If the length of a pendulum is made 9 times and mass of the bob is made 4 times, then the value of time period will become:

1. 3T

2. 3/2T

3. 4T

4. 2T

Subtopic: Angular SHM |

82%

From NCERT

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In a simple pendulum, the period of oscillation \(T\)* *is related to length of the pendulum \(L\) as:

1. \(\frac{L}{T}= \text{constant}\)

2. \(\frac{L^2}{T}= \text{constant}\)

3. \(\frac{L}{T^2}= \text{constant}\)

4. \(\frac{L^2}{T^2}= \text{constant}\)

1. \(\frac{L}{T}= \text{constant}\)

2. \(\frac{L^2}{T}= \text{constant}\)

3. \(\frac{L}{T^2}= \text{constant}\)

4. \(\frac{L^2}{T^2}= \text{constant}\)

Subtopic: Angular SHM |

84%

From NCERT

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A pendulum has time period *T*. If it is taken on to another planet having acceleration due to gravity half and mass 9 times that of the earth, then its time period on the other planet will be:

1. | \(\sqrt{\mathrm{T}} \) | 2. | \(T \) |

3. | \(\mathrm{T}^{1 / 3} \) | 4. | \(\sqrt{2} \mathrm{~T}\) |

Subtopic: Angular SHM |

82%

From NCERT

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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 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 |

67%

From NCERT

AIPMT - 1999

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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%

From NCERT

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The period of oscillation of a simple pendulum of length *\(\mathrm{L}\)* suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $\theta $, is given by:

1. $2\mathrm{\pi}\sqrt{\frac{\mathrm{L}}{\mathrm{gcos\theta}}}$

2. $2\mathrm{\pi}\sqrt{\frac{\mathrm{L}}{\mathrm{gsin}\theta}}$

3. $2\mathrm{\pi}\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$

4. $2\mathrm{\pi}\sqrt{\frac{\mathrm{L}}{\mathrm{gtan\theta}}}$

Subtopic: Angular SHM |

60%

From NCERT

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A simple pendulum hanging from the ceiling of a stationary lift has a time period T_{1}. When the lift moves downward with constant velocity, then the time period becomes T_{2}. It can be concluded that:

1. | \(T_2 ~\text{is infinity} \) | 2. | \(\mathrm{T}_2>\mathrm{T}_1 \) |

3. | \(\mathrm{T}_2<\mathrm{T}_1 \) | 4. | \(T_2=T_1\) |

Subtopic: Angular SHM |

61%

From NCERT

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Two simple pendulums of length 1 m and 16 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. $\frac{3\mathrm{T}}{2}$

2. $\frac{3\mathrm{T}}{4}$

3. $\frac{2\mathrm{T}}{3}$

4. $\frac{4\mathrm{T}}{3}$

Subtopic: Angular SHM |

From NCERT

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A simple pendulum of mass m swings about point B between extreme positions A and C. Net force acting on the bob at these three points is correctly shown by:

1. | 2. | ||

3. | 4. |

Subtopic: Angular SHM |

From NCERT

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