2

{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}

3

{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}

4

2 (I_{1} + I_{2})

Explanation:

D Resultant intensity of two periodic waves is given by
$I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2} \cos ϕ}$
Where $ϕ$ is the phase difference between two waves for maximum intensity
$ϕ = 2 π n (n = 0, 1, 2 \dots \dots \dots)$
For maxima
$\cos ϕ = 1$
$I_{max} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}}$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$I_{min} when \cos 180^{\circ} = - 1$
$I_{min} = I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}}$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
So, the sum of the maximum and minimum intensity
$I_{max} + I_{min} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} + {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$I_{max} + I_{min} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} + I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}}$
$I_{max} + I_{min} = 2 (I_{1} + I_{2})$

AIPMT-2008

WAVES

172366 Two waves represented by $y = a \sin (ω t - k x)$ and $y = a \cos (ω t - k x)$ are superimposed. The resultant wave will have an amplitude

1 a

2

\sqrt{2} a

3

2 a

4 zero

Explanation:

B Given,
$y_{1} = a \sin (ω t - k x)$
$y_{2} = a \cos (ω t - k x)$
Then, $y = y_{1} + y_{2}$
$y = a \sin (ω t - k x) + \cos (ω t - k x)$
$y = a [\sin (ω t - k x) + \sin {\frac{π}{2} - (ω t - k x)}]$
$= a [2 \sin {\frac{ω t - k x + \frac{π}{2} - (ω t - k x)}{2}} \cos {\frac{ω t - k x - \frac{π}{2} + (ω t - k x)}{2}}]$
$= a [2 \sin \frac{π}{4} \times \cos \frac{{2 (ω t - kx) - \frac{π}{2}}}{2}]$
$= 2 a \times \frac{1}{\sqrt{2}} \cos [\frac{2 (ω t - kx) - \frac{π}{2}}{2}]$
$y = \sqrt{2} a \cos [(ω t - k x) - \frac{π}{4}]$
Hence, the resultant wave will have an amplitude $= \sqrt{2} a$

UPSEE - 2008

WAVES

172367 Three sinusoidal waves of the same frequency travel along a string in the positive $x$ -direction. Their amplitudes are $y, y / 2$ and $y / 3$ and their phase constants are $0, π / 2$ and $π$ respectively. What is the amplitude of the resultant wave?

1

0.63 y

2

0.72 y

3

0.83 y

4

0.52 y

Explanation:

C Equation for sinusoidal waves
$Y = A \sin (ω t + ϕ)$
Given that, $A_{1} = y, A_{2} = y / 2 A_{3} = y / 3$
$and ϕ_{1} = 0, ϕ_{2} = π / 2, ϕ_{3} = π$
$ω = Constant$
Hence three sinusoidal waves are given by
$Y_{1} = y \sin (ω t + 0)$
$Y_{2} = y / 2 \sin (ω t + π / 2)$
$Y_{3} = y / 3 \sin (ω t + π)$
According to superposition principle
$Y = Y_{1} + Y_{2} + Y_{3}$
$= y \sin ω t + \frac{y}{2} \sin (ω t + \frac{π}{2}) + \frac{y}{3} \sin (ω t + π)$
$= y \sin ω t + \frac{y}{2} \cos ω t + \frac{y}{3} (- \sin ω t)$
$= y \sin ω t - \frac{y}{3} \sin ω t + \frac{y}{2} \cos ω t$
$= \frac{2 y}{3} \sin ω t + \frac{y}{2} \cos ω t$
$= y (\frac{2}{3} \sin ω t + \frac{1}{2} \cos ω t)$
$= \frac{y}{6} (4 \sin ω t + 3 \cos ω t)$
$y = \frac{4 y}{6} \sin ω t + \frac{3 y}{6} \cos ω t$
$A_{1} = \frac{4 y}{6}, A_{2} = \frac{3 y}{6}$
$∴ A_{1} = \frac{2 y}{3}, A_{2} = \frac{y}{2}$
Net amplitude $A = \sqrt{A_{1}^{2} + A_{2}^{2}}$
$= \sqrt{{(\frac{2 y}{3})}^{2} + {(\frac{y}{2})}^{2}}$
$= y \sqrt{\frac{4}{9} + \frac{1}{4}}$
$= y \sqrt{\frac{25}{36}}$
$= \frac{5 y}{6}$
$= 0.83 y$

