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In a spring pendulum, in place of mass, a liquid is used. If liquid leaks out continuously, then the time period of the spring pendulum:

1. | Decreases continuously |

2. | Increases continuously |

3. | First increases and then decreases |

4. | First decreases and then increases |

Subtopic: Spring mass system |

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Equation of a simple harmonic motion is given by \(x= a\sin \omega t\). For which value of \(x\), kinetic energy is equal to the potential energy?

1. \(x = \pm a\)

2. \(x = \pm \frac{a}{2}\)

3. \(x = \pm \frac{a}{\sqrt{2}}\)

4. \(x = \pm \frac{\sqrt{3}a}{2}\)

1. \(x = \pm a\)

2. \(x = \pm \frac{a}{2}\)

3. \(x = \pm \frac{a}{\sqrt{2}}\)

4. \(x = \pm \frac{\sqrt{3}a}{2}\)

Subtopic: Energy of SHM |

82%

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The displacement \( x\) of a particle varies with time \(t\) as \(x = A sin\left (\frac{2\pi t}{T} +\frac{\pi}{3} \right)\)$\mathrm{}$. The time taken by the particle to reach from \(x = \frac{A}{2} \) to \(x = -\frac{A}{2} \) will be:

1. | \(\frac{T}{2}\) | 2. | \(\frac{T}{3}\) |

3. | \(\frac{T}{12}\) | 4. | \(\frac{T}{6}\) |

Subtopic: Phasor Diagram |

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Force on a particle \(F\) varies with time \(t\) as shown in the given graph. The displacement \(x\) vs time \(t\) graph corresponding to the force-time graph will be:

1. | 2. | ||

3. | 4. |

Subtopic: Linear SHM |

66%

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A particle executes linear SHM between \(x=A.\) The time taken to go from \(0\) to \(A/2\) is \(T_1\) and to go from \(A/2\) to \(A\) is \(T_2\) then:

1. | \(T_1<T_2\) | 2. | \(T_1>T_2\) |

3. | \(T_1=T_2\) |
4. | \(T_1= 2T_2\) |

Subtopic: Linear SHM |

71%

From NCERT

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Two simple pendulums of length \(1~\text{m}\) and \(16~\text{m}\) are in the same phase at the mean position at any instant. If \(T\) is the time period of the smaller pendulum, then the minimum time after which they will again be in the same phase will be:

1. \(\frac{3T}{2}\)

2. \(\frac{3T}{4}\)

3. \(\frac{2T}{3}\)

4. \(\frac{4T}{3}\)

1. \(\frac{3T}{2}\)

2. \(\frac{3T}{4}\)

3. \(\frac{2T}{3}\)

4. \(\frac{4T}{3}\)

Subtopic: Angular SHM |

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A particle executes SHM with a time period of \(4~\text{s}\). The time taken by the particle to go directly from its mean position to half of its amplitude will be:

1. \(\frac{1}{3}~\text{s}\)

2. \(1~\text{s}\)

3. \(\frac{1}{2}~\text{s}\)

4. \(2~\text{s}\)

1. \(\frac{1}{3}~\text{s}\)

2. \(1~\text{s}\)

3. \(\frac{1}{2}~\text{s}\)

4. \(2~\text{s}\)

Subtopic: Linear SHM |

75%

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The graph between the velocity \((v)\) of a particle executing SHM and its displacement \((x)\) is shown in the figure. The time period of oscillation for this SHM will be:

1. \(\sqrt{\frac{\alpha}{\beta}}\)

2. \(2\pi\sqrt{\frac{\alpha}{\beta}}\)

3. \(2\pi\left(\frac{\beta}{\alpha}\right)\)

4. \(2\pi\left(\frac{\alpha}{\beta}\right)\)

Subtopic: Simple Harmonic Motion |

65%

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A simple pendulum attached to the ceiling of a stationary lift has a time period of 1 s. The distance y covered by the lift moving downward varies with time as y = 3.75 ${\mathrm{t}}^{2}$, where y is in meters and t is in seconds. If g = 10 $\mathrm{m}/{\mathrm{s}}^{2}$, then the time period of the pendulum will be:

1. | 4 s | 2. | 6 s |

3. | 2 s | 4. | 12 s |

Subtopic: Types of Motion | Simple Harmonic Motion | Angular SHM |

61%

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The time period of the spring-mass system depends upon:

1. | the gravity of the earth | 2. | the mass of the block |

3. | spring constant | 4. | both (2) & (3) |

Subtopic: Spring mass system |

88%

From NCERT

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