QBManager

WAVES

172370 Two waves are given by $y_{1} = a \sin (ω t - k x)$ and $y_{2} = a \cos (ω t - k x)$ . The phase difference between the two waves is

1

π / 4

2

π

3

π / 8

4

π / 2

Explanation:

D Given that,
$y_{1} = a \sin (ω t - kx)$
$y_{2} = a \cos (ω t - kx)$
Equation (ii) we can write as
$y_{2} = a \sin (ω t - kx + \frac{π}{2})$
Hence, phase difference, $Δ ϕ = \frac{π}{2}$

CG PET- 2006

WAVES

172371 If two waves represented by $y_{1} = 4 \sin ω t$ and $y_{2} = 3 \sin (ω t + \frac{π}{3})$ interfere at a point.
The amplitude of the resulting wave will be about

1 7

2 6

3 5

4 3.5

Explanation:

B Given that,
$y_{1} = 4 \sin ω t$
$y_{2} = 3 \sin (ω t + \frac{π}{3})$
From equation (i) and (ii) we get -
Phase difference $(ϕ) = \frac{π}{3}$
And $A_{1} = 4, A_{2} = 3$
$Now, A_{Resultant} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
$A_{Resultant} = \sqrt{4^{2} + 3^{2} + 2 \times 4 \times 3 \times \cos \frac{π}{3}}$
$A_{Resultant} = \sqrt{25 + 24 \times \frac{1}{2}}$
$A_{Resultant} = \sqrt{37}$
$A_{Resultant} \approx 6$

Manipal UGET-2010

WAVES

172373 Two periodic waves of intensities $I_{1}, I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

1

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

2

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

3

I_{1} + I_{2}

4

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

Explanation:

B Let,
$y_{1} = a_{1} \sin (ω t - kx)$
$y_{2} = a_{2} \sin (ω t - kx + ϕ)$
As we know that,
$I \propto a^{2}$
$I = {ka}^{2}$
$I = I_{1} + I_{2} + 2 \sqrt{I_{1} \cdot I_{2}} \cos ϕ$
Then, $I_{max} = I_{1} + I_{2} + 2 \sqrt{I_{1} \cdot I_{2}}$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
And, $I_{min} = I_{1} + I_{2} - 2 \sqrt{I_{1} \cdot I_{2}}$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
Hence, $I_{max} + I_{min} = 2 I_{1} + 2 I_{2}$
$= 2 (I_{1} + I_{2})$

COMEDK 2016

WAVES

172375 Two waves represented by $x_{1} = 2 \sin (20 π t) m$ and $x_{2} = 2 \sin (20 π t + \frac{π}{3}) m$ are superimposed then the amplitude of the resultant wave.

1

2 \sqrt{3} m

2

3 \sqrt{3} m

3

4 \sqrt{3} m

4

5 \sqrt{3} m

Explanation:

A The two waves are given by
$x_{1} = 2 \sin (20 π t) m$
$x_{2} = 2 \sin (20 π t + \frac{π}{3}) m$
$A_{1} = 2, A_{2} = 2, ϕ = \frac{π}{3} = 60^{\circ}$
The resultant wave amplitude
$A = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
$= \sqrt{(2)^{2} + (2)^{2} + 2 \times 2 \times 2 \cos 60^{\circ}}$
$= \sqrt{4 + 4 + 8 \times \frac{1}{2}}$
$= \sqrt{8 + 4}$
$A = 2 \sqrt{3} m$

AP EAMCET-28.04.2017

WAVES

172370 Two waves are given by $y_{1} = a \sin (ω t - k x)$ and $y_{2} = a \cos (ω t - k x)$ . The phase difference between the two waves is

1

π / 4

2

π

3

π / 8

4

π / 2

Explanation:

D Given that,
$y_{1} = a \sin (ω t - kx)$
$y_{2} = a \cos (ω t - kx)$
Equation (ii) we can write as
$y_{2} = a \sin (ω t - kx + \frac{π}{2})$
Hence, phase difference, $Δ ϕ = \frac{π}{2}$

CG PET- 2006

WAVES

172371 If two waves represented by $y_{1} = 4 \sin ω t$ and $y_{2} = 3 \sin (ω t + \frac{π}{3})$ interfere at a point.
The amplitude of the resulting wave will be about

1 7

2 6

3 5

4 3.5

Explanation:

B Given that,
$y_{1} = 4 \sin ω t$
$y_{2} = 3 \sin (ω t + \frac{π}{3})$
From equation (i) and (ii) we get -
Phase difference $(ϕ) = \frac{π}{3}$
And $A_{1} = 4, A_{2} = 3$
$Now, A_{Resultant} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
$A_{Resultant} = \sqrt{4^{2} + 3^{2} + 2 \times 4 \times 3 \times \cos \frac{π}{3}}$
$A_{Resultant} = \sqrt{25 + 24 \times \frac{1}{2}}$
$A_{Resultant} = \sqrt{37}$
$A_{Resultant} \approx 6$

Manipal UGET-2010

WAVES

172373 Two periodic waves of intensities $I_{1}, I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

1

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

2

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

3

I_{1} + I_{2}

4

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

Explanation:

B Let,
$y_{1} = a_{1} \sin (ω t - kx)$
$y_{2} = a_{2} \sin (ω t - kx + ϕ)$
As we know that,
$I \propto a^{2}$
$I = {ka}^{2}$
$I = I_{1} + I_{2} + 2 \sqrt{I_{1} \cdot I_{2}} \cos ϕ$
Then, $I_{max} = I_{1} + I_{2} + 2 \sqrt{I_{1} \cdot I_{2}}$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
And, $I_{min} = I_{1} + I_{2} - 2 \sqrt{I_{1} \cdot I_{2}}$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
Hence, $I_{max} + I_{min} = 2 I_{1} + 2 I_{2}$
$= 2 (I_{1} + I_{2})$

COMEDK 2016

WAVES

172375 Two waves represented by $x_{1} = 2 \sin (20 π t) m$ and $x_{2} = 2 \sin (20 π t + \frac{π}{3}) m$ are superimposed then the amplitude of the resultant wave.

1

2 \sqrt{3} m

2

3 \sqrt{3} m

3

4 \sqrt{3} m

4

5 \sqrt{3} m

Explanation:

A The two waves are given by
$x_{1} = 2 \sin (20 π t) m$
$x_{2} = 2 \sin (20 π t + \frac{π}{3}) m$
$A_{1} = 2, A_{2} = 2, ϕ = \frac{π}{3} = 60^{\circ}$
The resultant wave amplitude
$A = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
$= \sqrt{(2)^{2} + (2)^{2} + 2 \times 2 \times 2 \cos 60^{\circ}}$
$= \sqrt{4 + 4 + 8 \times \frac{1}{2}}$
$= \sqrt{8 + 4}$
$A = 2 \sqrt{3} m$

AP EAMCET-28.04.2017

WAVES

172370 Two waves are given by $y_{1} = a \sin (ω t - k x)$ and $y_{2} = a \cos (ω t - k x)$ . The phase difference between the two waves is

1

π / 4

2

π

3

π / 8

4

π / 2

Explanation:

D Given that,
$y_{1} = a \sin (ω t - kx)$
$y_{2} = a \cos (ω t - kx)$
Equation (ii) we can write as
$y_{2} = a \sin (ω t - kx + \frac{π}{2})$
Hence, phase difference, $Δ ϕ = \frac{π}{2}$

CG PET- 2006

WAVES

172371 If two waves represented by $y_{1} = 4 \sin ω t$ and $y_{2} = 3 \sin (ω t + \frac{π}{3})$ interfere at a point.
The amplitude of the resulting wave will be about

1 7

2 6

3 5

4 3.5

Explanation:

B Given that,
$y_{1} = 4 \sin ω t$
$y_{2} = 3 \sin (ω t + \frac{π}{3})$
From equation (i) and (ii) we get -
Phase difference $(ϕ) = \frac{π}{3}$
And $A_{1} = 4, A_{2} = 3$
$Now, A_{Resultant} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
$A_{Resultant} = \sqrt{4^{2} + 3^{2} + 2 \times 4 \times 3 \times \cos \frac{π}{3}}$
$A_{Resultant} = \sqrt{25 + 24 \times \frac{1}{2}}$
$A_{Resultant} = \sqrt{37}$
$A_{Resultant} \approx 6$

Manipal UGET-2010

WAVES

172373 Two periodic waves of intensities $I_{1}, I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

1

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

2

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

3

I_{1} + I_{2}

4

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

Explanation:

B Let,
$y_{1} = a_{1} \sin (ω t - kx)$
$y_{2} = a_{2} \sin (ω t - kx + ϕ)$
As we know that,
$I \propto a^{2}$
$I = {ka}^{2}$
$I = I_{1} + I_{2} + 2 \sqrt{I_{1} \cdot I_{2}} \cos ϕ$
Then, $I_{max} = I_{1} + I_{2} + 2 \sqrt{I_{1} \cdot I_{2}}$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
And, $I_{min} = I_{1} + I_{2} - 2 \sqrt{I_{1} \cdot I_{2}}$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
Hence, $I_{max} + I_{min} = 2 I_{1} + 2 I_{2}$
$= 2 (I_{1} + I_{2})$

COMEDK 2016

WAVES

172375 Two waves represented by $x_{1} = 2 \sin (20 π t) m$ and $x_{2} = 2 \sin (20 π t + \frac{π}{3}) m$ are superimposed then the amplitude of the resultant wave.