AMU-2011

WAVES

172368 Two waves $y_{1} = 2 \sin ω t$ and $y_{2} = 4 \sin (ω t + δ)$ superimpose. The ratio of the maximum to the minimum intensity of the resultant wave is

1 9

2 3

3 infinity

4 zero

Explanation:

A Given that,
$y_{1} = 2 \sin ω t$
$y_{2} = 4 \sin (ω t + δ)$
Comparing the resultant wave equation,
$y_{1} = A_{1} \sin ω t$
$y_{2} = A_{2} \sin (ω t + ϕ)$
Then, $A_{1} = 2, A_{2} = 4$
So, $\frac{I_{max}}{I_{min}} = \frac{{(A_{1} + A_{2})}^{2}}{{(A_{1} - A_{2})}^{2}}$
$\frac{I_{max}}{I_{min}} = \frac{(2 + 4)^{2}}{(2 - 4)^{2}}$
$\frac{I_{max}}{I_{min}} = \frac{36}{4} = 9$

UPSEE - 2011

WAVES

172364 Two harmonic travelling waves are described by the equations $y_{1} = a \sin (k x - ω t)$ and $y_{2} = a$ $\sin (- k x + ω t + ϕ)$ . The amplitude of the superimposed wave is

1

2 acos \frac{ϕ}{2}

2

2 asin ϕ

3

2 acos ϕ

4

2 asin \frac{ϕ}{2}

Explanation:

D Given,
$y_{1} = a \sin (kx - ω t)$ and $y_{2} = a \sin (- k x + ω t + ϕ)$
Let, $k x - ω t = α$
$y_{1} = a \sin α, y_{2} = a \sin (α - ϕ)$
Superimposed wave, $y = y_{1} + y_{2}$
$y = [a \sin α + a \sin ϕ \cos α - a \cos ϕ \sin α]$
$y = [a (1 - \cos ϕ) \sin α + (a \cos ϕ) \cos α]$
Let $R \cos θ = a (1 - \cos ϕ)$ and $R \sin θ = a \sin ϕ$ .
Then, magnitude of $R$ will be the magnitude of resultant wave,
$R^{2} = a^{2} [\sin^{2} ϕ + (1 - \cos ϕ)^{2}]$
$R^{2} = a^{2} [1 + 1 - 2 \cos ϕ]$
$R^{2} = 4 a^{2} \sin^{2} \frac{ϕ}{2}$
Hence, $R = 2 a \sin \frac{ϕ}{2}$

TS- EAMCET-04.05.2018

WAVES

172365 Two periodic waves of intensities $I_{1}$ and $I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

1

I_{1} + I_{2}

2

{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}

3

{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}

4

2 (I_{1} + I_{2})

Explanation:

D Resultant intensity of two periodic waves is given by
$I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2} \cos ϕ}$
Where $ϕ$ is the phase difference between two waves for maximum intensity
$ϕ = 2 π n (n = 0, 1, 2 \dots \dots \dots)$
For maxima
$\cos ϕ = 1$
$I_{max} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}}$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$I_{min} when \cos 180^{\circ} = - 1$
$I_{min} = I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}}$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
So, the sum of the maximum and minimum intensity
$I_{max} + I_{min} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} + {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$I_{max} + I_{min} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} + I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}}$
$I_{max} + I_{min} = 2 (I_{1} + I_{2})$

AIPMT-2008

WAVES

172366 Two waves represented by $y = a \sin (ω t - k x)$ and $y = a \cos (ω t - k x)$ are superimposed. The resultant wave will have an amplitude

1 a

2

\sqrt{2} a

3

2 a

4 zero

Explanation:

B Given,
$y_{1} = a \sin (ω t - k x)$
$y_{2} = a \cos (ω t - k x)$
Then, $y = y_{1} + y_{2}$
$y = a \sin (ω t - k x) + \cos (ω t - k x)$
$y = a [\sin (ω t - k x) + \sin {\frac{π}{2} - (ω t - k x)}]$
$= a [2 \sin {\frac{ω t - k x + \frac{π}{2} - (ω t - k x)}{2}} \cos {\frac{ω t - k x - \frac{π}{2} + (ω t - k x)}{2}}]$
$= a [2 \sin \frac{π}{4} \times \cos \frac{{2 (ω t - kx) - \frac{π}{2}}}{2}]$
$= 2 a \times \frac{1}{\sqrt{2}} \cos [\frac{2 (ω t - kx) - \frac{π}{2}}{2}]$
$y = \sqrt{2} a \cos [(ω t - k x) - \frac{π}{4}]$
Hence, the resultant wave will have an amplitude $= \sqrt{2} a$

UPSEE - 2008

WAVES

172367 Three sinusoidal waves of the same frequency travel along a string in the positive $x$ -direction. Their amplitudes are $y, y / 2$ and $y / 3$ and their phase constants are $0, π / 2$ and $π$ respectively. What is the amplitude of the resultant wave?

1

0.63 y

2

0.72 y

3

0.83 y

4

0.52 y

Explanation:

C Equation for sinusoidal waves
$Y = A \sin (ω t + ϕ)$
Given that, $A_{1} = y, A_{2} = y / 2 A_{3} = y / 3$
$and ϕ_{1} = 0, ϕ_{2} = π / 2, ϕ_{3} = π$
$ω = Constant$
Hence three sinusoidal waves are given by
$Y_{1} = y \sin (ω t + 0)$
$Y_{2} = y / 2 \sin (ω t + π / 2)$
$Y_{3} = y / 3 \sin (ω t + π)$
According to superposition principle
$Y = Y_{1} + Y_{2} + Y_{3}$
$= y \sin ω t + \frac{y}{2} \sin (ω t + \frac{π}{2}) + \frac{y}{3} \sin (ω t + π)$
$= y \sin ω t + \frac{y}{2} \cos ω t + \frac{y}{3} (- \sin ω t)$
$= y \sin ω t - \frac{y}{3} \sin ω t + \frac{y}{2} \cos ω t$
$= \frac{2 y}{3} \sin ω t + \frac{y}{2} \cos ω t$
$= y (\frac{2}{3} \sin ω t + \frac{1}{2} \cos ω t)$
$= \frac{y}{6} (4 \sin ω t + 3 \cos ω t)$
$y = \frac{4 y}{6} \sin ω t + \frac{3 y}{6} \cos ω t$
$A_{1} = \frac{4 y}{6}, A_{2} = \frac{3 y}{6}$
$∴ A_{1} = \frac{2 y}{3}, A_{2} = \frac{y}{2}$
Net amplitude $A = \sqrt{A_{1}^{2} + A_{2}^{2}}$
$= \sqrt{{(\frac{2 y}{3})}^{2} + {(\frac{y}{2})}^{2}}$
$= y \sqrt{\frac{4}{9} + \frac{1}{4}}$
$= y \sqrt{\frac{25}{36}}$
$= \frac{5 y}{6}$
$= 0.83 y$

AMU-2011

WAVES

172368 Two waves $y_{1} = 2 \sin ω t$ and $y_{2} = 4 \sin (ω t + δ)$ superimpose. The ratio of the maximum to the minimum intensity of the resultant wave is

1 9

2 3

3 infinity

4 zero

Explanation:

A Given that,
$y_{1} = 2 \sin ω t$
$y_{2} = 4 \sin (ω t + δ)$
Comparing the resultant wave equation,
$y_{1} = A_{1} \sin ω t$
$y_{2} = A_{2} \sin (ω t + ϕ)$
Then, $A_{1} = 2, A_{2} = 4$
So, $\frac{I_{max}}{I_{min}} = \frac{{(A_{1} + A_{2})}^{2}}{{(A_{1} - A_{2})}^{2}}$
$\frac{I_{max}}{I_{min}} = \frac{(2 + 4)^{2}}{(2 - 4)^{2}}$
$\frac{I_{max}}{I_{min}} = \frac{36}{4} = 9$

UPSEE - 2011