An ideal spring with spring-constant K is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is -

(1) 4 Mg/K         

(2) 2 Mg/K

(3) Mg/K             

(4) Mg/2K

Subtopic:  Combination of Springs |
 56%
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A particle of mass m is attached to three identical springs A, B and C each of force constant k a shown in figure. If the particle of mass m is pushed slightly against the spring A and released then the time period of oscillations is -

(a) 2π2mk          (b) 2πm2k

(c) 2πmk            (d) 2πm3k

              

                 

Subtopic:  Combination of Springs |
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The graph shows the variation of displacement of a particle executing SHM with time. We infer from this graph that:

   
 

1. the force is zero at the time \(T/8\).
2. the velocity is maximum at the time \(T/4\).
3. the acceleration is maximum at the time \(T\).
4. the P.E. is equal to the total energy at the time \(T/4\).

Subtopic:  Energy of SHM |
 65%
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For a particle executing SHM the displacement \(x \) is given by, \(A\cos \omega t.\)  Identify the graph which represents the variation of potential energy (P.E.) as a function of time \(t\) and displacement \(x.\)

   
1. I, III
2. II, IV
3. II, III
4. I, IV

Subtopic:  Energy of SHM |
 56%
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The velocity-time diagram of a harmonic oscillator is shown in the adjoining figure. The frequency of oscillation is

                                     

(1) 25 Hz           

(2) 50 Hz

(3) 12.25 Hz       

(4) 33.3 Hz

Subtopic:  Simple Harmonic Motion |
 73%
From NCERT
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The variation of potential energy of harmonic oscillator is as shown in figure. The spring constant is

                                                      

(1) 1 ×102 N/m               

(2) 150 N/m

(3) 0.667 × 102 N/m       

(4) 3 × 102 N/m

Subtopic:  Energy of SHM |
 63%
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A body performs S.H.M. . Its kinetic energy K varies with time t as indicated by graph

(a)    (b) 

(c)      (d) 

 

Subtopic:  Energy of SHM |
 75%
From NCERT
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The displacement of a body executing SHM is given by x= A sin (2πt + π/3). The first time from t = 0 when the velocity is maximum is :
1. 0.33 sec
2. 0.16 sec
3. 0.25 sec
4. 0.5 sec

Subtopic:  Phasor Diagram |
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The time period of a particle executing SHM is 8 sec. At t = 0 it is at the mean position. The ratio of the distance covered by the particle in the 1st second to the 2nd second is :

1. 12 + 1

2. 2

3. 12 

4. 2 + 1

Subtopic:  Phasor Diagram |
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Two particles are in SHM in a straight line. Amplitude A and time period T of both the particles are equal. At time t = 0, one particle is at displacement Y1 = + A and the other at Y2 = -A/2, and they are approaching towards each other. After what time they cross each other?
1. T/3
2. T/4
3. 5T/6
4. T/6

Subtopic:  Phasor Diagram |
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