QBManager

Oscillations

140398 The angular velocity and the amplitude of a simple pendulum is $ω$ and $A$ respectively. At a displacement $x$ from the mean position, if its kinetic energy is $T$ and potential energy is $U$ , then the ratio of $T$ to $U$ is

1

(\frac{A^{2} - x^{2} ω^{2}}{x^{2} ω^{2}})

2

(\frac{x^{2} ω^{2}}{A^{2} - x^{2} ω^{2}})

3

\frac{(A^{2} - x^{2})}{x^{2}}

4

\frac{x^{2}}{(A^{2} - x^{2})}

Explanation:

C For a simple pendulum, expression for kinetic energy $(KE)$ and potential energy $(PE)$ is-
$KE = T = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$PE = U = \frac{1}{2} m ω^{2} x^{2}$
Then the ratio of $PE$ and K.E,
$\frac{T}{U} = \frac{\frac{1}{2} m ω^{2} (A^{2} - x^{2})}{\frac{1}{2} m ω^{2} x^{2}}$
$\frac{T}{U} = \frac{A^{2} - x^{2}}{x^{2}}$

AP EAMCET (Medical)-07.10.2020

Oscillations

140399 In $SHM$ restoring force is $F = - kx$ , where $k$ is force constant, $x$ is displacement and $A$ is amplitude of motion, then total energy depends upon

1

k, A

and

m

2

k, x, m

3

k, A

4

k, x

Explanation:

C We know that, K.E. $= \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$P.E. = \frac{1}{2} {kx}^{2}$
$P.E. = \frac{1}{2} m^{2} x^{2} [∴ ω = \sqrt{\frac{k}{m}}]$
In S.H.M, total energy $=$ Potential energy + Kinetic energy
$T.E. = \frac{1}{2} m ω^{2} x^{2} + \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$T.E. = \frac{1}{2} m ω^{2} A^{2}$
$T.E. = \frac{1}{2} kA A^{2}$
Where, $k =$ Force constant $= m ω^{2}$
From this we can say that, total energy depends on $k$ and a.

AIPMT-2001

Oscillations

140400 The oscillation of a body on a smooth horizontal surface is represented by the equation,
$X = A \cos (ω t)$
Where, $X =$ displacement at time $t$
$ω = frequency of oscillation$
Which one of the following graphs shows correctly the variation a with $t$ ?

1

2

3

4

Here,

a =

acceleration at time

t

T =

Time period

Explanation:

C Given, $x = A \cos ω t$
$∴ Velocity (v) = \frac{dx}{dt} = \frac{d}{dt} (A \cos ω t)$
$v = - A ω \sin ω t$
Acceleration $(a) = \frac{dv}{dt} = \frac{d}{dt} (- A ω \sin ω t)$
$a = - A ω^{2} \cos ω t$

Hence, graph given in option (c) shows the correct variation a with $t$ .

AIPMT-2014

Oscillations

140401 The total mechanical energy of a harmonic oscillator of $A = 1 m$ and force constant 200 ${Nm}^{- 1}$ is $150 J$ . Then

1 the minimum PE is zero

2 the minimum PE is

100 J

3 the minimum PE is

50 J

4 the maximum

KE

is

150 J

Explanation:

C Given, force constant $(K) = 200 N / m$ , amplitude $(A) = 1 m, T . E . = 150 J$
$∵$ Total energy $=$ mechanical energy
T.E. $=$ K. $E_{max} +$ P. $E_{min}$
$K. E_{max} = \frac{1}{2} {KA}^{2} = \frac{1}{2} \times 200 \times 1^{2} = 100 J$
$150 = 100 + P . E_{min}$
$or P. E_{min} = 50 J$ .

