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Two pendulums suspended from the same point have lengths of \(2\) m and \(0.5\) m. If they are displaced slightly and released, then they will be in the same phase when the small pendulum has completed:

1. \(2\) oscillations

2. \(4\) oscillations

3. \(3\) oscillations

4. \(5\) oscillations

Subtopic: Simple Harmonic Motion |

65%

From NCERT

AIPMT - 1998

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If the time of mean position from amplitude (extreme) position is 6 seconds, then the frequency of SHM will be:

1. | \(0.01\) Hz | 2. | \(0.02\) Hz |

3. | \(0.03\) Hz | 4. | \(0.04\) Hz |

Subtopic: Simple Harmonic Motion |

68%

From NCERT

AIPMT - 1998

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Hints

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The frequency of a spring is \(n\) after suspending mass \(M.\) Now, after mass \(4M\) mass is suspended from the spring, the frequency will be:

1. | \(2n\) | 2. | \(n/2\) |

3. | \(n\) | 4. | none of the above |

Subtopic: Spring mass system |

81%

From NCERT

AIPMT - 1998

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Hints

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A spring elongates by a length 'L' when a mass 'M' is suspended to it. Now a tiny mass 'm' is attached to the mass 'M' and then released. The new time period of oscillation will be:

1. \(2 \pi \sqrt{\frac{\left(\right. M + m \left.\right) l}{Mg}}\)

2. \(2 \pi \sqrt{\frac{ml}{Mg}}\)

3. \(2 \pi \sqrt{L / g}\)

4. \(2 \pi \sqrt{\frac{Ml}{\left(\right. m + M \left.\right) g}}\)

Subtopic: Spring mass system |

59%

From NCERT

AIPMT - 1999

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The frequency of a simple pendulum in a free-falling lift will be:

1. zero

2. infinite

3. can't say

4. finite

Subtopic: Angular SHM |

66%

From NCERT

AIPMT - 1999

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Two spherical bobs of masses \(M_A\) and \(M_B\) are hung vertically from two strings of length \(l_A\)* *and \(l_B\) respectively. If they are executing SHM with frequency as per the relation \(f_A=2f_B,\) Then:

1. ${l}_{A}=\frac{{l}_{B}}{4}$

2. ${l}_{A}=4{l}_{B}$

3. ${l}_{A}=2{l}_{B}$ $\&$ ${M}_{A}=2{M}_{B}$

4. ${l}_{A}=\frac{{l}_{B}}{2}$ $\&$ ${M}_{A}=\frac{{M}_{B}}{2}$

Subtopic: Angular SHM |

72%

From NCERT

AIPMT - 2000

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The total energy of the particle performing SHM depends on:

1. \(k,\) \(a,\) \(m\)

2. \(k,\) \(a\)

3. \(k,\) \(a\), \(x \)

4. \(k,\) \(x \)

Subtopic: Energy of SHM |

68%

From NCERT

AIPMT - 2001

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When a mass is suspended separately by two different springs, in successive order, then the time period of oscillations is \(t _1\) and \(t_2\) respectively. If it is connected by both springs as shown in the figure below, then the time period of oscillation becomes \(t_0.\) The correct relation between \(t_0,\) \(t_1\) & \(t_2\) is:

1. ${{\mathrm{t}}_{0}}^{2}={{\mathrm{t}}_{1}}^{2}+{{\mathrm{t}}_{2}}^{2}$

2. ${{\mathrm{t}}_{0}}^{-2}={{\mathrm{t}}_{1}}^{-2}+{{\mathrm{t}}_{2}}^{-2}$

3. ${{\mathrm{t}}_{0}}^{-1}={{\mathrm{t}}_{1}}^{-1}+{{\mathrm{t}}_{2}}^{-1}$

4. ${\mathrm{t}}_{0}={\mathrm{t}}_{1}+{\mathrm{t}}_{2}$

Subtopic: Combination of Springs |

68%

From NCERT

AIPMT - 2002

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The displacement between the maximum potential energy position and maximum kinetic energy position for a particle executing simple harmonic motion is:

1. $\pm \frac{\mathrm{a}}{2}$

2. $+\mathrm{a}$

3. $\pm \mathrm{a}$

4. $-1$

Subtopic: Energy of SHM |

73%

From NCERT

AIPMT - 2002

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The time period of a mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be:

1. T/4

2. T

3. T/2

4. 2T

Subtopic: Spring mass system |

72%

From NCERT

AIPMT - 2003

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Hints

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