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PHXI14:OSCILLATIONS

364210 The displacement-time equation of particle executing SHM is: $x = A \sin (ω t + ϕ)$ . At time $t = 0$ , position of the particle is $x = A / 2$ and it is moving along negative $x$ -direction. Then the angle $ϕ$ can be:

1

\frac{π}{6}

2

\frac{π}{4}

3

\frac{5 π}{6}

4

\frac{7 π}{6}

Explanation:

At $t = 0, x = \frac{A}{2}$
$\frac{A}{2} = A \sin ϕ or ϕ = \frac{π}{6} or ϕ = \frac{5 π}{6}$
$v = \frac{d x}{d t} = A ω \cos (ω t + ϕ)$
At $t = 0, v = A ω \cos ϕ$
Now, $\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ and $\cos \frac{5 π}{6} = - \frac{\sqrt{3}}{2}$
As $v$ is negative at $t = 0, ϕ$ must be $5 π / 6$ .

PHXI14:OSCILLATIONS

364211 Two particles execute SHM with same amplitudes and same angular frequency on same straight line with same mean position. Given that during oscillation they cross each other in opposite direction when at a distance $A / 2$ from mean position. Find phase difference in the two $S H M s$ .

1

80^{\circ}

2

120^{\circ}

3

55^{\circ}

4

150^{\circ}

Explanation:

Figure shows that two respective particles $P$ and $Q^{'}$ in $S H M$ along with their corresponding particles in circular motion. Let $P$ moves in upward direction when crossing $Q^{'}$ at $A / 2$ as shown in Fig. (a), at this instant phase of $P$ is
$ϕ_{1} = \sin^{- 1} (\frac{1}{2}) = \frac{π}{6} (1)$
supporting img
Similarly, as shown in Fig. (b) we can take particle $Q^{'}$ is moving in downward direction (opposite $P^{'}$ ) at $A / 2$ , this implies its circular motion particle is in second quadrant thus its phase angle is
$ϕ_{2} = π - \sin^{- 1} (\frac{1}{2} = π - \frac{π}{6} = \frac{5 π}{6}) (2)$
As both are oscillating at same angular frequency their phase difference remains constant which can be given from Eqs. (1) and (2) as
$Δ ϕ = ϕ_{2} - ϕ_{1} = \frac{5 π}{6} - \frac{π}{6} = \frac{2 π}{3} r a d = 120^{\circ} .$

PHXI14:OSCILLATIONS

364212 Two pendulums have time periods T and $\frac{5 T}{4}$ .
They start S.H.M. at the same time from the mean position. What will be the phase difference between them after the bigger pendulum
complete one oscillation?

1

90^{\circ}

2

45^{\circ}

3

30^{\circ}

4

60^{\circ}

Explanation:

$\frac{5 T}{4} = T + \frac{T}{4}$
supporting img
By the time, the bigger pendulum makes one full oscillation, the smaller pendulum will make $(1 + \frac{1}{4})$ oscillation. The bigger pendulum will be in the mean position and the smaller one will be in the positive extreme position. Thus, phase difference $= 90^{\circ}$

PHXI14:OSCILLATIONS

364213 The equation of S.H.M is $y = a \sin (2 π n t + α)$ , then its phase at time $t$ is

1

α

2

2 π n t

3

2 π t

4

2 π n t + α

Explanation:

$y = a \sin (2 π n t + α)$ .
Its phase at time
$t = 2 π n t + α$

PHXI14:OSCILLATIONS

364214 Two simple harmonic motions exhibited by two particles $A$ and $B$ are given respectively by the following equations.
$y_{1} = a \sin [ω t + \frac{π}{6}]$
$y_{2} = a \sin [ω t + \frac{3 π}{6}]$
Phase difference between them is

1

\frac{π}{2}

2

\frac{π}{6}

3

\frac{π}{3}

4 Zero

Explanation:

Phase difference $Δ ϕ = \frac{3 π}{2} - \frac{π}{6} = \frac{π}{3}$ .

