Lagrangian and Hamiltonian Formalism MCQ Quiz - Objective Question with Answer for Lagrangian and Hamiltonian Formalism - Download Free PDF

Calculation:

We are given a particle of mass \( m \) sliding under the influence of gravity without friction along a parabolic path described by \( y = ax^2 \), where \( a \) is a constant.

To determine the Lagrangian \( L \) for the particle, we need to consider the kinetic and potential energies of the system.

The kinetic energy \( T \) of the particle can be expressed as:

\[ T = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 \right) \]

Since \( y = ax^2 \), differentiating with respect to time \( t \) gives:

\[ \dot{y} = \frac{dy}{dt} = \frac{d}{dt} (ax^2) = 2ax \dot{x} \]

Substituting \( \dot{y} \) into the expression for kinetic energy, we get:

\[ T = \frac{1}{2} m \left( \dot{x}^2 + (2ax \dot{x})^2 \right) = \frac{1}{2} m \left( \dot{x}^2 + 4a^2 x^2 \dot{x}^2 \right) = \frac{1}{2} m \left( 1 + 4a^2 x^2 \right) \dot{x}^2 \]

The potential energy \( V \) of the particle due to gravity is given by:

\[ V = mgy = mg(ax^2) = mgax^2 \]

The Lagrangian \( L \) is the difference between the kinetic and potential energies:

\[ L = T - V = \frac{1}{2} m \left( 1 + 4a^2 x^2 \right) \dot{x}^2 - mgax^2 \]

Final Answer: The correct Lagrangian for the particle is given by option 2:

\[ \mathrm{L} = \frac{1}{2} \mathrm{~m}\left(1+4 \mathrm{a}^{2} \mathrm{x}^{2}\right) \dot{\mathrm{x}}^{2}-\mathrm{mg\ ax}^{2} \]

Explanation:

Corresponding Hamiltonian is given as :

\(H=\frac{P^2}{2m}+ax^2+V_0e^{−10x}\)

Now, clearly, this is not an elliptical relation between p and x. So, we can eliminate options 3 and 4.

• Potential energy \(V(x)=ax^2+V_0e^{−10x}\) has its minimum at x>0.
• When we decrease energy, the phase space closed curve shrinks at a potential minimum or we can say stable equilibrium.
• First graph is to shrink at \(x_0<0\).
• The second graph is to shrink at \(x_0>0\). Hence, is the answer.

Explanation:

The equations of motion for the system in terms of the normal coordinates are obtained by applying Euler-Lagrange equations for i = 1, and i = 2.

Euler-Lagrange equations are: \(\frac d{dt} (\frac{∂L}{∂ẋᵢ}) - \frac{∂L}{∂xᵢ} = 0\) for i = 1, 2. So the equations of motion are \( \ddot x₁ - x₁ - 0.5x₂ = 0\) ...(1) & \(4\ddot x₂ - 0.5x₁ - x₂ = 0\) ....(2) 

Solve the characteristic equation for eigenvalues (which correspond to the squares of the normal mode frequencies), which is in the form of \( |m - λI| = 0\), where m is the 2x2 matrix of coefficients of terms linear in x, I is the 2x2 identity matrix and λ are the eigenvalues.

This equation gives us a quadratic equation for the eigenvalues: \((1 - λ)(4 - λ) - (0.5)(0.5) = λ² - 5λ + 4 - 0.25 = λ² - 5λ + 3.75 = 0\)

Solving this quadratic equation for λ gives: \(λ = \frac {[5 ± \sqrt{(5)² - 15}]}{(2\times1)} = \frac {[5 ± \sqrt{(25 - 15)]}} { 2} = \frac{[5 ± \sqrt{10}]}{ 2} = \frac {[5 ± √10]} { 2 }\)

The corresponding frequencies are the square roots of these eigenvalues: \(ω = \sqrt{(λ)} = \sqrt{\frac{5 ± √10}{ 2}}\)

So we have two frequencies, ω₁ and ω₂: \( ω₁ = \sqrt{5 - √10}\times{ 2} = 0.4\) (approximately, when rounding) \(ω₂ = \sqrt{\frac {[5 + √10]}{ 2} }= 1\)

Concept:

The Lagranges equation of motion of a system is given by

\({d \over dt} {\partial L \over \partial \dot{q}} - {\partial L \over \partial q} = 0\)

Calculation:

The Lagrangian L depends on 

L(x,y,\(̇ x\),\(̇ y\))

\({d \over dt} {\partial L \over \partial \dot{x}} - {\partial L \over \partial x} = 0\)

\({d \over dt} {\partial L \over \partial \dot{ y}} - {\partial L \over \partial y} = 0 \)

L^' = L(x,y,\(\dot x\),\(\dot{y}\))

\({d' \over dt'} {\partial L' \over \partial x} - {\partial L' \over \partial x} = ​​{d \over dt} {\partial L \over \partial x} - {\partial L \over \partial x}+ \ddot{y} = 0\)

⇒\(\dot{y} = c_1\)

Similarly \(\dot{x} = c_2\)

The correct answer is option (2).

