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PHXI14:OSCILLATIONS

364520 Consider two SHMs along the same straight line $x_{1} = A_{1} \sin (ω t + ϕ_{1}), x_{2} = A_{2} \sin (ω t + ϕ_{2})$ , where $A_{1}$ and $A_{2}$ are their amplitudes and $ϕ_{1}$ and $ϕ_{2}$ are their initial phase angle. If the two SHMs meet simultaneously and ' $R$ ' is the resultant amplitude, match column I with column II.
supporting img

1

A - R, B - P, C - S, D - Q

2

A - S, B - R, C - Q, D - P

3

A - P, B - R, C - Q, D - S

4

A - R, B - S, C - P, D - Q

Explanation:

$x_{1} = A_{1} \sin (ω t + ϕ_{1})$
$x_{2} = A_{2} \sin (ω t + ϕ_{2})$
where $A_{1}, A_{2}$ are amplitudes of two SHM's and $ϕ_{1}, ϕ_{2}$ are phase angles.
Resultant amplitude,
$R = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos (ϕ_{1} - ϕ_{2})} (1)$
(i) When two SHM's are in phase and, $A_{1} = A_{2} = A$ , then from equation (1) $\Rightarrow R = \sqrt{A^{2} + A^{2} + 2 A^{2}} = 2 A$
(ii) When two SHM's are in phase but $A_{1} \neq A_{2}$ , then
$R = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2}} = \sqrt{{(A_{1} + A_{2})}^{2}}$
$\Rightarrow R = A_{1} + A_{2}$
(iii) When two $SHMs$ are $90^{\circ}$ out of phase, then
$\Rightarrow R = \sqrt{A^{2} + A^{2} + 0} = \sqrt{2} A$
(iv) When two SHMs are $180^{\circ}$ out of phase and $A_{1} = A_{2}$ ,
$R = \sqrt{A_{1}^{2} + A_{1}^{2} - 2 A_{1}^{2}}$
$\Rightarrow R = 0$
Hence option (1) is correct answer.

MHTCET - 2022

PHXI14:OSCILLATIONS

364521 For a periodic motion represented by the equation $y = \sin ω t + \cos ω t$ , the amplitude of the motion is

1

\sqrt{2}

2 2

3 0.5

4 1

Explanation:

As, $y = \sin ω t + \cos ω t$
$= \sqrt{2} (\sin ω t \cdot \frac{1}{\sqrt{2}} + \cos ω t \cdot \frac{1}{\sqrt{2}})$
$\Rightarrow y = \sqrt{2} \sin (ω t + π / 4)$ .
The above equation represents on SHM with initial phase $π / 4$ and amplitude $\sqrt{2}$ .

JEE - 2023

PHXI14:OSCILLATIONS

364522 The composition of two simple harmonic motions of equal periods at right angle to each others and with a phase difference of $π$ results in the displacement of the particle along.

1 Circle

2 Figure of Eight

3 Straight line

4 Ellipse

Explanation:

$x = A \sin ω t &$
$\begin{matrix} y = \sin (ω + \frac{π}{2}) = - A \sin (ω t) \\ \Rightarrow \frac{x}{y} = - 1 \Rightarrow x = - y (straight line) \end{matrix}$

PHXI14:OSCILLATIONS

364523 The resultant amplitude due to superposition of three simple harmonic motions $y_{1} = 3 \sin ω t y_{2} = 5 \sin ω t + 37^{o}$ and $y_{3} = - 15 \cos ω t$ is ____.

1

\sqrt{193}

2

\sqrt{73}

3

8 \sqrt{2}

4

7 \sqrt{3}

Explanation:

$y_{3} = - 15 \cos ω t = - 15 \sin (\frac{π}{2} + ω t)$
$= 15 \sin (ω t + \frac{3 π}{2}) = 15 \sin (ω t + 270^{\circ})$
$∴$ Vector diagram of amplitudes is:
supporting img
$Σ A_{x} = 3 + 5 \cos 37^{\circ} = 7$
$Σ A_{y} = - 15 + 5 \sin 37^{\circ} = - 12$
$∴ A = \sqrt{{(Σ A_{x})}^{2} + {(Σ A_{y})}^{2}} = \sqrt{7^{2} + 12^{2}} = \sqrt{193}$ units