PHXI14:OSCILLATIONS

364210 The displacement-time equation of particle executing SHM is: $x = A \sin (ω t + ϕ)$ . At time $t = 0$ , position of the particle is $x = A / 2$ and it is moving along negative $x$ -direction. Then the angle $ϕ$ can be:

1

\frac{π}{6}

2

\frac{π}{4}

3

\frac{5 π}{6}

4

\frac{7 π}{6}

Explanation:

At $t = 0, x = \frac{A}{2}$
$\frac{A}{2} = A \sin ϕ or ϕ = \frac{π}{6} or ϕ = \frac{5 π}{6}$
$v = \frac{d x}{d t} = A ω \cos (ω t + ϕ)$
At $t = 0, v = A ω \cos ϕ$
Now, $\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ and $\cos \frac{5 π}{6} = - \frac{\sqrt{3}}{2}$
As $v$ is negative at $t = 0, ϕ$ must be $5 π / 6$ .

PHXI14:OSCILLATIONS

364211 Two particles execute SHM with same amplitudes and same angular frequency on same straight line with same mean position. Given that during oscillation they cross each other in opposite direction when at a distance $A / 2$ from mean position. Find phase difference in the two $S H M s$ .

1

80^{\circ}

2

120^{\circ}

3

55^{\circ}

4

150^{\circ}

Explanation:

Figure shows that two respective particles $P$ and $Q^{'}$ in $S H M$ along with their corresponding particles in circular motion. Let $P$ moves in upward direction when crossing $Q^{'}$ at $A / 2$ as shown in Fig. (a), at this instant phase of $P$ is
$ϕ_{1} = \sin^{- 1} (\frac{1}{2}) = \frac{π}{6} (1)$
supporting img
Similarly, as shown in Fig. (b) we can take particle $Q^{'}$ is moving in downward direction (opposite $P^{'}$ ) at $A / 2$ , this implies its circular motion particle is in second quadrant thus its phase angle is
$ϕ_{2} = π - \sin^{- 1} (\frac{1}{2} = π - \frac{π}{6} = \frac{5 π}{6}) (2)$
As both are oscillating at same angular frequency their phase difference remains constant which can be given from Eqs. (1) and (2) as
$Δ ϕ = ϕ_{2} - ϕ_{1} = \frac{5 π}{6} - \frac{π}{6} = \frac{2 π}{3} r a d = 120^{\circ} .$

PHXI14:OSCILLATIONS

364212 Two pendulums have time periods T and $\frac{5 T}{4}$ .
They start S.H.M. at the same time from the mean position. What will be the phase difference between them after the bigger pendulum
complete one oscillation?

1

90^{\circ}

2

45^{\circ}

3

30^{\circ}

4

60^{\circ}

Explanation:

$\frac{5 T}{4} = T + \frac{T}{4}$
supporting img
By the time, the bigger pendulum makes one full oscillation, the smaller pendulum will make $(1 + \frac{1}{4})$ oscillation. The bigger pendulum will be in the mean position and the smaller one will be in the positive extreme position. Thus, phase difference $= 90^{\circ}$

PHXI14:OSCILLATIONS

364213 The equation of S.H.M is $y = a \sin (2 π n t + α)$ , then its phase at time $t$ is

1

α

2

2 π n t

3

2 π t

4

2 π n t + α

Explanation:

$y = a \sin (2 π n t + α)$ .
Its phase at time
$t = 2 π n t + α$

PHXI14:OSCILLATIONS

364214 Two simple harmonic motions exhibited by two particles $A$ and $B$ are given respectively by the following equations.
$y_{1} = a \sin [ω t + \frac{π}{6}]$
$y_{2} = a \sin [ω t + \frac{3 π}{6}]$
Phase difference between them is

1

\frac{π}{2}

2

\frac{π}{6}

3

\frac{π}{3}

4 Zero

Explanation:

Phase difference $Δ ϕ = \frac{3 π}{2} - \frac{π}{6} = \frac{π}{3}$ .