AP EAMCET-1991

Oscillations

140404 A body is executing S.H.M. Its potential energy is $E_{1}$ and $E_{2}$ at displacement $x$ and $y$ respectively. The potential energy at displacement $(x + y)$ is

1

E_{1} - E_{2} = E

2

\sqrt{E_{1}} - \sqrt{E_{2}} = \sqrt{E}

3

E_{1} + E_{2} = E

4

\sqrt{E_{1}} + \sqrt{E_{2}} = \sqrt{E}

Explanation:

D The body is executing S.H.M. Its potential energy $E_{1}$ and $E_{2}$ at displacement $x$ and $y$ respectively the potential energy at displacement $(x + y)$ is Potential energy at $x = E_{1} = \frac{1}{2} {kx}^{2}$
$x = \sqrt{\frac{2 E_{1}}{k}}$
Potential energy at $y = E_{2} = \frac{1}{2} {ky}^{2}$
$y = \sqrt{\frac{2 E_{2}}{k}}$
Potential energy at $x + y = E = \frac{1}{2} k (x + y)^{2}$
$E = \frac{1}{2} k {(\sqrt{\frac{2 E_{1}}{k}} + \sqrt{\frac{2 E_{2}}{k}})}^{2}$
$\sqrt{E} = \sqrt{E_{1}} + \sqrt{E_{2}}$

MHT-CET-2020

Oscillations

140398 The angular velocity and the amplitude of a simple pendulum is $ω$ and $A$ respectively. At a displacement $x$ from the mean position, if its kinetic energy is $T$ and potential energy is $U$ , then the ratio of $T$ to $U$ is

1

(\frac{A^{2} - x^{2} ω^{2}}{x^{2} ω^{2}})

2

(\frac{x^{2} ω^{2}}{A^{2} - x^{2} ω^{2}})

3

\frac{(A^{2} - x^{2})}{x^{2}}

4

\frac{x^{2}}{(A^{2} - x^{2})}

Explanation:

C For a simple pendulum, expression for kinetic energy $(KE)$ and potential energy $(PE)$ is-
$KE = T = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$PE = U = \frac{1}{2} m ω^{2} x^{2}$
Then the ratio of $PE$ and K.E,
$\frac{T}{U} = \frac{\frac{1}{2} m ω^{2} (A^{2} - x^{2})}{\frac{1}{2} m ω^{2} x^{2}}$
$\frac{T}{U} = \frac{A^{2} - x^{2}}{x^{2}}$

AP EAMCET (Medical)-07.10.2020

Oscillations

140399 In $SHM$ restoring force is $F = - kx$ , where $k$ is force constant, $x$ is displacement and $A$ is amplitude of motion, then total energy depends upon

1

k, A

and

m

2

k, x, m

3

k, A

4

k, x

Explanation:

C We know that, K.E. $= \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$P.E. = \frac{1}{2} {kx}^{2}$
$P.E. = \frac{1}{2} m^{2} x^{2} [∴ ω = \sqrt{\frac{k}{m}}]$
In S.H.M, total energy $=$ Potential energy + Kinetic energy
$T.E. = \frac{1}{2} m ω^{2} x^{2} + \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$T.E. = \frac{1}{2} m ω^{2} A^{2}$
$T.E. = \frac{1}{2} kA A^{2}$
Where, $k =$ Force constant $= m ω^{2}$
From this we can say that, total energy depends on $k$ and a.

AIPMT-2001

Oscillations

140400 The oscillation of a body on a smooth horizontal surface is represented by the equation,
$X = A \cos (ω t)$
Where, $X =$ displacement at time $t$
$ω = frequency of oscillation$
Which one of the following graphs shows correctly the variation a with $t$ ?

1

2

3

4

Here,

a =

acceleration at time

t

T =

Time period

Explanation:

C Given, $x = A \cos ω t$
$∴ Velocity (v) = \frac{dx}{dt} = \frac{d}{dt} (A \cos ω t)$
$v = - A ω \sin ω t$
Acceleration $(a) = \frac{dv}{dt} = \frac{d}{dt} (- A ω \sin ω t)$
$a = - A ω^{2} \cos ω t$

Hence, graph given in option (c) shows the correct variation a with $t$ .

AIPMT-2014

Oscillations

140401 The total mechanical energy of a harmonic oscillator of $A = 1 m$ and force constant 200 ${Nm}^{- 1}$ is $150 J$ . Then

1 the minimum PE is zero

2 the minimum PE is

100 J

3 the minimum PE is

50 J

4 the maximum

KE

is

150 J

Explanation:

C Given, force constant $(K) = 200 N / m$ , amplitude $(A) = 1 m, T . E . = 150 J$
$∵$ Total energy $=$ mechanical energy
T.E. $=$ K. $E_{max} +$ P. $E_{min}$
$K. E_{max} = \frac{1}{2} {KA}^{2} = \frac{1}{2} \times 200 \times 1^{2} = 100 J$
$150 = 100 + P . E_{min}$
$or P. E_{min} = 50 J$ .