PHXI14:OSCILLATIONS

364520 Consider two SHMs along the same straight line $x_{1} = A_{1} \sin (ω t + ϕ_{1}), x_{2} = A_{2} \sin (ω t + ϕ_{2})$ , where $A_{1}$ and $A_{2}$ are their amplitudes and $ϕ_{1}$ and $ϕ_{2}$ are their initial phase angle. If the two SHMs meet simultaneously and ' $R$ ' is the resultant amplitude, match column I with column II.
supporting img

1

A - R, B - P, C - S, D - Q

2

A - S, B - R, C - Q, D - P

3

A - P, B - R, C - Q, D - S

4

A - R, B - S, C - P, D - Q

Explanation:

$x_{1} = A_{1} \sin (ω t + ϕ_{1})$
$x_{2} = A_{2} \sin (ω t + ϕ_{2})$
where $A_{1}, A_{2}$ are amplitudes of two SHM's and $ϕ_{1}, ϕ_{2}$ are phase angles.
Resultant amplitude,
$R = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos (ϕ_{1} - ϕ_{2})} (1)$
(i) When two SHM's are in phase and, $A_{1} = A_{2} = A$ , then from equation (1) $\Rightarrow R = \sqrt{A^{2} + A^{2} + 2 A^{2}} = 2 A$
(ii) When two SHM's are in phase but $A_{1} \neq A_{2}$ , then
$R = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2}} = \sqrt{{(A_{1} + A_{2})}^{2}}$
$\Rightarrow R = A_{1} + A_{2}$
(iii) When two $SHMs$ are $90^{\circ}$ out of phase, then
$\Rightarrow R = \sqrt{A^{2} + A^{2} + 0} = \sqrt{2} A$
(iv) When two SHMs are $180^{\circ}$ out of phase and $A_{1} = A_{2}$ ,
$R = \sqrt{A_{1}^{2} + A_{1}^{2} - 2 A_{1}^{2}}$
$\Rightarrow R = 0$
Hence option (1) is correct answer.

MHTCET - 2022

PHXI14:OSCILLATIONS

364521 For a periodic motion represented by the equation $y = \sin ω t + \cos ω t$ , the amplitude of the motion is

1

\sqrt{2}

2 2

3 0.5

4 1

Explanation:

As, $y = \sin ω t + \cos ω t$
$= \sqrt{2} (\sin ω t \cdot \frac{1}{\sqrt{2}} + \cos ω t \cdot \frac{1}{\sqrt{2}})$
$\Rightarrow y = \sqrt{2} \sin (ω t + π / 4)$ .
The above equation represents on SHM with initial phase $π / 4$ and amplitude $\sqrt{2}$ .

JEE - 2023

PHXI14:OSCILLATIONS

364522 The composition of two simple harmonic motions of equal periods at right angle to each others and with a phase difference of $π$ results in the displacement of the particle along.

1 Circle

2 Figure of Eight

3 Straight line

4 Ellipse

Explanation:

$x = A \sin ω t &$
$\begin{matrix} y = \sin (ω + \frac{π}{2}) = - A \sin (ω t) \\ \Rightarrow \frac{x}{y} = - 1 \Rightarrow x = - y (straight line) \end{matrix}$

PHXI14:OSCILLATIONS

364523 The resultant amplitude due to superposition of three simple harmonic motions $y_{1} = 3 \sin ω t y_{2} = 5 \sin ω t + 37^{o}$ and $y_{3} = - 15 \cos ω t$ is ____.

1

\sqrt{193}

2

\sqrt{73}

3

8 \sqrt{2}

4

7 \sqrt{3}

Explanation:

$y_{3} = - 15 \cos ω t = - 15 \sin (\frac{π}{2} + ω t)$
$= 15 \sin (ω t + \frac{3 π}{2}) = 15 \sin (ω t + 270^{\circ})$
$∴$ Vector diagram of amplitudes is:
supporting img
$Σ A_{x} = 3 + 5 \cos 37^{\circ} = 7$
$Σ A_{y} = - 15 + 5 \sin 37^{\circ} = - 12$
$∴ A = \sqrt{{(Σ A_{x})}^{2} + {(Σ A_{y})}^{2}} = \sqrt{7^{2} + 12^{2}} = \sqrt{193}$ units