1

2 \sqrt{3} m

2

3 \sqrt{3} m

3

4 \sqrt{3} m

4

5 \sqrt{3} m

Explanation:

A The two waves are given by
$x_{1} = 2 \sin (20 π t) m$
$x_{2} = 2 \sin (20 π t + \frac{π}{3}) m$
$A_{1} = 2, A_{2} = 2, ϕ = \frac{π}{3} = 60^{\circ}$
The resultant wave amplitude
$A = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
$= \sqrt{(2)^{2} + (2)^{2} + 2 \times 2 \times 2 \cos 60^{\circ}}$
$= \sqrt{4 + 4 + 8 \times \frac{1}{2}}$
$= \sqrt{8 + 4}$
$A = 2 \sqrt{3} m$

AP EAMCET-28.04.2017

WAVES

172370 Two waves are given by $y_{1} = a \sin (ω t - k x)$ and $y_{2} = a \cos (ω t - k x)$ . The phase difference between the two waves is

1

π / 4

2

π

3

π / 8

4

π / 2

Explanation:

D Given that,
$y_{1} = a \sin (ω t - kx)$
$y_{2} = a \cos (ω t - kx)$
Equation (ii) we can write as
$y_{2} = a \sin (ω t - kx + \frac{π}{2})$
Hence, phase difference, $Δ ϕ = \frac{π}{2}$

CG PET- 2006

WAVES

172371 If two waves represented by $y_{1} = 4 \sin ω t$ and $y_{2} = 3 \sin (ω t + \frac{π}{3})$ interfere at a point.
The amplitude of the resulting wave will be about

1 7

2 6

3 5

4 3.5

Explanation:

B Given that,
$y_{1} = 4 \sin ω t$
$y_{2} = 3 \sin (ω t + \frac{π}{3})$
From equation (i) and (ii) we get -
Phase difference $(ϕ) = \frac{π}{3}$
And $A_{1} = 4, A_{2} = 3$
$Now, A_{Resultant} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
$A_{Resultant} = \sqrt{4^{2} + 3^{2} + 2 \times 4 \times 3 \times \cos \frac{π}{3}}$
$A_{Resultant} = \sqrt{25 + 24 \times \frac{1}{2}}$
$A_{Resultant} = \sqrt{37}$
$A_{Resultant} \approx 6$

Manipal UGET-2010

WAVES

172373 Two periodic waves of intensities $I_{1}, I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

1

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

2

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

3

I_{1} + I_{2}

4

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

Explanation:

B Let,
$y_{1} = a_{1} \sin (ω t - kx)$
$y_{2} = a_{2} \sin (ω t - kx + ϕ)$
As we know that,
$I \propto a^{2}$
$I = {ka}^{2}$
$I = I_{1} + I_{2} + 2 \sqrt{I_{1} \cdot I_{2}} \cos ϕ$
Then, $I_{max} = I_{1} + I_{2} + 2 \sqrt{I_{1} \cdot I_{2}}$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
And, $I_{min} = I_{1} + I_{2} - 2 \sqrt{I_{1} \cdot I_{2}}$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
Hence, $I_{max} + I_{min} = 2 I_{1} + 2 I_{2}$
$= 2 (I_{1} + I_{2})$

COMEDK 2016

WAVES

172375 Two waves represented by $x_{1} = 2 \sin (20 π t) m$ and $x_{2} = 2 \sin (20 π t + \frac{π}{3}) m$ are superimposed then the amplitude of the resultant wave.

1

2 \sqrt{3} m

2

3 \sqrt{3} m

3

4 \sqrt{3} m

4

5 \sqrt{3} m

Explanation:

A The two waves are given by
$x_{1} = 2 \sin (20 π t) m$
$x_{2} = 2 \sin (20 π t + \frac{π}{3}) m$
$A_{1} = 2, A_{2} = 2, ϕ = \frac{π}{3} = 60^{\circ}$
The resultant wave amplitude
$A = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
$= \sqrt{(2)^{2} + (2)^{2} + 2 \times 2 \times 2 \cos 60^{\circ}}$
$= \sqrt{4 + 4 + 8 \times \frac{1}{2}}$
$= \sqrt{8 + 4}$
$A = 2 \sqrt{3} m$

AP EAMCET-28.04.2017

Question Bank Series

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