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Oscillations

140621 One end of a massless spring of spring constant $k$ and natural length $l_{0}$ is fixed while the other end is connected to a small object of mass $m$ lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $ω$ about an axis passing through fixed end, then the elongation of the spring will be:

1

\frac{k - m ω^{2} l_{0}}{m ω^{2}}

2

\frac{m ω^{2} l_{0}}{k + m ω^{2}}

3

\frac{m ω^{2} l_{0}}{k - m ω^{2}}

4

\frac{k + m ω^{2} l_{0}}{m ω^{2}}

Explanation:

C Given, Spring constant $= k$ , natural length $= l_{o}$ Angular velocity $= ω$
We know that,
Centripetal force $= k l$
$∴ m ω^{2} r = k l$
$k l = m ω^{2} (l + l_{o})$
$l (k - m ω^{2}) = m^{2} l_{o}$
$l = \frac{m ω^{2} l_{o}}{k - m ω^{2}}$

JEE Main-27.06.2022

Oscillations

140622 As per the given figure, two blocks each of mass $250 g$ are connected to a spring of spring constant $2 {Nm}^{- 1}$ . If both are given velocity $v$ in opposite directions, then maximum elongation of the spring is:

1

\frac{V}{2 \sqrt{2}}

2

\frac{V}{2}

3

\frac{v}{4}

4

\frac{v}{\sqrt{2}}

Explanation:

B Given, mass of each block $(m) = 250 gm =$ $0.25 kg$ , spring constant $(k) = 2 N / m$
According to work energy theorem-
Gain in spring energy $=$ Loss in K.E
$\frac{1}{2} {kx}_{m}^{2} = 2 \times \frac{1}{2} {mv}^{2}$
$2 x_{m}^{2} = 2 \times 0.25 \times v^{2}$
$x_{m}^{2} = 0.25 v^{2}$
$x_{m}^{2} = \frac{25}{100} v^{2}$
$x_{m}^{2} = \frac{1}{4} v^{2}$
$x_{m} = \frac{v}{2}$

JEE Main-26.07.2022

Oscillations

140623 In figure (A), mass ' $2 m$ ' is fixed on mass ' $m$ ' which is attached to two springs of spring constant $k$ . In figure (B) mass ' $m$ ' is attached to two spring of spring constant ' $k$ ' and ' $2 k$ '. If mass ' $m$ ' in (A) and (B) are displaced by distance ' $x$ ' horizontally and then released, then time period $T_{1}$ and $T_{2}$ corresponding to (A) and (B) respectively follow the relation.

1

\frac{T_{1}}{T_{2}} = \frac{3}{\sqrt{2}}

2

\frac{T_{1}}{T_{2}} = \sqrt{\frac{3}{2}}

3

\frac{T_{1}}{T_{2}} = \sqrt{\frac{2}{3}}

4

\frac{T_{1}}{T_{2}} = \frac{\sqrt{2}}{3}

Explanation:

A In figure (A),
Total mass $= 2 m + m = 3 m$
Time period $(T_{1}) = 2 π \sqrt{\frac{3 m}{2 k}}$
In figure $(B)$ , total mass $= m$
Time period $(T_{2}) = 2 π \sqrt{\frac{m}{3 k}}$
Dividing equation (i) by (ii), we get -
$∴ \frac{T_{1}}{T_{2}} = \frac{2 π \sqrt{\frac{3 m}{2 k}}}{2 π \sqrt{\frac{m}{3 k}}} = \sqrt{\frac{9}{2}} \frac{T_{1}}{T_{2}} = \frac{3}{\sqrt{2}}$

JEE Main-25.07.2022

Oscillations

140624 The restoring force of a spring with a block attached to the free end of the spring is represented by:

1 a

2 b

3 d

4 d

Explanation:

D According to Hooke's law-
$F = - kx$
Where, $k =$ spring constant
$x =$ displacement of the spring from its mean position.
The negative sign indicates that the restoring force is proportional to the negative of the displacement. Hence, the graph (d) between spring force and positive displacement of the spring.