PHXI14:OSCILLATIONS

364520 Consider two SHMs along the same straight line $x_{1} = A_{1} \sin (ω t + ϕ_{1}), x_{2} = A_{2} \sin (ω t + ϕ_{2})$ , where $A_{1}$ and $A_{2}$ are their amplitudes and $ϕ_{1}$ and $ϕ_{2}$ are their initial phase angle. If the two SHMs meet simultaneously and ' $R$ ' is the resultant amplitude, match column I with column II.
supporting img

1

A - R, B - P, C - S, D - Q

2

A - S, B - R, C - Q, D - P

3

A - P, B - R, C - Q, D - S

4

A - R, B - S, C - P, D - Q

Explanation:

$x_{1} = A_{1} \sin (ω t + ϕ_{1})$
$x_{2} = A_{2} \sin (ω t + ϕ_{2})$
where $A_{1}, A_{2}$ are amplitudes of two SHM's and $ϕ_{1}, ϕ_{2}$ are phase angles.
Resultant amplitude,
$R = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos (ϕ_{1} - ϕ_{2})} (1)$
(i) When two SHM's are in phase and, $A_{1} = A_{2} = A$ , then from equation (1) $\Rightarrow R = \sqrt{A^{2} + A^{2} + 2 A^{2}} = 2 A$
(ii) When two SHM's are in phase but $A_{1} \neq A_{2}$ , then
$R = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2}} = \sqrt{{(A_{1} + A_{2})}^{2}}$
$\Rightarrow R = A_{1} + A_{2}$
(iii) When two $SHMs$ are $90^{\circ}$ out of phase, then
$\Rightarrow R = \sqrt{A^{2} + A^{2} + 0} = \sqrt{2} A$
(iv) When two SHMs are $180^{\circ}$ out of phase and $A_{1} = A_{2}$ ,
$R = \sqrt{A_{1}^{2} + A_{1}^{2} - 2 A_{1}^{2}}$
$\Rightarrow R = 0$
Hence option (1) is correct answer.

MHTCET - 2022

PHXI14:OSCILLATIONS

364521 For a periodic motion represented by the equation $y = \sin ω t + \cos ω t$ , the amplitude of the motion is

1

\sqrt{2}

2 2

3 0.5

4 1

Explanation:

As, $y = \sin ω t + \cos ω t$
$= \sqrt{2} (\sin ω t \cdot \frac{1}{\sqrt{2}} + \cos ω t \cdot \frac{1}{\sqrt{2}})$
$\Rightarrow y = \sqrt{2} \sin (ω t + π / 4)$ .
The above equation represents on SHM with initial phase $π / 4$ and amplitude $\sqrt{2}$ .

JEE - 2023

PHXI14:OSCILLATIONS

364522 The composition of two simple harmonic motions of equal periods at right angle to each others and with a phase difference of $π$ results in the displacement of the particle along.

1 Circle

2 Figure of Eight

3 Straight line

4 Ellipse

Explanation:

$x = A \sin ω t &$
$\begin{matrix} y = \sin (ω + \frac{π}{2}) = - A \sin (ω t) \\ \Rightarrow \frac{x}{y} = - 1 \Rightarrow x = - y (straight line) \end{matrix}$

PHXI14:OSCILLATIONS

364523 The resultant amplitude due to superposition of three simple harmonic motions $y_{1} = 3 \sin ω t y_{2} = 5 \sin ω t + 37^{o}$ and $y_{3} = - 15 \cos ω t$ is ____.

1

\sqrt{193}

2

\sqrt{73}

3

8 \sqrt{2}

4

7 \sqrt{3}

Explanation:

$y_{3} = - 15 \cos ω t = - 15 \sin (\frac{π}{2} + ω t)$
$= 15 \sin (ω t + \frac{3 π}{2}) = 15 \sin (ω t + 270^{\circ})$
$∴$ Vector diagram of amplitudes is:
supporting img
$Σ A_{x} = 3 + 5 \cos 37^{\circ} = 7$
$Σ A_{y} = - 15 + 5 \sin 37^{\circ} = - 12$
$∴ A = \sqrt{{(Σ A_{x})}^{2} + {(Σ A_{y})}^{2}} = \sqrt{7^{2} + 12^{2}} = \sqrt{193}$ units

