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PHXI14:OSCILLATIONS

364141 The potential energy of a particle $(U_{x})$ executing $S . H . M .$ is given by

1

U_{x} = \frac{k}{2} (x - a)^{2}

2

U_{x} = k_{1} x + k_{2} x^{2} + k_{3} x^{3}

3

U_{x} = A e^{- b x}

4

U_{x} = a c o n s t a n t

Explanation:

$P . E .$ of body in $S . H . M .$ at an instant,
$U = \frac{1}{2} m ω^{2} y^{2} = \frac{1}{2} k y^{2}$
If the displacement, $y = (a - x)$ then $U = \frac{1}{2} k (a - x)^{2} = \frac{1}{2} k (x - a)^{2}$

PHXI14:OSCILLATIONS

364142 The average kinetic energy of a simple harmonic oscillator is 2 joule and its total energy is 5 joule. Its minimum potential energy is

1

1 J

2

1.5 J

3

2 J

4

3 J

Explanation:

$(K E)_{a v} = 2$
$\Rightarrow \frac{(K E)_{max}}{2} = 2 J$
$\Rightarrow (K E)_{max} = 4 J$
Applying energy conservation
$P E_{min} + K E_{max} = P E_{max}$
$\Rightarrow P E_{min} + 4 = 5$
$\Rightarrow P E_{min} = 1 J$

PHXI14:OSCILLATIONS

364143 The total energy of a simple harmonic oscillator is proportional to

1 Square of the amplitude

2 Square root of displacement

3 Amplitude

4 Frequency

Explanation:

Total energy of simple harmonic oscillator is given by
$E = \frac{1}{2} m ω^{2} A^{2}$
where, $m =$ mass of body performing SHM
$ω =$ angular velocity and $A =$ amplitude
$∴ E \propto A^{2}$

MHTCET - 2019

PHXI14:OSCILLATIONS

364144 The physical quantity conserved in simple harmonic motion is

1 time period

2 total energy

3 displacement

4 force

PHXI14:OSCILLATIONS

364145 The displacement of a particle of mass $3 g m$ executing simple harmonic motion is given by $y = 3 \sin (0.2 t)$ in SI units. The kinetic energy of the particle at a point which is at a distance equal to $\frac{1}{3}$ of its amplitude from its mean position is

1

12 \times 10^{- 3} J

2

25 \times 10^{- 3} J

3

0.48 \times 10^{- 3} J

A

4

0.24 \times 10^{- 3} J

Explanation:

$y = 3 \sin (0.2 t) \Rightarrow ω = 0.2$
$K E = \frac{1}{2} m ω^{2} (A^{2} - y^{2})$
$= \frac{1}{2} m ω^{2} (A^{2} - \frac{A^{2}}{9})$
$= \frac{1}{2} m ω^{2} \frac{8 A^{2}}{9}$
$= \frac{4}{9} m ω^{2} A^{2}$
$= \frac{4}{9} \times 3 \times 10^{- 3} \times 4 \times 10^{- 2} \times 9$
$= 48 \times 10^{- 5} J$
$= 0.48 \times 10^{- 3} J$

PHXI14:OSCILLATIONS

364141 The potential energy of a particle $(U_{x})$ executing $S . H . M .$ is given by

1

U_{x} = \frac{k}{2} (x - a)^{2}

2

U_{x} = k_{1} x + k_{2} x^{2} + k_{3} x^{3}

3

U_{x} = A e^{- b x}

4

U_{x} = a c o n s t a n t

Explanation:

$P . E .$ of body in $S . H . M .$ at an instant,
$U = \frac{1}{2} m ω^{2} y^{2} = \frac{1}{2} k y^{2}$
If the displacement, $y = (a - x)$ then $U = \frac{1}{2} k (a - x)^{2} = \frac{1}{2} k (x - a)^{2}$