Concept:

The Lagranges equation of motion of a system is given by

\({d \over dt} {\partial L \over \partial \dot{q}} - {\partial L \over \partial q} = 0\)

Calculation:

The Lagrangian L depends on 

L(x,y,\(̇ x\),\(̇ y\))

\({d \over dt} {\partial L \over \partial \dot{x}} - {\partial L \over \partial x} = 0\)

\({d \over dt} {\partial L \over \partial \dot{ y}} - {\partial L \over \partial y} = 0 \)

L^' = L(x,y,\(\dot x\),\(\dot{y}\))

\({d' \over dt'} {\partial L' \over \partial x} - {\partial L' \over \partial x} = ​​{d \over dt} {\partial L \over \partial x} - {\partial L \over \partial x}+ \ddot{y} = 0\)

⇒\(\dot{y} = c_1\)

Similarly \(\dot{x} = c_2\)

The correct answer is option (2).

Concept:

The Lagranges equation of motion of a system is given by

\({d \over dt} {\partial L \over \partial \dot{q}} - {\partial L \over \partial q} = 0\)

Calculation:

The Lagrangian L depends on 

L(x,y,\(̇ x\),\(̇ y\))

\({d \over dt} {\partial L \over \partial \dot{x}} - {\partial L \over \partial x} = 0\)

\({d \over dt} {\partial L \over \partial \dot{ y}} - {\partial L \over \partial y} = 0 \)

L^' = L(x,y,\(\dot x\),\(\dot{y}\))

\({d' \over dt'} {\partial L' \over \partial x} - {\partial L' \over \partial x} = ​​{d \over dt} {\partial L \over \partial x} - {\partial L \over \partial x}+ \ddot{y} = 0\)

⇒\(\dot{y} = c_1\)

Similarly \(\dot{x} = c_2\)

The correct answer is option (2).

Calculation:

We are given a particle of mass \( m \) sliding under the influence of gravity without friction along a parabolic path described by \( y = ax^2 \), where \( a \) is a constant.

To determine the Lagrangian \( L \) for the particle, we need to consider the kinetic and potential energies of the system.

The kinetic energy \( T \) of the particle can be expressed as:

\[ T = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 \right) \]

Since \( y = ax^2 \), differentiating with respect to time \( t \) gives:

\[ \dot{y} = \frac{dy}{dt} = \frac{d}{dt} (ax^2) = 2ax \dot{x} \]

Substituting \( \dot{y} \) into the expression for kinetic energy, we get:

\[ T = \frac{1}{2} m \left( \dot{x}^2 + (2ax \dot{x})^2 \right) = \frac{1}{2} m \left( \dot{x}^2 + 4a^2 x^2 \dot{x}^2 \right) = \frac{1}{2} m \left( 1 + 4a^2 x^2 \right) \dot{x}^2 \]

The potential energy \( V \) of the particle due to gravity is given by:

\[ V = mgy = mg(ax^2) = mgax^2 \]

The Lagrangian \( L \) is the difference between the kinetic and potential energies:

\[ L = T - V = \frac{1}{2} m \left( 1 + 4a^2 x^2 \right) \dot{x}^2 - mgax^2 \]

Final Answer: The correct Lagrangian for the particle is given by option 2:

\[ \mathrm{L} = \frac{1}{2} \mathrm{~m}\left(1+4 \mathrm{a}^{2} \mathrm{x}^{2}\right) \dot{\mathrm{x}}^{2}-\mathrm{mg\ ax}^{2} \]

Explanation:

Corresponding Hamiltonian is given as :

\(H=\frac{P^2}{2m}+ax^2+V_0e^{−10x}\)

Now, clearly, this is not an elliptical relation between p and x. So, we can eliminate options 3 and 4.

• Potential energy \(V(x)=ax^2+V_0e^{−10x}\) has its minimum at x>0.
• When we decrease energy, the phase space closed curve shrinks at a potential minimum or we can say stable equilibrium.
• First graph is to shrink at \(x_0<0\).
• The second graph is to shrink at \(x_0>0\). Hence, is the answer.

Explanation:

The equations of motion for the system in terms of the normal coordinates are obtained by applying Euler-Lagrange equations for i = 1, and i = 2.

Euler-Lagrange equations are: \(\frac d{dt} (\frac{∂L}{∂ẋᵢ}) - \frac{∂L}{∂xᵢ} = 0\) for i = 1, 2. So the equations of motion are \( \ddot x₁ - x₁ - 0.5x₂ = 0\) ...(1) & \(4\ddot x₂ - 0.5x₁ - x₂ = 0\) ....(2) 

Solve the characteristic equation for eigenvalues (which correspond to the squares of the normal mode frequencies), which is in the form of \( |m - λI| = 0\), where m is the 2x2 matrix of coefficients of terms linear in x, I is the 2x2 identity matrix and λ are the eigenvalues.

This equation gives us a quadratic equation for the eigenvalues: \((1 - λ)(4 - λ) - (0.5)(0.5) = λ² - 5λ + 4 - 0.25 = λ² - 5λ + 3.75 = 0\)

Solving this quadratic equation for λ gives: \(λ = \frac {[5 ± \sqrt{(5)² - 15}]}{(2\times1)} = \frac {[5 ± \sqrt{(25 - 15)]}} { 2} = \frac{[5 ± \sqrt{10}]}{ 2} = \frac {[5 ± √10]} { 2 }\)

The corresponding frequencies are the square roots of these eigenvalues: \(ω = \sqrt{(λ)} = \sqrt{\frac{5 ± √10}{ 2}}\)

So we have two frequencies, ω₁ and ω₂: \( ω₁ = \sqrt{5 - √10}\times{ 2} = 0.4\) (approximately, when rounding) \(ω₂ = \sqrt{\frac {[5 + √10]}{ 2} }= 1\)