There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is T. If the resultant acceleration becomes g/4, then the new time period of the pendulum is 

(1) 0.8 T

(2) 0.25 T

(3) 2 T

(4) 4 T

Subtopic:  Angular SHM |
 83%
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When two displacements represented by y1=asin(ωt) and y2=bcos(ωt) are superimposed,the motion is -

(1) not a simple harmonic

(2) simple harmonic with amplitude a/b

(3) simple harmonic with amplitude a2 + b2

(4) simple harmonic with amplitude (a+b)/2

Subtopic:  Simple Harmonic Motion |
 94%
From NCERT
NEET - 2015
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A S.H.M. has amplitude ‘a’ and  time period T. The maximum velocity will be -

(1) 4aT           

(2) 2aT

(3) 2πaT        

(4) 2πaT                       

Subtopic:  Simple Harmonic Motion |
 89%
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Two particles P and Q start from origin and execute Simple Harmonic Motion along X-axis with same amplitude but with periods 3 seconds and 6 seconds respectively. The ratio of the velocities of P and Q when they meet is -

(1)  1 : 2                 

(2)  2 : 1

(3)  2 : 3                 

(4)  3 : 2

Subtopic:  Simple Harmonic Motion |
 89%
From NCERT
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The angular velocities of three bodies in simple harmonic motion are \(\omega_1, \omega_2, \omega_3\) with their respective amplitudes as \(A_1, A_2, A_3.\) If all the three bodies have the same mass and maximum velocity, then:
1. \(A_1 \omega_1=A_2 \omega_2=A_3 \omega_3\)
2. \(A_1 \omega_1^2=A_2 \omega_2^2=A_3 \omega_3^2\)
3. \(A_1^2 \omega_1=A_2^2 \omega_2=A_3^2 \omega_3\)
4. \(A_1^2 \omega_1^2=A_2^2 \omega_2^2=A^2\)
Subtopic:  Simple Harmonic Motion |
 91%
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The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes 83cm/sec
 will be

(1) 23cm             

(2) 3cm

(3) 1 cm                   

(4) 2 cm

Subtopic:  Simple Harmonic Motion |
 70%
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The maximum velocity of a simple harmonic motion represented by y=3 sin 100t+π6 is given by

(1)   300       

(2)   3π6

(3)   100         

(4)   π6

Subtopic:  Simple Harmonic Motion |
 94%
From NCERT
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The displacement equation of a particle is x=3sin 2t+4cos 2t  The amplitude and maximum velocity will be respectively

(a)       5, 10       (b)         3, 2

c)         4, 2        (d)         3, 4

Subtopic:  Simple Harmonic Motion |
 92%
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The instantaneous displacement of a simple pendulum oscillator is given by x=A cos ωt+π4 . Its speed will be maximum at time

(1) π4ω                   

(2) π2ω

(3) πω                   

(4) 2πω        

Subtopic:  Simple Harmonic Motion |
 59%
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The displacement of a particle moving in S.H.M. at any instant is given by y=a sinωt . The acceleration after time t=T4 (where T is the time period) -

1. aω                         

2.-aω

3. aω2                         

4.  -aω2

 

Subtopic:  Simple Harmonic Motion |
 88%
From NCERT
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