A particle executes simple harmonic motion along a straight line with an amplitude A. The potential energy is maximum when the displacement is

(1)     ±A          

(2)    Zero

(3)     ±A2         

(4)   ±A2

Subtopic:  Energy of SHM |
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The potential energy of a particle with displacement X depends as U(X). The motion is simple harmonic, when (K is a positive constant)

(1) U=KX22                 

(2) U=KX2   

(3)  U=K                       

(4) U=KX 

Subtopic:  Energy of SHM |
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The angular velocity and the amplitude of a simple pendulum is ω and a respectively. At a displacement X from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is 

(1)     X2ω2a2-X2ω2       

(2)    X2/a2-x2

(3)   a2-X2ω2/X2ω2       

(4)   (a2-x2)/X2

Subtopic:  Energy of SHM |
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A particle is executing simple harmonic motion with frequency f. The frequency at which its kinetic energy changes into potential energy, will be:

1.   f/2         

2.  f

3.   2 f        

4.  4 f

Subtopic:  Energy of SHM |
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There is a body having mass m and performing S.H.M. with amplitude a. There is a restoring force ,F=-Kx where x is the displacement. The total energy of body depends upon -

(1)   K, x         

(2)  K, a

(3)   K, a, x    

(4)  K, a, v

Subtopic:  Energy of SHM |
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The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where E is the total energy)

(1)    18E       

(2)        14E

(3)    12E       

(4)        23E

Subtopic:  Energy of SHM |
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A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of displacement x. Which of the following statements is true ?

(1)  P.E. is maximum when x = 0

(2)  K.E. is maximum when x = 0

(3)  T.E. is zero when x = 0

(4)  K.E. is maximum when x is maximum

Subtopic:  Energy of SHM |
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­­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 |
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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 |
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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 |
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