QBManager

PHXI14:OSCILLATIONS

364099 The particle executing simple harmonic motion has a kinetic energy $K_{0} \cos^{2} ω t$ . The maximum values of the potential energy and the total energy are respectively:

1 0 and

2 K_{0}

2

\frac{K_{0}}{2}

and

K_{0}

3

K_{0}

and

2 K_{0}

4

K_{0}

and

K_{0}

Explanation:

In simple harmonic motion, the total energy of the particle is constant at all instants which is totally kinetic when particle is passing through the mean position and is totally potential when particle is passing through the extreme position.
Now, $E_{k} = K_{0} \cos^{2} ω t$
$∴ {(E_{K})}_{M a x} = K_{0}$
So, ${(E_{p})}_{Max} = K_{0}$ and $(E)_{total} = K_{0}$

PHXI14:OSCILLATIONS

364100 The average energy in one time period in simple harmonic motion is

1

\frac{1}{2} m ω^{2} a^{2}

2

m ω^{2} a^{2}

3

1.5 m ω^{2} a^{2}

4 None of these

Explanation:

Total energy in SHM is always constant and it is equal to $\frac{1}{2} m ω^{2} a^{2}$ .
So $(T . E)_{avg} = \frac{1}{2} m ω^{2} a^{2}$

PHXI14:OSCILLATIONS

364101 The time period of the variation of potential energy of a particle executing SHM with period $T$ is

1

T

2

T / 4

3

T / 2

4

2 T

Explanation:

The potential energy of a particle executing SHM is periodic with time period $\frac{T}{2}$ .

PHXI14:OSCILLATIONS

364102 For a particle executing S.H.M. The
displacement $x$ is given by $x = A \cos ω t$ . Identity the graph which represents the variation of potential energy (P.E.) as a function of time $t$ and displacement $x$
supporting img

1 II, IV

2 I, III

3 I, IV

4 II, III

Explanation:

At time $t = 0, x = A,$ hence potential energy should be maximum. Therefore graph I is correct. Further in graph III. Potential energy is mimimum at $x = 0,$ hence this is also correct.

PHXI14:OSCILLATIONS

364099 The particle executing simple harmonic motion has a kinetic energy $K_{0} \cos^{2} ω t$ . The maximum values of the potential energy and the total energy are respectively:

1 0 and

2 K_{0}

2

\frac{K_{0}}{2}

and

K_{0}

3

K_{0}

and

2 K_{0}

4

K_{0}

and

K_{0}

Explanation:

In simple harmonic motion, the total energy of the particle is constant at all instants which is totally kinetic when particle is passing through the mean position and is totally potential when particle is passing through the extreme position.
Now, $E_{k} = K_{0} \cos^{2} ω t$
$∴ {(E_{K})}_{M a x} = K_{0}$
So, ${(E_{p})}_{Max} = K_{0}$ and $(E)_{total} = K_{0}$

PHXI14:OSCILLATIONS

364100 The average energy in one time period in simple harmonic motion is

1

\frac{1}{2} m ω^{2} a^{2}

2

m ω^{2} a^{2}

3

1.5 m ω^{2} a^{2}

4 None of these

Explanation:

Total energy in SHM is always constant and it is equal to $\frac{1}{2} m ω^{2} a^{2}$ .
So $(T . E)_{avg} = \frac{1}{2} m ω^{2} a^{2}$

PHXI14:OSCILLATIONS

364101 The time period of the variation of potential energy of a particle executing SHM with period $T$ is

1

T

2

T / 4

3

T / 2

4

2 T

Explanation:

The potential energy of a particle executing SHM is periodic with time period $\frac{T}{2}$ .

PHXI14:OSCILLATIONS

364102 For a particle executing S.H.M. The
displacement $x$ is given by $x = A \cos ω t$ . Identity the graph which represents the variation of potential energy (P.E.) as a function of time $t$ and displacement $x$
supporting img

1 II, IV

2 I, III

3 I, IV

4 II, III

Explanation:

At time $t = 0, x = A,$ hence potential energy should be maximum. Therefore graph I is correct. Further in graph III. Potential energy is mimimum at $x = 0,$ hence this is also correct.