PHXI14:OSCILLATIONS

364210 The displacement-time equation of particle executing SHM is: $x = A \sin (ω t + ϕ)$ . At time $t = 0$ , position of the particle is $x = A / 2$ and it is moving along negative $x$ -direction. Then the angle $ϕ$ can be:

1

\frac{π}{6}

2

\frac{π}{4}

3

\frac{5 π}{6}

4

\frac{7 π}{6}

Explanation:

At $t = 0, x = \frac{A}{2}$
$\frac{A}{2} = A \sin ϕ or ϕ = \frac{π}{6} or ϕ = \frac{5 π}{6}$
$v = \frac{d x}{d t} = A ω \cos (ω t + ϕ)$
At $t = 0, v = A ω \cos ϕ$
Now, $\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ and $\cos \frac{5 π}{6} = - \frac{\sqrt{3}}{2}$
As $v$ is negative at $t = 0, ϕ$ must be $5 π / 6$ .

PHXI14:OSCILLATIONS

364211 Two particles execute SHM with same amplitudes and same angular frequency on same straight line with same mean position. Given that during oscillation they cross each other in opposite direction when at a distance $A / 2$ from mean position. Find phase difference in the two $S H M s$ .

1

80^{\circ}

2

120^{\circ}

3

55^{\circ}

4

150^{\circ}

Explanation:

Figure shows that two respective particles $P$ and $Q^{'}$ in $S H M$ along with their corresponding particles in circular motion. Let $P$ moves in upward direction when crossing $Q^{'}$ at $A / 2$ as shown in Fig. (a), at this instant phase of $P$ is
$ϕ_{1} = \sin^{- 1} (\frac{1}{2}) = \frac{π}{6} (1)$
supporting img
Similarly, as shown in Fig. (b) we can take particle $Q^{'}$ is moving in downward direction (opposite $P^{'}$ ) at $A / 2$ , this implies its circular motion particle is in second quadrant thus its phase angle is
$ϕ_{2} = π - \sin^{- 1} (\frac{1}{2} = π - \frac{π}{6} = \frac{5 π}{6}) (2)$
As both are oscillating at same angular frequency their phase difference remains constant which can be given from Eqs. (1) and (2) as
$Δ ϕ = ϕ_{2} - ϕ_{1} = \frac{5 π}{6} - \frac{π}{6} = \frac{2 π}{3} r a d = 120^{\circ} .$

PHXI14:OSCILLATIONS

364212 Two pendulums have time periods T and $\frac{5 T}{4}$ .
They start S.H.M. at the same time from the mean position. What will be the phase difference between them after the bigger pendulum
complete one oscillation?

1

90^{\circ}

2

45^{\circ}

3

30^{\circ}

4

60^{\circ}

Explanation:

$\frac{5 T}{4} = T + \frac{T}{4}$
supporting img
By the time, the bigger pendulum makes one full oscillation, the smaller pendulum will make $(1 + \frac{1}{4})$ oscillation. The bigger pendulum will be in the mean position and the smaller one will be in the positive extreme position. Thus, phase difference $= 90^{\circ}$

PHXI14:OSCILLATIONS

364213 The equation of S.H.M is $y = a \sin (2 π n t + α)$ , then its phase at time $t$ is

1

α

2

2 π n t

3

2 π t

4

2 π n t + α

Explanation:

$y = a \sin (2 π n t + α)$ .
Its phase at time
$t = 2 π n t + α$

PHXI14:OSCILLATIONS

364214 Two simple harmonic motions exhibited by two particles $A$ and $B$ are given respectively by the following equations.
$y_{1} = a \sin [ω t + \frac{π}{6}]$
$y_{2} = a \sin [ω t + \frac{3 π}{6}]$
Phase difference between them is

1

\frac{π}{2}

2

\frac{π}{6}

3

\frac{π}{3}

4 Zero

Explanation:

Phase difference $Δ ϕ = \frac{3 π}{2} - \frac{π}{6} = \frac{π}{3}$ .

