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

Solution:

L = (15/2) m x² + 6mxẏ + 3my² − mg(x + 2y)

⇒  (∂L/∂ẋ) − ∂L/∂x = 0 ⇔ 15mẍ + 6mÿ + mg = 0 ........(1)

⇒  (∂L/∂ẏ) − ∂L/∂y = 0 ⇔ 6mẍ + 6mÿ + 2mg = 0 .......(2)

Use operation 2(1) − (2)

24mẍ + 6ẏ = 0 ⇒ d/dt (4x + y) = 0 ⇒ 4ẋ + ẏ = 0 ⇒ 12ẋ + 3ẏ = c

Calculation:

We are given a particle of mass m" id="MathJax-Element-174-Frame" role="presentation" style="position: relative;" tabindex="0">mm $m$ sliding under the influence of gravity without friction along a parabolic path described by y=ax2" id="MathJax-Element-175-Frame" role="presentation" style="position: relative;" tabindex="0">y=ax2y=ax2 $y = ax^2$ , where a" id="MathJax-Element-176-Frame" role="presentation" style="position: relative;" tabindex="0">aa $a$ is a constant.

To determine the Lagrangian L" id="MathJax-Element-177-Frame" role="presentation" style="position: relative;" tabindex="0">LL $L$ for the particle, we need to consider the kinetic and potential energies of the system.

The kinetic energy T" id="MathJax-Element-178-Frame" role="presentation" style="position: relative;" tabindex="0">TT $T$ of the particle can be expressed as:

 

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

Since y=ax2" id="MathJax-Element-180-Frame" role="presentation" style="position: relative;" tabindex="0">y=ax2y=ax2 $y = ax^2$ , differentiating with respect to time t" id="MathJax-Element-181-Frame" role="presentation" style="position: relative;" tabindex="0">tt $t$ gives:

 

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

Substituting y˙" id="MathJax-Element-183-Frame" role="presentation" style="position: relative;" tabindex="0">y˙y˙ $\dot{y}$ 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" id="MathJax-Element-185-Frame" role="presentation" style="position: relative;" tabindex="0">VV $V$ of the particle due to gravity is given by:

 

$$V = mgy = mg(ax^2) = mgax^2$$

The Lagrangian L" id="MathJax-Element-187-Frame" role="presentation" style="position: relative;" tabindex="0">LL $L$ 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).

Solution:

L = (15/2) m x² + 6mxẏ + 3my² − mg(x + 2y)

⇒  (∂L/∂ẋ) − ∂L/∂x = 0 ⇔ 15mẍ + 6mÿ + mg = 0 ........(1)

⇒  (∂L/∂ẏ) − ∂L/∂y = 0 ⇔ 6mẍ + 6mÿ + 2mg = 0 .......(2)

Use operation 2(1) − (2)

24mẍ + 6ẏ = 0 ⇒ d/dt (4x + y) = 0 ⇒ 4ẋ + ẏ = 0 ⇒ 12ẋ + 3ẏ = c

Calculation:

We are given a particle of mass m" id="MathJax-Element-174-Frame" role="presentation" style="position: relative;" tabindex="0">mm $m$ sliding under the influence of gravity without friction along a parabolic path described by y=ax2" id="MathJax-Element-175-Frame" role="presentation" style="position: relative;" tabindex="0">y=ax2y=ax2 $y = ax^2$ , where a" id="MathJax-Element-176-Frame" role="presentation" style="position: relative;" tabindex="0">aa $a$ is a constant.

To determine the Lagrangian L" id="MathJax-Element-177-Frame" role="presentation" style="position: relative;" tabindex="0">LL $L$ for the particle, we need to consider the kinetic and potential energies of the system.

The kinetic energy T" id="MathJax-Element-178-Frame" role="presentation" style="position: relative;" tabindex="0">TT $T$ of the particle can be expressed as:

 

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

Since y=ax2" id="MathJax-Element-180-Frame" role="presentation" style="position: relative;" tabindex="0">y=ax2y=ax2 $y = ax^2$ , differentiating with respect to time t" id="MathJax-Element-181-Frame" role="presentation" style="position: relative;" tabindex="0">tt $t$ gives:

 

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

Substituting y˙" id="MathJax-Element-183-Frame" role="presentation" style="position: relative;" tabindex="0">y˙y˙ $\dot{y}$ 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" id="MathJax-Element-185-Frame" role="presentation" style="position: relative;" tabindex="0">VV $V$ of the particle due to gravity is given by:

 

$$V = mgy = mg(ax^2) = mgax^2$$

The Lagrangian L" id="MathJax-Element-187-Frame" role="presentation" style="position: relative;" tabindex="0">LL $L$ 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) \(ω₂ = \sqrt{\frac {[5 + √10]}{ 2} }= 1\)