PHXI14:OSCILLATIONS

364142 The average kinetic energy of a simple harmonic oscillator is 2 joule and its total energy is 5 joule. Its minimum potential energy is

1

1 J

2

1.5 J

3

2 J

4

3 J

Explanation:

$(K E)_{a v} = 2$
$\Rightarrow \frac{(K E)_{max}}{2} = 2 J$
$\Rightarrow (K E)_{max} = 4 J$
Applying energy conservation
$P E_{min} + K E_{max} = P E_{max}$
$\Rightarrow P E_{min} + 4 = 5$
$\Rightarrow P E_{min} = 1 J$

PHXI14:OSCILLATIONS

364143 The total energy of a simple harmonic oscillator is proportional to

1 Square of the amplitude

2 Square root of displacement

3 Amplitude

4 Frequency

Explanation:

Total energy of simple harmonic oscillator is given by
$E = \frac{1}{2} m ω^{2} A^{2}$
where, $m =$ mass of body performing SHM
$ω =$ angular velocity and $A =$ amplitude
$∴ E \propto A^{2}$

MHTCET - 2019

PHXI14:OSCILLATIONS

364144 The physical quantity conserved in simple harmonic motion is

1 time period

2 total energy

3 displacement

4 force

PHXI14:OSCILLATIONS

364145 The displacement of a particle of mass $3 g m$ executing simple harmonic motion is given by $y = 3 \sin (0.2 t)$ in SI units. The kinetic energy of the particle at a point which is at a distance equal to $\frac{1}{3}$ of its amplitude from its mean position is

1

12 \times 10^{- 3} J

2

25 \times 10^{- 3} J

3

0.48 \times 10^{- 3} J

A

4

0.24 \times 10^{- 3} J

Explanation:

$y = 3 \sin (0.2 t) \Rightarrow ω = 0.2$
$K E = \frac{1}{2} m ω^{2} (A^{2} - y^{2})$
$= \frac{1}{2} m ω^{2} (A^{2} - \frac{A^{2}}{9})$
$= \frac{1}{2} m ω^{2} \frac{8 A^{2}}{9}$
$= \frac{4}{9} m ω^{2} A^{2}$
$= \frac{4}{9} \times 3 \times 10^{- 3} \times 4 \times 10^{- 2} \times 9$
$= 48 \times 10^{- 5} J$
$= 0.48 \times 10^{- 3} J$

PHXI14:OSCILLATIONS

364141 The potential energy of a particle $(U_{x})$ executing $S . H . M .$ is given by

1

U_{x} = \frac{k}{2} (x - a)^{2}

2

U_{x} = k_{1} x + k_{2} x^{2} + k_{3} x^{3}

3

U_{x} = A e^{- b x}

4

U_{x} = a c o n s t a n t

Explanation:

$P . E .$ of body in $S . H . M .$ at an instant,
$U = \frac{1}{2} m ω^{2} y^{2} = \frac{1}{2} k y^{2}$
If the displacement, $y = (a - x)$ then $U = \frac{1}{2} k (a - x)^{2} = \frac{1}{2} k (x - a)^{2}$

PHXI14:OSCILLATIONS

364142 The average kinetic energy of a simple harmonic oscillator is 2 joule and its total energy is 5 joule. Its minimum potential energy is

1

1 J

2

1.5 J

3

2 J

4

3 J

Explanation:

$(K E)_{a v} = 2$
$\Rightarrow \frac{(K E)_{max}}{2} = 2 J$
$\Rightarrow (K E)_{max} = 4 J$
Applying energy conservation
$P E_{min} + K E_{max} = P E_{max}$
$\Rightarrow P E_{min} + 4 = 5$
$\Rightarrow P E_{min} = 1 J$

PHXI14:OSCILLATIONS

364143 The total energy of a simple harmonic oscillator is proportional to

1 Square of the amplitude

2 Square root of displacement

3 Amplitude

4 Frequency

Explanation:

Total energy of simple harmonic oscillator is given by
$E = \frac{1}{2} m ω^{2} A^{2}$
where, $m =$ mass of body performing SHM
$ω =$ angular velocity and $A =$ amplitude
$∴ E \propto A^{2}$

MHTCET - 2019

PHXI14:OSCILLATIONS

364144 The physical quantity conserved in simple harmonic motion is

1 time period

2 total energy

3 displacement

4 force

PHXI14:OSCILLATIONS

364145 The displacement of a particle of mass $3 g m$ executing simple harmonic motion is given by $y = 3 \sin (0.2 t)$ in SI units. The kinetic energy of the particle at a point which is at a distance equal to $\frac{1}{3}$ of its amplitude from its mean position is

1

12 \times 10^{- 3} J

2

25 \times 10^{- 3} J

3

0.48 \times 10^{- 3} J

A

4

0.24 \times 10^{- 3} J

Explanation:

$y = 3 \sin (0.2 t) \Rightarrow ω = 0.2$
$K E = \frac{1}{2} m ω^{2} (A^{2} - y^{2})$
$= \frac{1}{2} m ω^{2} (A^{2} - \frac{A^{2}}{9})$
$= \frac{1}{2} m ω^{2} \frac{8 A^{2}}{9}$
$= \frac{4}{9} m ω^{2} A^{2}$
$= \frac{4}{9} \times 3 \times 10^{- 3} \times 4 \times 10^{- 2} \times 9$
$= 48 \times 10^{- 5} J$
$= 0.48 \times 10^{- 3} J$

PHXI14:OSCILLATIONS

364141 The potential energy of a particle $(U_{x})$ executing $S . H . M .$ is given by

1

U_{x} = \frac{k}{2} (x - a)^{2}

2

U_{x} = k_{1} x + k_{2} x^{2} + k_{3} x^{3}

3

U_{x} = A e^{- b x}

4

U_{x} = a c o n s t a n t

Explanation:

$P . E .$ of body in $S . H . M .$ at an instant,
$U = \frac{1}{2} m ω^{2} y^{2} = \frac{1}{2} k y^{2}$
If the displacement, $y = (a - x)$ then $U = \frac{1}{2} k (a - x)^{2} = \frac{1}{2} k (x - a)^{2}$

PHXI14:OSCILLATIONS

364142 The average kinetic energy of a simple harmonic oscillator is 2 joule and its total energy is 5 joule. Its minimum potential energy is

1

1 J

2

1.5 J

3

2 J

4

3 J

Explanation:

$(K E)_{a v} = 2$
$\Rightarrow \frac{(K E)_{max}}{2} = 2 J$
$\Rightarrow (K E)_{max} = 4 J$
Applying energy conservation
$P E_{min} + K E_{max} = P E_{max}$
$\Rightarrow P E_{min} + 4 = 5$
$\Rightarrow P E_{min} = 1 J$

PHXI14:OSCILLATIONS

364143 The total energy of a simple harmonic oscillator is proportional to

1 Square of the amplitude

2 Square root of displacement

3 Amplitude

4 Frequency

Explanation:

Total energy of simple harmonic oscillator is given by
$E = \frac{1}{2} m ω^{2} A^{2}$
where, $m =$ mass of body performing SHM
$ω =$ angular velocity and $A =$ amplitude
$∴ E \propto A^{2}$

MHTCET - 2019

PHXI14:OSCILLATIONS

364144 The physical quantity conserved in simple harmonic motion is

1 time period

2 total energy

3 displacement

4 force

PHXI14:OSCILLATIONS

364145 The displacement of a particle of mass $3 g m$ executing simple harmonic motion is given by $y = 3 \sin (0.2 t)$ in SI units. The kinetic energy of the particle at a point which is at a distance equal to $\frac{1}{3}$ of its amplitude from its mean position is

