A particle of mass m oscillates with simple harmonic motion between points x1 and x2, the equilibrium position being O. Its potential energy is plotted. It will be as given below in the graph:

1. 2.
3. 4.
Subtopic:  Energy of SHM |
 84%
From NCERT
AIPMT - 2003
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Equation of a simple harmonic motion is  given by x = asinωt. For which value of x, kinetic energy is equal to the potential energy?

1.  x = ± a

2.  x = ± a2

3.  x = ± a2

4.  x = ± 3a2

Subtopic:  Energy of SHM |
 82%
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The potential energy of a simple harmonic oscillator, when the particle is halfway to its endpoint, will be:
1. \(\frac{2E}{3}\)
2. \(\frac{E}{8}\)
3. \(\frac{E}{4}\)
4. \(\frac{E}{2}\)

Subtopic:  Energy of SHM |
 79%
From NCERT
AIPMT - 2003
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When the displacement is half the amplitude in an SHM, the ratio of potential energy to the total energy is:
1. 1 / 2

2. 1 / 4

3. 1

4. 1 / 8

Subtopic:  Energy of SHM |
 81%
From NCERT
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The kinetic energy (K) of a simple harmonic oscillator varies with displacement (x) as shown. The period of the oscillation will be: (mass of oscillator is 1 kg)

                     

1. π2 sec 2.  12 sec
3.  π sec 4. 1 sec
Subtopic:  Energy of SHM |
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A block of mass \(4~\text{kg}\) hangs from a spring of spring constant \(k = 400~\text{N/m}\). The block is pulled down through \(15~\text{cm}\) below the equilibrium position and released. What is its kinetic energy when the block is \(10~\text{cm}\) below the equilibrium position? [Ignore gravity]
1. \(5~\text{J}\)
2. \(2.5~\text{J}\)
3. \(1~\text{J}\)
4. \(1.9~\text{J}\)

Subtopic:  Energy of SHM |
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Kinetic energy of a particle executing simple harmonic motion in straight line is \(pv^2\) and potential energy is \(qx^2,\) where \(v\) is speed at distance \(x\) from the mean position. The time period of the SHM is given by the expression:

1. 2πqp

2. 2πpq

3. 2πqp+q

4. 2πpp+q

Subtopic:  Energy of 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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If the potential energy U (in J) of a body executing SHM is given by U = 20 + 10 (sin2100πt), then the minimum potential energy of the body will be:

1. Zero 2. 30 J
3. 20 J 4. 40 J
Subtopic:  Energy of SHM |
 72%
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The displacement between the maximum potential energy position and maximum kinetic energy position for a particle executing simple harmonic motion is:

1. ±a2

2. +a

3. ±a

4. -1

Subtopic:  Energy of SHM |
 73%
From NCERT
AIPMT - 2002
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