­­A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. If the lift accelerates upwards with an acceleration g4 , then the period of the pendulum will be

(1) T

(2) T4

(3) 2T5

(4) 2T5

Subtopic:  Angular SHM |
 83%
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The total energy of a particle, executing simple harmonic motion is

(1)     x                 

(2)     x2

(3)   Independent of x 

(4)     x1/2

Subtopic:  Energy of SHM |
 76%
From NCERT
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The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle θ and then released. The bob will then pass through its equilibrium position with a speed v, where v equals

(1) 2gl(1-sinθ)

(2) 2gl(1+cosθ)

(3) 2gl(1-cosθ)

(4) 2gl(1+sinθ)

Subtopic:  Angular SHM |
 76%
From NCERT
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A body is executing Simple Harmonic Motion. At a displacement x its potential energy is E1 and at a displacement y its potential energy is E2. The potential energy E at displacement x+y is 

(1)  E=E1+E2   

(2)  E=E1+E2

(3)   E=E1+E2           

(4)  None of these.

Subtopic:  Energy of SHM |
 56%
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. lT= constant
2. l2T= constant
3. lT2= constant
4. l2T2= constant

Subtopic:  Angular SHM |
 84%
From NCERT
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The equation of motion of a particle is d2ydt2+Ky=0 where K is positive constant. The time period of the motion is given by

(1) 2πK             

(2) 2πK

(3) 2πK           

(4)  2πK

Subtopic:  Simple Harmonic Motion |
 77%
From NCERT
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The kinetic energy of a particle executing S.H.M. is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of its oscillation is -

(1) π5sec       

(2) 2π sec

(3) 20π sec   

(4) 5π sec

Subtopic:  Energy of SHM |
 66%
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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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 |
 60%
From NCERT
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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 |
 83%
From NCERT
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