PHXI14:OSCILLATIONS

364520 Consider two SHMs along the same straight line $x_{1} = A_{1} \sin (ω t + ϕ_{1}), x_{2} = A_{2} \sin (ω t + ϕ_{2})$ , where $A_{1}$ and $A_{2}$ are their amplitudes and $ϕ_{1}$ and $ϕ_{2}$ are their initial phase angle. If the two SHMs meet simultaneously and ' $R$ ' is the resultant amplitude, match column I with column II.
supporting img

1

A - R, B - P, C - S, D - Q

2

A - S, B - R, C - Q, D - P

3

A - P, B - R, C - Q, D - S

4

A - R, B - S, C - P, D - Q

Explanation:

$x_{1} = A_{1} \sin (ω t + ϕ_{1})$
$x_{2} = A_{2} \sin (ω t + ϕ_{2})$
where $A_{1}, A_{2}$ are amplitudes of two SHM's and $ϕ_{1}, ϕ_{2}$ are phase angles.
Resultant amplitude,
$R = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos (ϕ_{1} - ϕ_{2})} (1)$
(i) When two SHM's are in phase and, $A_{1} = A_{2} = A$ , then from equation (1) $\Rightarrow R = \sqrt{A^{2} + A^{2} + 2 A^{2}} = 2 A$
(ii) When two SHM's are in phase but $A_{1} \neq A_{2}$ , then
$R = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2}} = \sqrt{{(A_{1} + A_{2})}^{2}}$
$\Rightarrow R = A_{1} + A_{2}$
(iii) When two $SHMs$ are $90^{\circ}$ out of phase, then
$\Rightarrow R = \sqrt{A^{2} + A^{2} + 0} = \sqrt{2} A$
(iv) When two SHMs are $180^{\circ}$ out of phase and $A_{1} = A_{2}$ ,
$R = \sqrt{A_{1}^{2} + A_{1}^{2} - 2 A_{1}^{2}}$
$\Rightarrow R = 0$
Hence option (1) is correct answer.

MHTCET - 2022

PHXI14:OSCILLATIONS

364521 For a periodic motion represented by the equation $y = \sin ω t + \cos ω t$ , the amplitude of the motion is

1

\sqrt{2}

2 2

3 0.5

4 1

Explanation:

As, $y = \sin ω t + \cos ω t$
$= \sqrt{2} (\sin ω t \cdot \frac{1}{\sqrt{2}} + \cos ω t \cdot \frac{1}{\sqrt{2}})$
$\Rightarrow y = \sqrt{2} \sin (ω t + π / 4)$ .
The above equation represents on SHM with initial phase $π / 4$ and amplitude $\sqrt{2}$ .

JEE - 2023

PHXI14:OSCILLATIONS

364522 The composition of two simple harmonic motions of equal periods at right angle to each others and with a phase difference of $π$ results in the displacement of the particle along.

1 Circle

2 Figure of Eight

3 Straight line

4 Ellipse

Explanation:

$x = A \sin ω t &$
$\begin{matrix} y = \sin (ω + \frac{π}{2}) = - A \sin (ω t) \\ \Rightarrow \frac{x}{y} = - 1 \Rightarrow x = - y (straight line) \end{matrix}$

PHXI14:OSCILLATIONS

364523 The resultant amplitude due to superposition of three simple harmonic motions $y_{1} = 3 \sin ω t y_{2} = 5 \sin ω t + 37^{o}$ and $y_{3} = - 15 \cos ω t$ is ____.

1

\sqrt{193}

2

\sqrt{73}

3

8 \sqrt{2}

4

7 \sqrt{3}

Explanation:

$y_{3} = - 15 \cos ω t = - 15 \sin (\frac{π}{2} + ω t)$
$= 15 \sin (ω t + \frac{3 π}{2}) = 15 \sin (ω t + 270^{\circ})$
$∴$ Vector diagram of amplitudes is:
supporting img
$Σ A_{x} = 3 + 5 \cos 37^{\circ} = 7$
$Σ A_{y} = - 15 + 5 \sin 37^{\circ} = - 12$
$∴ A = \sqrt{{(Σ A_{x})}^{2} + {(Σ A_{y})}^{2}} = \sqrt{7^{2} + 12^{2}} = \sqrt{193}$ units

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