1

12 \times 10^{- 3} J

2

25 \times 10^{- 3} J

3

0.48 \times 10^{- 3} J

A

4

0.24 \times 10^{- 3} J

Explanation:

$y = 3 \sin (0.2 t) \Rightarrow ω = 0.2$
$K E = \frac{1}{2} m ω^{2} (A^{2} - y^{2})$
$= \frac{1}{2} m ω^{2} (A^{2} - \frac{A^{2}}{9})$
$= \frac{1}{2} m ω^{2} \frac{8 A^{2}}{9}$
$= \frac{4}{9} m ω^{2} A^{2}$
$= \frac{4}{9} \times 3 \times 10^{- 3} \times 4 \times 10^{- 2} \times 9$
$= 48 \times 10^{- 5} J$
$= 0.48 \times 10^{- 3} J$

PHXI14:OSCILLATIONS

364141 The potential energy of a particle $(U_{x})$ executing $S . H . M .$ is given by

1

U_{x} = \frac{k}{2} (x - a)^{2}

2

U_{x} = k_{1} x + k_{2} x^{2} + k_{3} x^{3}

3

U_{x} = A e^{- b x}

4

U_{x} = a c o n s t a n t

Explanation:

$P . E .$ of body in $S . H . M .$ at an instant,
$U = \frac{1}{2} m ω^{2} y^{2} = \frac{1}{2} k y^{2}$
If the displacement, $y = (a - x)$ then $U = \frac{1}{2} k (a - x)^{2} = \frac{1}{2} k (x - a)^{2}$

PHXI14:OSCILLATIONS

364142 The average kinetic energy of a simple harmonic oscillator is 2 joule and its total energy is 5 joule. Its minimum potential energy is

1

1 J

2

1.5 J

3

2 J

4

3 J

Explanation:

$(K E)_{a v} = 2$
$\Rightarrow \frac{(K E)_{max}}{2} = 2 J$
$\Rightarrow (K E)_{max} = 4 J$
Applying energy conservation
$P E_{min} + K E_{max} = P E_{max}$
$\Rightarrow P E_{min} + 4 = 5$
$\Rightarrow P E_{min} = 1 J$

PHXI14:OSCILLATIONS

364143 The total energy of a simple harmonic oscillator is proportional to

1 Square of the amplitude

2 Square root of displacement

3 Amplitude

4 Frequency

Explanation:

Total energy of simple harmonic oscillator is given by
$E = \frac{1}{2} m ω^{2} A^{2}$
where, $m =$ mass of body performing SHM
$ω =$ angular velocity and $A =$ amplitude
$∴ E \propto A^{2}$

MHTCET - 2019

PHXI14:OSCILLATIONS

364144 The physical quantity conserved in simple harmonic motion is

1 time period

2 total energy

3 displacement

4 force

PHXI14:OSCILLATIONS

364145 The displacement of a particle of mass $3 g m$ executing simple harmonic motion is given by $y = 3 \sin (0.2 t)$ in SI units. The kinetic energy of the particle at a point which is at a distance equal to $\frac{1}{3}$ of its amplitude from its mean position is

1

12 \times 10^{- 3} J

2

25 \times 10^{- 3} J

3

0.48 \times 10^{- 3} J

A

4

0.24 \times 10^{- 3} J

Explanation:

$y = 3 \sin (0.2 t) \Rightarrow ω = 0.2$
$K E = \frac{1}{2} m ω^{2} (A^{2} - y^{2})$
$= \frac{1}{2} m ω^{2} (A^{2} - \frac{A^{2}}{9})$
$= \frac{1}{2} m ω^{2} \frac{8 A^{2}}{9}$
$= \frac{4}{9} m ω^{2} A^{2}$
$= \frac{4}{9} \times 3 \times 10^{- 3} \times 4 \times 10^{- 2} \times 9$
$= 48 \times 10^{- 5} J$
$= 0.48 \times 10^{- 3} J$