PHXI14:OSCILLATIONS

364210 The displacement-time equation of particle executing SHM is: $x = A \sin (ω t + ϕ)$ . At time $t = 0$ , position of the particle is $x = A / 2$ and it is moving along negative $x$ -direction. Then the angle $ϕ$ can be:

1

\frac{π}{6}

2

\frac{π}{4}

3

\frac{5 π}{6}

4

\frac{7 π}{6}

Explanation:

At $t = 0, x = \frac{A}{2}$
$\frac{A}{2} = A \sin ϕ or ϕ = \frac{π}{6} or ϕ = \frac{5 π}{6}$
$v = \frac{d x}{d t} = A ω \cos (ω t + ϕ)$
At $t = 0, v = A ω \cos ϕ$
Now, $\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ and $\cos \frac{5 π}{6} = - \frac{\sqrt{3}}{2}$
As $v$ is negative at $t = 0, ϕ$ must be $5 π / 6$ .

PHXI14:OSCILLATIONS

364211 Two particles execute SHM with same amplitudes and same angular frequency on same straight line with same mean position. Given that during oscillation they cross each other in opposite direction when at a distance $A / 2$ from mean position. Find phase difference in the two $S H M s$ .

1

80^{\circ}

2

120^{\circ}

3

55^{\circ}

4

150^{\circ}

Explanation:

Figure shows that two respective particles $P$ and $Q^{'}$ in $S H M$ along with their corresponding particles in circular motion. Let $P$ moves in upward direction when crossing $Q^{'}$ at $A / 2$ as shown in Fig. (a), at this instant phase of $P$ is
$ϕ_{1} = \sin^{- 1} (\frac{1}{2}) = \frac{π}{6} (1)$
supporting img
Similarly, as shown in Fig. (b) we can take particle $Q^{'}$ is moving in downward direction (opposite $P^{'}$ ) at $A / 2$ , this implies its circular motion particle is in second quadrant thus its phase angle is
$ϕ_{2} = π - \sin^{- 1} (\frac{1}{2} = π - \frac{π}{6} = \frac{5 π}{6}) (2)$
As both are oscillating at same angular frequency their phase difference remains constant which can be given from Eqs. (1) and (2) as
$Δ ϕ = ϕ_{2} - ϕ_{1} = \frac{5 π}{6} - \frac{π}{6} = \frac{2 π}{3} r a d = 120^{\circ} .$

PHXI14:OSCILLATIONS

364212 Two pendulums have time periods T and $\frac{5 T}{4}$ .
They start S.H.M. at the same time from the mean position. What will be the phase difference between them after the bigger pendulum
complete one oscillation?

1

90^{\circ}

2

45^{\circ}

3

30^{\circ}

4

60^{\circ}

Explanation:

$\frac{5 T}{4} = T + \frac{T}{4}$
supporting img
By the time, the bigger pendulum makes one full oscillation, the smaller pendulum will make $(1 + \frac{1}{4})$ oscillation. The bigger pendulum will be in the mean position and the smaller one will be in the positive extreme position. Thus, phase difference $= 90^{\circ}$

PHXI14:OSCILLATIONS

364213 The equation of S.H.M is $y = a \sin (2 π n t + α)$ , then its phase at time $t$ is

1

α

2

2 π n t

3

2 π t

4

2 π n t + α

Explanation:

$y = a \sin (2 π n t + α)$ .
Its phase at time
$t = 2 π n t + α$

PHXI14:OSCILLATIONS

364214 Two simple harmonic motions exhibited by two particles $A$ and $B$ are given respectively by the following equations.
$y_{1} = a \sin [ω t + \frac{π}{6}]$
$y_{2} = a \sin [ω t + \frac{3 π}{6}]$
Phase difference between them is

1

\frac{π}{2}

2

\frac{π}{6}

3

\frac{π}{3}

4 Zero

Explanation:

Phase difference $Δ ϕ = \frac{3 π}{2} - \frac{π}{6} = \frac{π}{3}$ .