NEET 2022

Oscillations

140621 One end of a massless spring of spring constant $k$ and natural length $l_{0}$ is fixed while the other end is connected to a small object of mass $m$ lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $ω$ about an axis passing through fixed end, then the elongation of the spring will be:

1

\frac{k - m ω^{2} l_{0}}{m ω^{2}}

2

\frac{m ω^{2} l_{0}}{k + m ω^{2}}

3

\frac{m ω^{2} l_{0}}{k - m ω^{2}}

4

\frac{k + m ω^{2} l_{0}}{m ω^{2}}

Explanation:

C Given, Spring constant $= k$ , natural length $= l_{o}$ Angular velocity $= ω$
We know that,
Centripetal force $= k l$
$∴ m ω^{2} r = k l$
$k l = m ω^{2} (l + l_{o})$
$l (k - m ω^{2}) = m^{2} l_{o}$
$l = \frac{m ω^{2} l_{o}}{k - m ω^{2}}$

JEE Main-27.06.2022

Oscillations

140622 As per the given figure, two blocks each of mass $250 g$ are connected to a spring of spring constant $2 {Nm}^{- 1}$ . If both are given velocity $v$ in opposite directions, then maximum elongation of the spring is:

1

\frac{V}{2 \sqrt{2}}

2

\frac{V}{2}

3

\frac{v}{4}

4

\frac{v}{\sqrt{2}}

Explanation:

B Given, mass of each block $(m) = 250 gm =$ $0.25 kg$ , spring constant $(k) = 2 N / m$
According to work energy theorem-
Gain in spring energy $=$ Loss in K.E
$\frac{1}{2} {kx}_{m}^{2} = 2 \times \frac{1}{2} {mv}^{2}$
$2 x_{m}^{2} = 2 \times 0.25 \times v^{2}$
$x_{m}^{2} = 0.25 v^{2}$
$x_{m}^{2} = \frac{25}{100} v^{2}$
$x_{m}^{2} = \frac{1}{4} v^{2}$
$x_{m} = \frac{v}{2}$

JEE Main-26.07.2022

Oscillations

140623 In figure (A), mass ' $2 m$ ' is fixed on mass ' $m$ ' which is attached to two springs of spring constant $k$ . In figure (B) mass ' $m$ ' is attached to two spring of spring constant ' $k$ ' and ' $2 k$ '. If mass ' $m$ ' in (A) and (B) are displaced by distance ' $x$ ' horizontally and then released, then time period $T_{1}$ and $T_{2}$ corresponding to (A) and (B) respectively follow the relation.

1

\frac{T_{1}}{T_{2}} = \frac{3}{\sqrt{2}}

2

\frac{T_{1}}{T_{2}} = \sqrt{\frac{3}{2}}

3

\frac{T_{1}}{T_{2}} = \sqrt{\frac{2}{3}}

4

\frac{T_{1}}{T_{2}} = \frac{\sqrt{2}}{3}

Explanation:

A In figure (A),
Total mass $= 2 m + m = 3 m$
Time period $(T_{1}) = 2 π \sqrt{\frac{3 m}{2 k}}$
In figure $(B)$ , total mass $= m$
Time period $(T_{2}) = 2 π \sqrt{\frac{m}{3 k}}$
Dividing equation (i) by (ii), we get -
$∴ \frac{T_{1}}{T_{2}} = \frac{2 π \sqrt{\frac{3 m}{2 k}}}{2 π \sqrt{\frac{m}{3 k}}} = \sqrt{\frac{9}{2}} \frac{T_{1}}{T_{2}} = \frac{3}{\sqrt{2}}$

JEE Main-25.07.2022

Oscillations

140624 The restoring force of a spring with a block attached to the free end of the spring is represented by:

1 a

2 b

3 d

4 d

Explanation:

D According to Hooke's law-
$F = - kx$
Where, $k =$ spring constant
$x =$ displacement of the spring from its mean position.
The negative sign indicates that the restoring force is proportional to the negative of the displacement. Hence, the graph (d) between spring force and positive displacement of the spring.

