A point performs simple harmonic oscillation of period T and the equation of motion is given by x= a sin ωt +π/6.After the elapse of what fraction of the time period the velocity of the point will be equal to half to its maximum velocity?

(1)T8

(2) T6

(3) T3

(4) T12

Subtopic:  Simple Harmonic Motion |
 67%
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An SHM has an amplitude \(a\) and  a time period \(T.\) The maximum velocity will be:
1. \({4a \over T}\)       
2.  \({2a \over T}\)
3. \({2 \pi \over T}\)
4. \({2a \pi \over T}\)  

Subtopic:  Simple Harmonic Motion |
 90%
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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 |
 85%
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The angular velocities of three bodies in simple harmonic motion are ω1,ω2,ω3 with their respective amplitudes as A1,A2,A3. If all the three bodies have same mass and maximum velocity, then

(a) A1ω1=A2ω2=A3ω3        (b)  A1ω12=A2ω22=A3A32

(b) A12ω1=A22ω2=A32ω3   (d) A12ω12=A22ω22=A2  

Subtopic:  Simple Harmonic Motion |
 89%
From NCERT
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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%
From NCERT
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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 |
 93%
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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 |
 90%
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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 |
 61%
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The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is

(a)      144π2m/sec2        (b)     144m/sec2

c)      144π2m/sec2           (d)     288π2m/sec2   

Subtopic:  Simple Harmonic Motion |
 88%
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A particle moving along the x-axis executes simple harmonic motion, then the force acting on it is given by

(1)   – A Kx           

(2)   A cos (Kx)

(3)   A exp (– Kx

(4)   A Kx

Subtopic:  Simple Harmonic Motion |
 84%
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