AP EAMCET-1991

Oscillations

140404 A body is executing S.H.M. Its potential energy is $E_{1}$ and $E_{2}$ at displacement $x$ and $y$ respectively. The potential energy at displacement $(x + y)$ is

1

E_{1} - E_{2} = E

2

\sqrt{E_{1}} - \sqrt{E_{2}} = \sqrt{E}

3

E_{1} + E_{2} = E

4

\sqrt{E_{1}} + \sqrt{E_{2}} = \sqrt{E}

Explanation:

D The body is executing S.H.M. Its potential energy $E_{1}$ and $E_{2}$ at displacement $x$ and $y$ respectively the potential energy at displacement $(x + y)$ is Potential energy at $x = E_{1} = \frac{1}{2} {kx}^{2}$
$x = \sqrt{\frac{2 E_{1}}{k}}$
Potential energy at $y = E_{2} = \frac{1}{2} {ky}^{2}$
$y = \sqrt{\frac{2 E_{2}}{k}}$
Potential energy at $x + y = E = \frac{1}{2} k (x + y)^{2}$
$E = \frac{1}{2} k {(\sqrt{\frac{2 E_{1}}{k}} + \sqrt{\frac{2 E_{2}}{k}})}^{2}$
$\sqrt{E} = \sqrt{E_{1}} + \sqrt{E_{2}}$

MHT-CET-2020

Oscillations

140398 The angular velocity and the amplitude of a simple pendulum is $ω$ and $A$ respectively. At a displacement $x$ from the mean position, if its kinetic energy is $T$ and potential energy is $U$ , then the ratio of $T$ to $U$ is

1

(\frac{A^{2} - x^{2} ω^{2}}{x^{2} ω^{2}})

2

(\frac{x^{2} ω^{2}}{A^{2} - x^{2} ω^{2}})

3

\frac{(A^{2} - x^{2})}{x^{2}}

4

\frac{x^{2}}{(A^{2} - x^{2})}

Explanation:

C For a simple pendulum, expression for kinetic energy $(KE)$ and potential energy $(PE)$ is-
$KE = T = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$PE = U = \frac{1}{2} m ω^{2} x^{2}$
Then the ratio of $PE$ and K.E,
$\frac{T}{U} = \frac{\frac{1}{2} m ω^{2} (A^{2} - x^{2})}{\frac{1}{2} m ω^{2} x^{2}}$
$\frac{T}{U} = \frac{A^{2} - x^{2}}{x^{2}}$

AP EAMCET (Medical)-07.10.2020

Oscillations

140399 In $SHM$ restoring force is $F = - kx$ , where $k$ is force constant, $x$ is displacement and $A$ is amplitude of motion, then total energy depends upon

1

k, A

and

m

2

k, x, m

3

k, A

4

k, x

Explanation:

C We know that, K.E. $= \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$P.E. = \frac{1}{2} {kx}^{2}$
$P.E. = \frac{1}{2} m^{2} x^{2} [∴ ω = \sqrt{\frac{k}{m}}]$
In S.H.M, total energy $=$ Potential energy + Kinetic energy
$T.E. = \frac{1}{2} m ω^{2} x^{2} + \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$T.E. = \frac{1}{2} m ω^{2} A^{2}$
$T.E. = \frac{1}{2} kA A^{2}$
Where, $k =$ Force constant $= m ω^{2}$
From this we can say that, total energy depends on $k$ and a.

AIPMT-2001

Oscillations

140400 The oscillation of a body on a smooth horizontal surface is represented by the equation,
$X = A \cos (ω t)$
Where, $X =$ displacement at time $t$
$ω = frequency of oscillation$
Which one of the following graphs shows correctly the variation a with $t$ ?

1

2

3

4

Here,

a =

acceleration at time

t

T =

Time period

Explanation:

C Given, $x = A \cos ω t$
$∴ Velocity (v) = \frac{dx}{dt} = \frac{d}{dt} (A \cos ω t)$
$v = - A ω \sin ω t$
Acceleration $(a) = \frac{dv}{dt} = \frac{d}{dt} (- A ω \sin ω t)$
$a = - A ω^{2} \cos ω t$

Hence, graph given in option (c) shows the correct variation a with $t$ .

AIPMT-2014

Oscillations

140401 The total mechanical energy of a harmonic oscillator of $A = 1 m$ and force constant 200 ${Nm}^{- 1}$ is $150 J$ . Then

1 the minimum PE is zero

2 the minimum PE is

100 J

3 the minimum PE is

50 J

4 the maximum

KE

is

150 J

Explanation:

C Given, force constant $(K) = 200 N / m$ , amplitude $(A) = 1 m, T . E . = 150 J$
$∵$ Total energy $=$ mechanical energy
T.E. $=$ K. $E_{max} +$ P. $E_{min}$
$K. E_{max} = \frac{1}{2} {KA}^{2} = \frac{1}{2} \times 200 \times 1^{2} = 100 J$
$150 = 100 + P . E_{min}$
$or P. E_{min} = 50 J$ .

AP EAMCET-1991

Oscillations

140404 A body is executing S.H.M. Its potential energy is $E_{1}$ and $E_{2}$ at displacement $x$ and $y$ respectively. The potential energy at displacement $(x + y)$ is

1

E_{1} - E_{2} = E

2

\sqrt{E_{1}} - \sqrt{E_{2}} = \sqrt{E}

3

E_{1} + E_{2} = E

4

\sqrt{E_{1}} + \sqrt{E_{2}} = \sqrt{E}

Explanation:

D The body is executing S.H.M. Its potential energy $E_{1}$ and $E_{2}$ at displacement $x$ and $y$ respectively the potential energy at displacement $(x + y)$ is Potential energy at $x = E_{1} = \frac{1}{2} {kx}^{2}$
$x = \sqrt{\frac{2 E_{1}}{k}}$
Potential energy at $y = E_{2} = \frac{1}{2} {ky}^{2}$
$y = \sqrt{\frac{2 E_{2}}{k}}$
Potential energy at $x + y = E = \frac{1}{2} k (x + y)^{2}$
$E = \frac{1}{2} k {(\sqrt{\frac{2 E_{1}}{k}} + \sqrt{\frac{2 E_{2}}{k}})}^{2}$
$\sqrt{E} = \sqrt{E_{1}} + \sqrt{E_{2}}$

MHT-CET-2020

Oscillations

140398 The angular velocity and the amplitude of a simple pendulum is $ω$ and $A$ respectively. At a displacement $x$ from the mean position, if its kinetic energy is $T$ and potential energy is $U$ , then the ratio of $T$ to $U$ is

1

(\frac{A^{2} - x^{2} ω^{2}}{x^{2} ω^{2}})

2

(\frac{x^{2} ω^{2}}{A^{2} - x^{2} ω^{2}})

3

\frac{(A^{2} - x^{2})}{x^{2}}

4

\frac{x^{2}}{(A^{2} - x^{2})}

Explanation:

C For a simple pendulum, expression for kinetic energy $(KE)$ and potential energy $(PE)$ is-
$KE = T = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$PE = U = \frac{1}{2} m ω^{2} x^{2}$
Then the ratio of $PE$ and K.E,
$\frac{T}{U} = \frac{\frac{1}{2} m ω^{2} (A^{2} - x^{2})}{\frac{1}{2} m ω^{2} x^{2}}$
$\frac{T}{U} = \frac{A^{2} - x^{2}}{x^{2}}$

AP EAMCET (Medical)-07.10.2020

Oscillations

140399 In $SHM$ restoring force is $F = - kx$ , where $k$ is force constant, $x$ is displacement and $A$ is amplitude of motion, then total energy depends upon

1

k, A

and

m

2

k, x, m

3

k, A

4

k, x

Explanation:

C We know that, K.E. $= \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$P.E. = \frac{1}{2} {kx}^{2}$
$P.E. = \frac{1}{2} m^{2} x^{2} [∴ ω = \sqrt{\frac{k}{m}}]$
In S.H.M, total energy $=$ Potential energy + Kinetic energy
$T.E. = \frac{1}{2} m ω^{2} x^{2} + \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$T.E. = \frac{1}{2} m ω^{2} A^{2}$
$T.E. = \frac{1}{2} kA A^{2}$
Where, $k =$ Force constant $= m ω^{2}$
From this we can say that, total energy depends on $k$ and a.

AIPMT-2001

Oscillations

140400 The oscillation of a body on a smooth horizontal surface is represented by the equation,
$X = A \cos (ω t)$
Where, $X =$ displacement at time $t$
$ω = frequency of oscillation$
Which one of the following graphs shows correctly the variation a with $t$ ?

1

2

3

4

Here,

a =

acceleration at time

t

T =

Time period

Explanation:

C Given, $x = A \cos ω t$
$∴ Velocity (v) = \frac{dx}{dt} = \frac{d}{dt} (A \cos ω t)$
$v = - A ω \sin ω t$
Acceleration $(a) = \frac{dv}{dt} = \frac{d}{dt} (- A ω \sin ω t)$
$a = - A ω^{2} \cos ω t$

Hence, graph given in option (c) shows the correct variation a with $t$ .

AIPMT-2014

Oscillations

140401 The total mechanical energy of a harmonic oscillator of $A = 1 m$ and force constant 200 ${Nm}^{- 1}$ is $150 J$ . Then

1 the minimum PE is zero

2 the minimum PE is

100 J

3 the minimum PE is

50 J

4 the maximum

KE

is

150 J

Explanation:

C Given, force constant $(K) = 200 N / m$ , amplitude $(A) = 1 m, T . E . = 150 J$
$∵$ Total energy $=$ mechanical energy
T.E. $=$ K. $E_{max} +$ P. $E_{min}$
$K. E_{max} = \frac{1}{2} {KA}^{2} = \frac{1}{2} \times 200 \times 1^{2} = 100 J$
$150 = 100 + P . E_{min}$
$or P. E_{min} = 50 J$ .

AP EAMCET-1991

Oscillations

140404 A body is executing S.H.M. Its potential energy is $E_{1}$ and $E_{2}$ at displacement $x$ and $y$ respectively. The potential energy at displacement $(x + y)$ is

1

E_{1} - E_{2} = E

2

\sqrt{E_{1}} - \sqrt{E_{2}} = \sqrt{E}

3

E_{1} + E_{2} = E

4

\sqrt{E_{1}} + \sqrt{E_{2}} = \sqrt{E}

Explanation:

D The body is executing S.H.M. Its potential energy $E_{1}$ and $E_{2}$ at displacement $x$ and $y$ respectively the potential energy at displacement $(x + y)$ is Potential energy at $x = E_{1} = \frac{1}{2} {kx}^{2}$
$x = \sqrt{\frac{2 E_{1}}{k}}$
Potential energy at $y = E_{2} = \frac{1}{2} {ky}^{2}$
$y = \sqrt{\frac{2 E_{2}}{k}}$
Potential energy at $x + y = E = \frac{1}{2} k (x + y)^{2}$
$E = \frac{1}{2} k {(\sqrt{\frac{2 E_{1}}{k}} + \sqrt{\frac{2 E_{2}}{k}})}^{2}$
$\sqrt{E} = \sqrt{E_{1}} + \sqrt{E_{2}}$

MHT-CET-2020

Oscillations

140398 The angular velocity and the amplitude of a simple pendulum is $ω$ and $A$ respectively. At a displacement $x$ from the mean position, if its kinetic energy is $T$ and potential energy is $U$ , then the ratio of $T$ to $U$ is

1

(\frac{A^{2} - x^{2} ω^{2}}{x^{2} ω^{2}})

2

(\frac{x^{2} ω^{2}}{A^{2} - x^{2} ω^{2}})

3

\frac{(A^{2} - x^{2})}{x^{2}}

4

\frac{x^{2}}{(A^{2} - x^{2})}

Explanation:

C For a simple pendulum, expression for kinetic energy $(KE)$ and potential energy $(PE)$ is-
$KE = T = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$PE = U = \frac{1}{2} m ω^{2} x^{2}$
Then the ratio of $PE$ and K.E,
$\frac{T}{U} = \frac{\frac{1}{2} m ω^{2} (A^{2} - x^{2})}{\frac{1}{2} m ω^{2} x^{2}}$
$\frac{T}{U} = \frac{A^{2} - x^{2}}{x^{2}}$

AP EAMCET (Medical)-07.10.2020

Oscillations

140399 In $SHM$ restoring force is $F = - kx$ , where $k$ is force constant, $x$ is displacement and $A$ is amplitude of motion, then total energy depends upon

1

k, A

and

m

2

k, x, m

3

k, A

4

k, x

Explanation:

C We know that, K.E. $= \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$P.E. = \frac{1}{2} {kx}^{2}$
$P.E. = \frac{1}{2} m^{2} x^{2} [∴ ω = \sqrt{\frac{k}{m}}]$
In S.H.M, total energy $=$ Potential energy + Kinetic energy
$T.E. = \frac{1}{2} m ω^{2} x^{2} + \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
$T.E. = \frac{1}{2} m ω^{2} A^{2}$
$T.E. = \frac{1}{2} kA A^{2}$
Where, $k =$ Force constant $= m ω^{2}$
From this we can say that, total energy depends on $k$ and a.

AIPMT-2001

Oscillations

140400 The oscillation of a body on a smooth horizontal surface is represented by the equation,
$X = A \cos (ω t)$
Where, $X =$ displacement at time $t$
$ω = frequency of oscillation$
Which one of the following graphs shows correctly the variation a with $t$ ?

1

2

3

4

Here,

a =

acceleration at time

t

T =

Time period

Explanation:

C Given, $x = A \cos ω t$
$∴ Velocity (v) = \frac{dx}{dt} = \frac{d}{dt} (A \cos ω t)$
$v = - A ω \sin ω t$
Acceleration $(a) = \frac{dv}{dt} = \frac{d}{dt} (- A ω \sin ω t)$
$a = - A ω^{2} \cos ω t$

Hence, graph given in option (c) shows the correct variation a with $t$ .

AIPMT-2014

Oscillations

140401 The total mechanical energy of a harmonic oscillator of $A = 1 m$ and force constant 200 ${Nm}^{- 1}$ is $150 J$ . Then

1 the minimum PE is zero

2 the minimum PE is

100 J

3 the minimum PE is

50 J

4 the maximum

KE

is

150 J

Explanation:

C Given, force constant $(K) = 200 N / m$ , amplitude $(A) = 1 m, T . E . = 150 J$
$∵$ Total energy $=$ mechanical energy
T.E. $=$ K. $E_{max} +$ P. $E_{min}$
$K. E_{max} = \frac{1}{2} {KA}^{2} = \frac{1}{2} \times 200 \times 1^{2} = 100 J$
$150 = 100 + P . E_{min}$
$or P. E_{min} = 50 J$ .

AP EAMCET-1991

Oscillations

140404 A body is executing S.H.M. Its potential energy is $E_{1}$ and $E_{2}$ at displacement $x$ and $y$ respectively. The potential energy at displacement $(x + y)$ is

1

E_{1} - E_{2} = E

2

\sqrt{E_{1}} - \sqrt{E_{2}} = \sqrt{E}

3

E_{1} + E_{2} = E

4

\sqrt{E_{1}} + \sqrt{E_{2}} = \sqrt{E}

Explanation:

D The body is executing S.H.M. Its potential energy $E_{1}$ and $E_{2}$ at displacement $x$ and $y$ respectively the potential energy at displacement $(x + y)$ is Potential energy at $x = E_{1} = \frac{1}{2} {kx}^{2}$
$x = \sqrt{\frac{2 E_{1}}{k}}$
Potential energy at $y = E_{2} = \frac{1}{2} {ky}^{2}$
$y = \sqrt{\frac{2 E_{2}}{k}}$
Potential energy at $x + y = E = \frac{1}{2} k (x + y)^{2}$
$E = \frac{1}{2} k {(\sqrt{\frac{2 E_{1}}{k}} + \sqrt{\frac{2 E_{2}}{k}})}^{2}$
$\sqrt{E} = \sqrt{E_{1}} + \sqrt{E_{2}}$

MHT-CET-2020

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