PHXI14:OSCILLATIONS

364099 The particle executing simple harmonic motion has a kinetic energy $K_{0} \cos^{2} ω t$ . The maximum values of the potential energy and the total energy are respectively:

1 0 and

2 K_{0}

2

\frac{K_{0}}{2}

and

K_{0}

3

K_{0}

and

2 K_{0}

4

K_{0}

and

K_{0}

Explanation:

In simple harmonic motion, the total energy of the particle is constant at all instants which is totally kinetic when particle is passing through the mean position and is totally potential when particle is passing through the extreme position.
Now, $E_{k} = K_{0} \cos^{2} ω t$
$∴ {(E_{K})}_{M a x} = K_{0}$
So, ${(E_{p})}_{Max} = K_{0}$ and $(E)_{total} = K_{0}$

PHXI14:OSCILLATIONS

364100 The average energy in one time period in simple harmonic motion is

1

\frac{1}{2} m ω^{2} a^{2}

2

m ω^{2} a^{2}

3

1.5 m ω^{2} a^{2}

4 None of these

Explanation:

Total energy in SHM is always constant and it is equal to $\frac{1}{2} m ω^{2} a^{2}$ .
So $(T . E)_{avg} = \frac{1}{2} m ω^{2} a^{2}$

PHXI14:OSCILLATIONS

364101 The time period of the variation of potential energy of a particle executing SHM with period $T$ is

1

T

2

T / 4

3

T / 2

4

2 T

Explanation:

The potential energy of a particle executing SHM is periodic with time period $\frac{T}{2}$ .

PHXI14:OSCILLATIONS

364102 For a particle executing S.H.M. The
displacement $x$ is given by $x = A \cos ω t$ . Identity the graph which represents the variation of potential energy (P.E.) as a function of time $t$ and displacement $x$
supporting img

1 II, IV

2 I, III

3 I, IV

4 II, III

Explanation:

At time $t = 0, x = A,$ hence potential energy should be maximum. Therefore graph I is correct. Further in graph III. Potential energy is mimimum at $x = 0,$ hence this is also correct.

PHXI14:OSCILLATIONS

364099 The particle executing simple harmonic motion has a kinetic energy $K_{0} \cos^{2} ω t$ . The maximum values of the potential energy and the total energy are respectively:

1 0 and

2 K_{0}

2

\frac{K_{0}}{2}

and

K_{0}

3

K_{0}

and

2 K_{0}

4

K_{0}

and

K_{0}

Explanation:

In simple harmonic motion, the total energy of the particle is constant at all instants which is totally kinetic when particle is passing through the mean position and is totally potential when particle is passing through the extreme position.
Now, $E_{k} = K_{0} \cos^{2} ω t$
$∴ {(E_{K})}_{M a x} = K_{0}$
So, ${(E_{p})}_{Max} = K_{0}$ and $(E)_{total} = K_{0}$

PHXI14:OSCILLATIONS

364100 The average energy in one time period in simple harmonic motion is

1

\frac{1}{2} m ω^{2} a^{2}

2

m ω^{2} a^{2}

3

1.5 m ω^{2} a^{2}

4 None of these

Explanation:

Total energy in SHM is always constant and it is equal to $\frac{1}{2} m ω^{2} a^{2}$ .
So $(T . E)_{avg} = \frac{1}{2} m ω^{2} a^{2}$

PHXI14:OSCILLATIONS

364101 The time period of the variation of potential energy of a particle executing SHM with period $T$ is

1

T

2

T / 4

3

T / 2

4

2 T

Explanation:

The potential energy of a particle executing SHM is periodic with time period $\frac{T}{2}$ .

PHXI14:OSCILLATIONS

364102 For a particle executing S.H.M. The
displacement $x$ is given by $x = A \cos ω t$ . Identity the graph which represents the variation of potential energy (P.E.) as a function of time $t$ and displacement $x$
supporting img

1 II, IV

2 I, III

3 I, IV

4 II, III

Explanation:

At time $t = 0, x = A,$ hence potential energy should be maximum. Therefore graph I is correct. Further in graph III. Potential energy is mimimum at $x = 0,$ hence this is also correct.

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: