Classical Mechanics MCQ Quiz - Objective Question with Answer for Classical Mechanics - Download Free PDF

Last updated on May 14, 2025

Latest Classical Mechanics MCQ Objective Questions

Classical Mechanics Question 1:

A satellite of mass m orbits around earth in an elliptic trajectory of semi-major axis a. At a radial distance r = r0, measured from the centre of the earth, the kinetic energy is equal to half the magnitude of the total energy. If M denotes the mass of the earth and the total energy is \( - \frac{{{\rm{GMm}}}}{{{\rm{2a}}}}\), the value of r0 / a is nearest to

  1. 1.33
  2. 1.48
  3. 1.25
  4. 1.67
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 1.33

Classical Mechanics Question 1 Detailed Solution

CONCEPT:

As we know;

Total energy = K.E +P.E

CALCULATION:

Given: K.E = \(\frac{GMm}{4a}\)

and Total energy, \( E= - \frac{{{\rm{GMm}}}}{{{\rm{2a}}}}\)

Now, P.E = \(\frac{GMm}{4a}-​​\frac{GMm}{2a}​​\)

⇒  P.E = \(-\frac{3GMm}{4a}\)

When a particle is the to distance from the center of the earth, then potential energy is written as;

P.E = \(-\frac{GMm}{r_o}\)

Now compare both we have;

\(-\frac{3GMm}{4a}=-\frac{GMm}{r_o}\)

⇒ \(r_o= \frac{4}{3} = 1.33\)

Hence option 1) is the correct answer.

Classical Mechanics Question 2:

Two identical particles (p1 and p2) of mass m are involved in a perfectly elastic collision. The total energy is the E. Particle p1 is initially moving with speed v while p2 is stationary. After the collision, p1 is deflected by θ = 60°. Match the statements in Column-1 with their correct descriptions in Column-2.

Match the Following:

Column-1 Column-2
1) Speed of p2 after the collision (in v) a) 1 / 2
2) Kinetic energy of p2 (in E) b) 1/3 
3) Directions of motion of p1 and p2 c) Perpendicular to each other after the collision
4) Total kinetic energy of both particles

d) Remains unchanged

  e)Parallel to each other after the collision

  1. 1 - a, 2 - b, 3 - d, 4 - c
  2. 1 - b, 2 - a, 3 - c, 4 - e
  3. 1 - a, 2 - b, 3 - c, 4 - d
  4.  1 - d, 2 - b, 3 - a, 4 - c

Answer (Detailed Solution Below)

Option 3 : 1 - a, 2 - b, 3 - c, 4 - d

Classical Mechanics Question 2 Detailed Solution

Explanation:

1. Perfectly Elastic Collision :
    In a perfectly elastic collision, both kinetic energy and momentum are conserved.
    Since both particles are identical and the collision is elastic, conservation of momentum and energy will help determine the final velocities and directions.

2. Conservation of Momentum :
    Let m be the mass of each particle.
    Initially, particle pmoves with speed v , while p2 is at rest.
    After the collision, particle p1 moves at a certain angle of \(60^\circ\) to its original direction, and particle pwill also move at some velocity.

3. Velocity Relations After Collision :
    After the collision, the velocity of p1 changes direction, and a part of its momentum is transferred to p2 .
    For two identical particles in a perfectly elastic collision, if one is initially stationary, they move at an angle of \(90^\circ \) to each other after the collision.
    This means that the directions of motion of p1 and p2 are perpendicular to each other after the collision.

4. Speed of Particle p2 :
    Using conservation of kinetic energy and momentum, we can determine the velocities of both particles.
    For identical masses, the velocities of the particles after the collision can be shown to satisfy:
      The speed of \(p_1\) after the collision is \(v_1' = \frac{v}{2}\) .
      The speed of \(p_2\) after the collision is also \(v_2' = \frac{v\sqrt{3}}{2}\) .
    Thus, the speed of \(p_2\) after the collision is not \(\frac{v}{2}\) . The statement is incorrect.

5. Kinetic Energy of \(p_2\) :
    The kinetic energy of \(p_2\) after the collision can be found using the relation:
     
    \( K_{p_2} = \frac{1}{2} m v_2'^2 = \frac{1}{2} m \left( \frac{v\sqrt{3}}{2} \right)^2 = \frac{3}{8} m v^2 \)
    The total initial kinetic energy of the system was:
     
 \( K_{\text{total}} = \frac{1}{2} m v^2\)
     
    The fraction of the total energy possessed by \(p_2 \) after the collision is:
     
    \( \frac{K_{p_2}}{K_{\text{total}}} = \frac{\frac{3}{8} m v^2}{\frac{1}{2} m v^2} = \frac{3}{4}\)
     
    Therefore, the kinetic energy of \(p_2\) is not\( \frac{1}{3} \) of the total energy. This statement is incorrect.

6. Perpendicular Directions of Motion :
    As stated earlier, after the collision, the two particles move in directions that are perpendicular to each other.
    This statement is correct .

7. Total Kinetic Energy :
    Since the collision is perfectly elastic, the total kinetic energy of the system remains unchanged.
    This statement is correct .

∴ The correct answer is: Option 3 (1 - a, 2 - b, 3 - c, 4 - d).

Classical Mechanics Question 3:

A round disc of moment of inertia I2 about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia I1 rotating with an angular velocity ω about the same axis. The final angular velocity of the combination of discs is 

  1. \(\frac{\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega}{\mathrm{I}_{1}}\)
  2. \(\frac{\mathrm{I}_{2} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}\)
  3. ω 
  4. \(\frac{\mathrm{I}_{1} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{\mathrm{I}_{1} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}\)

Classical Mechanics Question 3 Detailed Solution

Concept:

When two rotating discs are combined, angular momentum is conserved. The total angular momentum before the discs are combined equals the total angular momentum after the discs are combined.

The initial angular momentum of the system is the angular momentum of the first disc:

Linitial = I1 × ω

After the discs are combined, the final angular momentum is:

Lfinal = (I1 + I2) × ωfinal

Using conservation of angular momentum: Linitial = Lfinal, we get:

I1 × ω = (I1 + I2) × ωfinal

Solving for ωfinal:

ωfinal = I1 × ω / (I1 + I2)

Conclusion:

The final angular velocity of the combination is ωfinal = I1 × ω / (I1 + I2). Option 4 is correct.

Classical Mechanics Question 4:

Four holes of radius 5cm are cut from a thin square plate of 20cm and mass 1kg. The moment of inertia of the remaining portion about Z -axis is

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  1. 15 kg − m2 
  2. 0.37 kg − m2  
  3. 0.0017 kg − m2 
  4. 0.08 kg − m2 

Answer (Detailed Solution Below)

Option 3 : 0.0017 kg − m2 

Classical Mechanics Question 4 Detailed Solution

Calculation:

Given:

Area mass density, σ = M / (16 R2) (∵ Area = 4R × 4R = 16R2)

Mass of each hole, m1 = σ × π R2

= (M / 16 R2) × π R2

= (π M) / 16

Distance between the center of the plate and the center of a hole:

x = √ [(2R)2 + (2R)2] / 2

= (2 √2 R) / 2

qImage6794e90a47f9c3869dd361bb

Moment of inertia of one hole about the Z-axis:

I1 = (1/2) m1 R2 + m1 x2

= (5 π M R2) / 32

Moment of inertia of the whole plate about the Z-axis:

I = (M (4 R)2) / 6

= (8 / 3) M R2

Required moment of inertia:

I0 = I - 4 I1

= [(8 / 3) - 4 (5 π / 32)] M R2

= [(8 / 3) - (5 π / 8)] M R2

Given, R = 5 cm and M = 1 kg

So,

I0 = [(8 / 3) - (5 π / 8)] × 1 × 25 × 10-4

= 0.0017 kg·m2

∴ The required moment of inertia is 0.0017 kg·m2.

Classical Mechanics Question 5:

A meter stick is at an angle of 45° to the x - axis in its rest frame. The rod moves with a speed of \(\frac{1}{\sqrt2}\) C along the +x - direction w.r.t. a frame S . The length of the rod in S is 

  1. \(\frac{\sqrt3}{2}\) metres
  2. \(\frac{\sqrt5}{3}\) metres
  3. \(\frac{\sqrt2}{3}\)
  4. \(\frac{2}{3}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{\sqrt3}{2}\) metres

Classical Mechanics Question 5 Detailed Solution

Concept:

L = L₀ √(1 - v²/C²)

  • In special relativity, an object moving at velocity v undergoes **length contraction** along the direction of motion.
  • The contracted length L is given by the formula:
  • The length component along the y-axis remains unchanged.
  • We resolve the length components along x and y and apply length contraction to the x-component.

 

Calculation:

Length of the meter stick in its rest frame, L₀ = 1 m

Angle with the x-axis, θ = 45°

Velocity of the rod, v = (1/√2) C

Speed of light, C

 

⇒ Components of length in rest frame:

L₀ₓ = L₀ cos(θ) = 1 × cos 45° = 1/√2

L₀ᵧ = L₀ sin(θ) = 1 × sin 45° = 1/√2

 

⇒ Length contraction in the x-direction:

Lₓ = L₀ₓ √(1 - v²/C²)

⇒ Lₓ = (1/√2) √(1 - (1/2))

⇒ Lₓ = (1/√2) × (√1/√2) = 1/2

 

⇒ Total length in frame S:

L = √(Lₓ² + L₀ᵧ²)

⇒ L = √((1/2)² + (1/√2)²)

⇒ L = √(1/4 + 1/2)

⇒ L = √(3/4)

⇒ L = √3 / 2

∴ The length of the rod in frame S is √3/2 meters.

Top Classical Mechanics MCQ Objective Questions

When an object undergoes acceleration

  1. A force always acts on it
  2. It always moves down
  3. It always moves up
  4. It always falls towards the earth

Answer (Detailed Solution Below)

Option 1 : A force always acts on it

Classical Mechanics Question 6 Detailed Solution

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The correct answer is Option 1.Key Points

  • When an object undergoes acceleration, it means there is a change in its velocity. This change in velocity can occur either in terms of speed, direction, or both.
  • A force always acts on it:
    • This statement is generally true.
    • According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma).
    • So, if there's acceleration, there must be a force acting on the object.

Additional Information

  • Acceleration is a fundamental concept in physics that describes the rate of change of velocity with respect to time.
  • Velocity is a vector quantity, meaning it has both magnitude (speed) and direction. Therefore, any change in speed, direction, or both constitutes acceleration.
  • The formula for acceleration (a) is a = F/m where a is acceleration. F is the net force acting on an object and m is the mass of the object.
  • This formula states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
  • In simpler terms, if you apply a force to an object, it will accelerate, and the acceleration will be larger if the force is stronger or if the object has less mass.
  • Acceleration can occur in various forms:
    • Linear Acceleration: Change in speed in a straight line.
    • Angular Acceleration: Change in rotational speed or direction.
    • Centripetal Acceleration: Acceleration directed towards the center of a circular path.
  • Acceleration can be positive or negative:
    • Positive Acceleration: Speeding up in the positive direction.
    • Negative Acceleration (Deceleration): Slowing down or moving in the opposite direction.
  • Gravity is a common force causing acceleration. Near the Earth's surface, objects in free fall experience acceleration due to gravity, denoted as g (approximately 9.8 m/s²).

A ball, initially at rest, is dropped from a height h above the floor bounces again and again vertically. If the coefficient of restitution between the ball and the floor is 0.5, the total distance travelled by the ball before it comes to rest is

  1. 8h/3
  2. 5h/3
  3. 3h
  4. 2h

Answer (Detailed Solution Below)

Option 2 : 5h/3

Classical Mechanics Question 7 Detailed Solution

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Concept:

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.

Calculation:

v = \(\sqrt{2gh} \) and v = e\(\sqrt{2gh} \)

0 = (ev)2 - 2gh1

h1 = \({e^2 × 2gh\over 2g}\) = e2h

Similarly, h2 = e4h

H = h + 2h1 + 2h2 +...∞ 

= h + 2(e2h + e4h + ... ∞)

= h + 2e2h(\({1\over 1-e^2}\))

= h × (\({1+ e^2 \over 1-e^2}\))

The coefficient of restitution between the ball and the floor is 0.5.

e = 0.5

H = 5h/3

The correct answer is option (2).

A uniform circular disc on the xy-plane with its centre at the origin has a moment of inertia I0 about the x- axis. If the disc is set in rotation about the origin with an angular velocity ω = ω0(ĵ + k̂), the direction of its angular momentum is along

  1. -î + ĵ + k̂
  2. -î + ĵ + 2k̂
  3. ĵ + 2k̂
  4. ĵ + k̂

Answer (Detailed Solution Below)

Option 3 : ĵ + 2k̂

Classical Mechanics Question 8 Detailed Solution

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Concept:

We are using the angular momentum formula which is \(\overrightarrow{L}=I\overrightarrow{\omega}\) and then using this formula for \(x,y,z,\) planes.

By using matrices for values of \(L,\omega\) and \(I\) we get the value of magnitude and direction of angular momentum.

Explanation:

 A circular disc is in rotation with the origin as center in \(xy\) plane.

Given,

  • \(\omega=\omega_0(\hat j+\hat k)\)  where \(\omega\) is angular velocity
  • \(I_0\) is the moment of inertia

 

We are using formula for angular momentum

  • \(\overrightarrow{L}=I\overrightarrow{\omega}\)
  • \(I_{xx}=I_{yy}=I_0\)

For denoting angular momentum in \(x,y,z,\)planes we will use vector notations for angular momentum \(L\),

  • \(\begin{bmatrix}L_x\\[0.3em]L_y\\[0.3em]L_z\end{bmatrix}\)\(=\begin{bmatrix}I_{xx} & I_{xy} & I_{xz} \\ I_{yx} &I_{yy} & I_{yz} \\[0.3em] I_{zx} & I_{zy} & I_{zz}\end{bmatrix}\)\(\begin{bmatrix}\omega_x\\[0.3em]\omega_y\\[0.3em]\omega_z\end{bmatrix}\)
  • \(I_{xx}=I_{yy}=I_0\)

Using the perpendicular axis theorem,

  • \(I_{xx}+I_{yy}=I_{zz}=I_0+I_0=2I_0\)
  • \(\omega=\omega_0(\hat j+\hat k)\)

putting values of \(I\) and \(\omega\) in matrix equation, we get

  • \(\begin{bmatrix}L_x\\[0.3em]L_y\\[0.3em]L_z\end{bmatrix}\)\(=\begin{bmatrix}I_0 & 0 & 0 \\ 0 &I_0 & 0 \\[0.3em] 0 & 0 & 2I_0\end{bmatrix}\)\(\begin{bmatrix}0\\[0.3em]\omega_0\\[0.3em]\omega_0\end{bmatrix}\)

By multiplication above matrices, we get

  • \(\begin{bmatrix}L_x\\[0.3em]L_y\\[0.3em]L_z\end{bmatrix}\)\(=\)\(\begin{bmatrix}0\\[0.3em]I_0\omega_0\\[0.3em]2I_0\omega_0\end{bmatrix}\)
  • \(\overrightarrow{L}=\overrightarrow{L_0}\hat j+2\overrightarrow{L_0}\hat k\)

This is the magnitude of angular momentum. The direction of angular momentum is \((\hat j +\hat 2k)\).

 

The minor axis of Earth's elliptical orbit divides the area within it into two halves. The eccentricity of the orbit is 0.0167. The difference in time spent by Earth in the two halves is closest to

  1. 3.9 days
  2. 4.8 days
  3. 12.3 days
  4. 0 days

Answer (Detailed Solution Below)

Option 1 : 3.9 days

Classical Mechanics Question 9 Detailed Solution

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Concept:

 We are using Kepler's law here which states that the radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time.

  • \(\frac{dA}{dT}=\frac{L}{2m}=\frac{A} {T}=constant\)

 

Explanation:

Using Kepler's second law, 

F1 Teaching Arbaz 23-10-23 D16

  • \(\frac{dA}{dT}=\frac{L}{2m}=\frac{A} {T}=constant\)
  • \(A_1=\frac {\pi ab}{2}+2\times\frac{1}{2}\times b\times c\)

eccentricity(e)\(=\frac {c}{a}\)=>\(c=ea\)

  • \(A_1=\frac {\pi ab}{2}+eba=ab(\frac {\pi} {2}+e)\)
  • \(A_2=\pi ab-ab(\frac {\pi}{2}+e)=ab(\frac {\pi} {2}-e)\)

Now,

  • \(\frac{T_1}{T_2}=\frac{A_1}{A_2}\)
  • \(\frac{T_1}{T_2}=\frac {ab(\frac {\pi}{2}+e)} {ab(\frac {\pi}{2}-e)}=\frac {(\frac {\pi}{2}+e)} {(\frac {\pi}{2}-e)}\)
  • \(T_1=\frac {\frac{\pi }{2}+e}{\pi}, T_2=\frac {\frac{\pi }{2}-e}{\pi}\)
  • \(T_1-T_2=\frac{2e}{\pi}=\frac {2\times0.0167\times 365}{3.14}\approx3.9 days.\)

 

For the transformation x → X = \(\frac{α p}{x},\) p → P = βx2 between conjugate pairs of a coordinate and its momentum, to be canonical, the constants α and β  must satisfy

  1. \(1+\frac{1}{2} \alpha \beta=0\)
  2. \(1-\frac{1}{2} \alpha \beta=0\)
  3. 1 + 2αβ = 0
  4. 1 - 2αβ = 0

Answer (Detailed Solution Below)

Option 3 : 1 + 2αβ = 0

Classical Mechanics Question 10 Detailed Solution

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Explanation:

  • Given the transformations \(X=\frac{α p}{x}\) and \(P = βx²\), we have to check them against the requirements of canonical transformations.
  • The main requirement, among others, is that the Poisson bracket {X, P} = 1.
  • In terms of partial derivatives, the Poisson bracket can be written as: \({X, P} = (\frac{∂X}{∂x})(\frac{∂P}{∂p}) - (\frac{∂X}{∂p})(\frac{∂P}{∂x})\)
  • Substituting X and P from the problem statement, we get: \({X, P} = (\frac{∂(αp/x)}{∂x})(\frac{∂(βx²)}{∂p}) - (\frac{∂(αp/x)}{∂p})(\frac{∂(βx²)}{∂x})\)
  • Computing partial derivatives, we get:\({X, P} = (\frac{-αp}{x²})(0) - (\frac{α}{x})(2βx) = -2αβ \)
  • Setting the above equation to 1: \(-2αβ = 1\)
  • Finally, \(1+2αβ = 0\)

Which of the following terms, when added to the Lagrangian L(x, y, \(\dot x\), \(\dot y\)) of a system with two degrees of freedom, will not change the equations of motion?

  1. \(x\ddot x - y\ddot y\)
  2. \(x\ddot y - y\ddot x\)
  3. \(x\dot y - y\dot x\)
  4. \(y{\dot x^2} + x{\dot y^2}\)

Answer (Detailed Solution Below)

Option 2 : \(x\ddot y - y\ddot x\)

Classical Mechanics Question 11 Detailed Solution

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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).

The trajectory of a particle moving in a plane is expressed in polar coordinates (r, θ) by the equations \(r=r_0 e^{β t} \text { and } \frac{d \theta}{d t}=ω\), where the parameters r0, β  and ω are positive. Let vr and ar denote the velocity and acceleration, respectively, in the radial direction. For this trajectory

  1. a< 0 at all times irrespective of the values of the parameters
  2. ar​ > 0 at all times irrespective of the values of the parameters
  3. \(\frac{d v_r}{d t}>0\) and ar > 0 for all choices of parameters
  4. \(\frac{d v_r}{d t}>0\) however, ar = 0 for some choices of parameters

Answer (Detailed Solution Below)

Option 4 : \(\frac{d v_r}{d t}>0\) however, ar = 0 for some choices of parameters

Classical Mechanics Question 12 Detailed Solution

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Explanation:

We will first write the velocity vector in polar co-ordinates and write the radial velocity and radial acceleration and by differentiating we get the desired solution.

Given,-\(r=r_0 e^{β t} \text { and } \frac{d \theta}{d t}=ω\)     ----------------1

Velocity in polar co-ordinates is given by the sum of radial and transverse velocity.

\(v=v_r+v_t=\dot r \hat{r}+r\dot \theta \hat{\theta}\)

Radial velocity \(v_r=\dot r\)

Radial acceleration \(a_r=(\ddot {r}-r\dot \theta^2)\)

Now, \(r=r_0 e^{β t} \) and \(\dot r=r_0 \beta e^{β t} \)

\(\frac{d v_r}{d t}=\)\(\beta ^2r_0 \beta e^{β t} \) => \(\frac{d v_r}{d t}\)\(>0\)

\(\ddot r=r_0 \beta^2 e^{β t} \) and \(\ddot \theta^2=\omega^2\)

\(a_r=\)\(\beta ^2r_0 \beta e^{β t} \) \(-r\omega^2\)\(=\)\(\beta ^2r_0 \ e^{β t} \)\(-r_0 \omega^2e^{β t}\)\(=\)\(r_0 \ e^{β t} (\beta^2-\omega^2)\)

If \(\beta=\omega\)\(a_r=0\) for some choices of parameters.

So, the correct answer is  -\(\frac{d v_r}{d t}>0\) however, \(a_r=0\) for some choices of parameters.

The Hamiltonian of a system with two degrees of freedom is H = q1p1 - q2p2\(aq_1^2\) , where a > 0 is a constant. The function q1q2 + λp1P2 is a constant of motion only if λ is

  1. 0
  2. 1
  3. -a
  4. a

Answer (Detailed Solution Below)

Option 1 : 0

Classical Mechanics Question 13 Detailed Solution

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Concept:

The Hamiltonian of a system specifies its total energy—i.e., the sum of its kinetic energy (that of motion) and its potential energy (that of position)—in terms of the Lagrangian function derived in earlier studies of dynamics and of the position and momentum of each of the particles.

Calculation:

H = q1p1 - q2p2 + \(aq_1^2\) , where a > 0 is a constant.

f = (q1q2 + λp1p2)

\({df \over dt}\) = [f,H] + \({\partial f \over \partial t }\)

\({\partial f \over \partial t }\) = 0 ⇒ \({df \over dt}\) = [f,H] = 0

[f,H] = \([{\partial f \over \partial q_1}{\partial H \over \partial p_1}-{\partial f \over \partial p_1}{\partial H \over \partial q_1}]+[{\partial f \over \partial q_2}{\partial H \over \partial p_2}-{\partial f \over \partial p_2}{\partial H \over \partial q_2}]= 0\)

q2 . q1 - λ p2 (p1 + 2aq1) + q1(-q2) - λ p1(-p2) = 0

∴ λ = 0

The correct answer is option (1).

The Hamiltonian of a two particle system is H = p1p2 + q1q2, where q1 and q2 are generalized coordinates and p1 and p2 are the respective canonical momenta. The Lagrangian of this system is

  1. \(\dot{q}_1 \dot{q}_2+q_1 q_2\)
  2. \(-\dot{q}_1 \dot{q}_2+q_1 q_2\)
  3. \(-\dot{q}_1 \dot{q}_2-q_1 q_2\)
  4. \(\dot{q}_1 \dot{q}_2-q_1 q_2\)

Answer (Detailed Solution Below)

Option 4 : \(\dot{q}_1 \dot{q}_2-q_1 q_2\)

Classical Mechanics Question 14 Detailed Solution

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Concept:

We will use the relationship of Hamiltonian and Lagrangian which is given by

  • \(H=\sum p_i \dot{q_i}-L\)
  • For a two-particle system, we can write the above equation as \(H=p_1\dot{q_1}+p_2\dot {q_2}-L\)

 

Explanation:

Given, \(H=p_1p_2+q_1q_2\)

  • \(H=p_1\dot{q_1}+p_2\dot {q_2}-L\)
  • \(H=p_1p_2+q_1q_2\) (given)

substitute value of H, we get,

  • \(L=p_1\dot{q_1}+p_2\dot {q_2}-p_1p_2-q_1q_2\)--------------------------------1
  • Using Hamilton equations, \(\dot p=\frac{-\partial H}{\partial q}, \dot{q}=\frac{\partial H}{\partial p}\)
  • \(\dot q_1=\frac {\partial H}{\partial p_1}=p_2\) and \(\dot q_2=\frac {\partial H}{\partial p_2}=p_1\)
  • Put the value of \(p_1\) and \(p_2\) in equation 1, to get the value of Lagrangian
  • \(L=p_1\dot{q_1}+p_2\dot {q_2}-p_1p_2-q_1q_2\) \(=\)\(L=\dot q_2\dot{q_1}+\dot q_1\dot {q_2}-\dot q_1\dot q_2-q_1q_2\)
  • \(L=\dot q_2 \dot q_1-q_1q_2\)

 

So, the correct answer is \(L=\dot q_2 \dot q_1-q_1q_2\).

 

A system of two identical masses connected by identical springs, as shown in the figure, oscillates along the vertical direction.

F1 Teaching Arbaz 23-10-23 D23
The ratio of the frequencies of the normal modes is

  1. \(\sqrt{3-\sqrt{5}}: \sqrt{3+\sqrt{5}}\)
  2. \(3-\sqrt{5}: 3+\sqrt{5}\)
  3. \(\sqrt{5-\sqrt{3}}: \sqrt{5+\sqrt{3}}\)
  4. \(5-\sqrt{3}: 5+\sqrt{3}\)

Answer (Detailed Solution Below)

Option 1 : \(\sqrt{3-\sqrt{5}}: \sqrt{3+\sqrt{5}}\)

Classical Mechanics Question 15 Detailed Solution

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Concept:

We will first write lagrangian for a given condition then we use the equation for normal nodes which is given by \(|\hat V-\omega^2\hat T|=0\)

Explanation:

Given m are two identical masses, k is spring constant and \(x_1\) and \(x_2\)are displacement of spring first and spring second respectively.

  • \(T=\frac {1}{2}mx_1^2+\frac{1}{2}mx_2^2\)
  • \(V=\frac {1}{2}k[x_1-0]^2+\frac{1}{2}[x_2-x_1]^2\)
  • \(V=\frac{1}{2}kx_1^2+\frac{1}{2}kx_1^2+\frac{1}{2}kx_2^2-kx_1x_2\)
  • \(V=kx_1^2+\frac{1}{2}kx_2^2-\frac{kx_1x_2}{2}-\frac{kx_1 x_2}{2}\)
  • Using matrix we can write operators of \(V\) and \(T\) as,
  • \(\hat T=\begin{bmatrix} \frac{m}{2} & 0 \\[0.3em] 0 & \frac{m}{2} \\[0.3em] \end{bmatrix} \) and \(\hat V=\begin{bmatrix} k & \frac{-k}{2} \\[0.3em] \frac{-k}{2} & \frac{k}{2} \\[0.3em] \end{bmatrix} \)
  • Put these values in equation-\(|\hat V-\omega^2\hat T|=0\) to get the ratio of normal modes of frequencies,
  • \(|\hat V-\omega^2\hat T|=0\)
  • \(k|\begin{bmatrix} (1-\frac{\omega^2}{2}) & \frac{-1}{2} \\[0.3em] \frac{-1}{2} &( \frac{1}{2}-\frac{\omega^2}{2}) \\[0.3em] \end{bmatrix} |=0\)
  • \(\det \begin{bmatrix} (1-\frac{\omega^2}{2}) & \frac{-1}{2} \\[0.3em] \frac{-1}{2} & (\frac{1}{2}-\frac{\omega^2}{2}) \\[0.3em] \end{bmatrix} =0\)
  • \((1-\frac{\omega ^2}{2})(\frac{1}{2}-\frac {\omega^2}{2})=0\)
  • \(\frac{1}{2}-\frac{\omega^2}{2}-\frac {\omega^2}{4}+\frac{\omega^4}{4}-\frac{1}{4}=0\)
  • \(2-2\omega^2-\omega^2+\omega^4-1=0 \)
  • \(\omega^4-3\omega^2+1=0\)
  • Solution of above equation is \(\omega^2=\frac {3\pm \sqrt5}{2}\)
  • \(\therefore \omega_+=\sqrt \frac{3+\sqrt 5}{2}, \omega_-=\sqrt\frac{3-\sqrt {5}}{2}\)
  • ratio of both frequencies \(=\frac {\omega_-}{\omega_+}=\frac{\sqrt {3-\sqrt 5}}{\sqrt{3+\sqrt 5}}\)

 

So, the correct answer is  \(=\frac {\omega_-}{\omega_+}=\frac{\sqrt {3-\sqrt 5}}{\sqrt{3+\sqrt 5}}\)

 

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