PHXI14:OSCILLATIONS

364210 The displacement-time equation of particle executing SHM is: $x = A \sin (ω t + ϕ)$ . At time $t = 0$ , position of the particle is $x = A / 2$ and it is moving along negative $x$ -direction. Then the angle $ϕ$ can be:

1

\frac{π}{6}

2

\frac{π}{4}

3

\frac{5 π}{6}

4

\frac{7 π}{6}

Explanation:

At $t = 0, x = \frac{A}{2}$
$\frac{A}{2} = A \sin ϕ or ϕ = \frac{π}{6} or ϕ = \frac{5 π}{6}$
$v = \frac{d x}{d t} = A ω \cos (ω t + ϕ)$
At $t = 0, v = A ω \cos ϕ$
Now, $\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ and $\cos \frac{5 π}{6} = - \frac{\sqrt{3}}{2}$
As $v$ is negative at $t = 0, ϕ$ must be $5 π / 6$ .

PHXI14:OSCILLATIONS

364211 Two particles execute SHM with same amplitudes and same angular frequency on same straight line with same mean position. Given that during oscillation they cross each other in opposite direction when at a distance $A / 2$ from mean position. Find phase difference in the two $S H M s$ .

1

80^{\circ}

2

120^{\circ}

3

55^{\circ}

4

150^{\circ}

Explanation:

Figure shows that two respective particles $P$ and $Q^{'}$ in $S H M$ along with their corresponding particles in circular motion. Let $P$ moves in upward direction when crossing $Q^{'}$ at $A / 2$ as shown in Fig. (a), at this instant phase of $P$ is
$ϕ_{1} = \sin^{- 1} (\frac{1}{2}) = \frac{π}{6} (1)$
supporting img
Similarly, as shown in Fig. (b) we can take particle $Q^{'}$ is moving in downward direction (opposite $P^{'}$ ) at $A / 2$ , this implies its circular motion particle is in second quadrant thus its phase angle is
$ϕ_{2} = π - \sin^{- 1} (\frac{1}{2} = π - \frac{π}{6} = \frac{5 π}{6}) (2)$
As both are oscillating at same angular frequency their phase difference remains constant which can be given from Eqs. (1) and (2) as
$Δ ϕ = ϕ_{2} - ϕ_{1} = \frac{5 π}{6} - \frac{π}{6} = \frac{2 π}{3} r a d = 120^{\circ} .$

PHXI14:OSCILLATIONS

364212 Two pendulums have time periods T and $\frac{5 T}{4}$ .
They start S.H.M. at the same time from the mean position. What will be the phase difference between them after the bigger pendulum
complete one oscillation?

1

90^{\circ}

2

45^{\circ}

3

30^{\circ}

4

60^{\circ}

Explanation:

$\frac{5 T}{4} = T + \frac{T}{4}$
supporting img
By the time, the bigger pendulum makes one full oscillation, the smaller pendulum will make $(1 + \frac{1}{4})$ oscillation. The bigger pendulum will be in the mean position and the smaller one will be in the positive extreme position. Thus, phase difference $= 90^{\circ}$

PHXI14:OSCILLATIONS

364213 The equation of S.H.M is $y = a \sin (2 π n t + α)$ , then its phase at time $t$ is

1

α

2

2 π n t

3

2 π t

4

2 π n t + α

Explanation:

$y = a \sin (2 π n t + α)$ .
Its phase at time
$t = 2 π n t + α$

PHXI14:OSCILLATIONS

364214 Two simple harmonic motions exhibited by two particles $A$ and $B$ are given respectively by the following equations.
$y_{1} = a \sin [ω t + \frac{π}{6}]$
$y_{2} = a \sin [ω t + \frac{3 π}{6}]$
Phase difference between them is

1

\frac{π}{2}

2

\frac{π}{6}

3

\frac{π}{3}

4 Zero

Explanation:

Phase difference $Δ ϕ = \frac{3 π}{2} - \frac{π}{6} = \frac{π}{3}$ .