NEET 2022

Oscillations

140621 One end of a massless spring of spring constant $k$ and natural length $l_{0}$ is fixed while the other end is connected to a small object of mass $m$ lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $ω$ about an axis passing through fixed end, then the elongation of the spring will be:

1

\frac{k - m ω^{2} l_{0}}{m ω^{2}}

2

\frac{m ω^{2} l_{0}}{k + m ω^{2}}

3

\frac{m ω^{2} l_{0}}{k - m ω^{2}}

4

\frac{k + m ω^{2} l_{0}}{m ω^{2}}

Explanation:

C Given, Spring constant $= k$ , natural length $= l_{o}$ Angular velocity $= ω$
We know that,
Centripetal force $= k l$
$∴ m ω^{2} r = k l$
$k l = m ω^{2} (l + l_{o})$
$l (k - m ω^{2}) = m^{2} l_{o}$
$l = \frac{m ω^{2} l_{o}}{k - m ω^{2}}$

JEE Main-27.06.2022

Oscillations

140622 As per the given figure, two blocks each of mass $250 g$ are connected to a spring of spring constant $2 {Nm}^{- 1}$ . If both are given velocity $v$ in opposite directions, then maximum elongation of the spring is:

1

\frac{V}{2 \sqrt{2}}

2

\frac{V}{2}

3

\frac{v}{4}

4

\frac{v}{\sqrt{2}}

Explanation:

B Given, mass of each block $(m) = 250 gm =$ $0.25 kg$ , spring constant $(k) = 2 N / m$
According to work energy theorem-
Gain in spring energy $=$ Loss in K.E
$\frac{1}{2} {kx}_{m}^{2} = 2 \times \frac{1}{2} {mv}^{2}$
$2 x_{m}^{2} = 2 \times 0.25 \times v^{2}$
$x_{m}^{2} = 0.25 v^{2}$
$x_{m}^{2} = \frac{25}{100} v^{2}$
$x_{m}^{2} = \frac{1}{4} v^{2}$
$x_{m} = \frac{v}{2}$

JEE Main-26.07.2022

Oscillations

140623 In figure (A), mass ' $2 m$ ' is fixed on mass ' $m$ ' which is attached to two springs of spring constant $k$ . In figure (B) mass ' $m$ ' is attached to two spring of spring constant ' $k$ ' and ' $2 k$ '. If mass ' $m$ ' in (A) and (B) are displaced by distance ' $x$ ' horizontally and then released, then time period $T_{1}$ and $T_{2}$ corresponding to (A) and (B) respectively follow the relation.

1

\frac{T_{1}}{T_{2}} = \frac{3}{\sqrt{2}}

2

\frac{T_{1}}{T_{2}} = \sqrt{\frac{3}{2}}

3

\frac{T_{1}}{T_{2}} = \sqrt{\frac{2}{3}}

4

\frac{T_{1}}{T_{2}} = \frac{\sqrt{2}}{3}

Explanation:

A In figure (A),
Total mass $= 2 m + m = 3 m$
Time period $(T_{1}) = 2 π \sqrt{\frac{3 m}{2 k}}$
In figure $(B)$ , total mass $= m$
Time period $(T_{2}) = 2 π \sqrt{\frac{m}{3 k}}$
Dividing equation (i) by (ii), we get -
$∴ \frac{T_{1}}{T_{2}} = \frac{2 π \sqrt{\frac{3 m}{2 k}}}{2 π \sqrt{\frac{m}{3 k}}} = \sqrt{\frac{9}{2}} \frac{T_{1}}{T_{2}} = \frac{3}{\sqrt{2}}$

JEE Main-25.07.2022

Oscillations

140624 The restoring force of a spring with a block attached to the free end of the spring is represented by:

1 a

2 b

3 d

4 d

Explanation:

D According to Hooke's law-
$F = - kx$
Where, $k =$ spring constant
$x =$ displacement of the spring from its mean position.
The negative sign indicates that the restoring force is proportional to the negative of the displacement. Hence, the graph (d) between spring force and positive displacement of the spring.

