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 |
 71%
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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 |
 79%
From NCERT
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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 |
 86%
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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 |
 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 |
 77%
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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%
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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%
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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 |
 78%
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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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