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. \(\frac{L^2}{T^2}= \text{constant}\)
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. lA=lB4

2. lA=4lB

3. lA=2lB & MA=2MB

4. lA=lB2 & MA=MB2

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 t2, where y is in meters and t is in seconds. If g = 10 m/s2, 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 θ, is given by:

1.   2πLgcosθ               

2.  2πLgsinθ

3.  2πLg                      

4. 2πLgtanθ

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 T1. When the lift moves downward with constant velocity, then the time period becomes T2. 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.  3T2

2.  3T4

3.  2T3

4.  4T3

Subtopic:  Angular SHM |
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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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