NEET 2022

Oscillations

140621 One end of a massless spring of spring constant $k$ and natural length $l_{0}$ is fixed while the other end is connected to a small object of mass $m$ lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $ω$ about an axis passing through fixed end, then the elongation of the spring will be:

1

\frac{k - m ω^{2} l_{0}}{m ω^{2}}

2

\frac{m ω^{2} l_{0}}{k + m ω^{2}}

3

\frac{m ω^{2} l_{0}}{k - m ω^{2}}

4

\frac{k + m ω^{2} l_{0}}{m ω^{2}}

Explanation:

C Given, Spring constant $= k$ , natural length $= l_{o}$ Angular velocity $= ω$
We know that,
Centripetal force $= k l$
$∴ m ω^{2} r = k l$
$k l = m ω^{2} (l + l_{o})$
$l (k - m ω^{2}) = m^{2} l_{o}$
$l = \frac{m ω^{2} l_{o}}{k - m ω^{2}}$

JEE Main-27.06.2022

Oscillations

140622 As per the given figure, two blocks each of mass $250 g$ are connected to a spring of spring constant $2 {Nm}^{- 1}$ . If both are given velocity $v$ in opposite directions, then maximum elongation of the spring is:

1

\frac{V}{2 \sqrt{2}}

2

\frac{V}{2}

3

\frac{v}{4}

4

\frac{v}{\sqrt{2}}

Explanation:

B Given, mass of each block $(m) = 250 gm =$ $0.25 kg$ , spring constant $(k) = 2 N / m$
According to work energy theorem-
Gain in spring energy $=$ Loss in K.E
$\frac{1}{2} {kx}_{m}^{2} = 2 \times \frac{1}{2} {mv}^{2}$
$2 x_{m}^{2} = 2 \times 0.25 \times v^{2}$
$x_{m}^{2} = 0.25 v^{2}$
$x_{m}^{2} = \frac{25}{100} v^{2}$
$x_{m}^{2} = \frac{1}{4} v^{2}$
$x_{m} = \frac{v}{2}$

JEE Main-26.07.2022

Oscillations

140623 In figure (A), mass ' $2 m$ ' is fixed on mass ' $m$ ' which is attached to two springs of spring constant $k$ . In figure (B) mass ' $m$ ' is attached to two spring of spring constant ' $k$ ' and ' $2 k$ '. If mass ' $m$ ' in (A) and (B) are displaced by distance ' $x$ ' horizontally and then released, then time period $T_{1}$ and $T_{2}$ corresponding to (A) and (B) respectively follow the relation.

1

\frac{T_{1}}{T_{2}} = \frac{3}{\sqrt{2}}

2

\frac{T_{1}}{T_{2}} = \sqrt{\frac{3}{2}}

3

\frac{T_{1}}{T_{2}} = \sqrt{\frac{2}{3}}

4

\frac{T_{1}}{T_{2}} = \frac{\sqrt{2}}{3}

Explanation:

A In figure (A),
Total mass $= 2 m + m = 3 m$
Time period $(T_{1}) = 2 π \sqrt{\frac{3 m}{2 k}}$
In figure $(B)$ , total mass $= m$
Time period $(T_{2}) = 2 π \sqrt{\frac{m}{3 k}}$
Dividing equation (i) by (ii), we get -
$∴ \frac{T_{1}}{T_{2}} = \frac{2 π \sqrt{\frac{3 m}{2 k}}}{2 π \sqrt{\frac{m}{3 k}}} = \sqrt{\frac{9}{2}} \frac{T_{1}}{T_{2}} = \frac{3}{\sqrt{2}}$

JEE Main-25.07.2022

Oscillations

140624 The restoring force of a spring with a block attached to the free end of the spring is represented by:

1 a

2 b

3 d

4 d

Explanation:

D According to Hooke's law-
$F = - kx$
Where, $k =$ spring constant
$x =$ displacement of the spring from its mean position.
The negative sign indicates that the restoring force is proportional to the negative of the displacement. Hence, the graph (d) between spring force and positive displacement of the spring.

NEET 2022

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