Physics Problems & Solutions: #43

Showing posts with label #43. Show all posts

Electromagnetism - Stokes Theorem

The line integral of u = yi − xj + zk around a circle of radius R in the xy-plane with center at the origin is equal to

A. 0
B. 2πR
C. 2πR²
D. πR²/4
E. 3R³

(GR9677 #43)

Solution:

Stokes' Theorem: The line integral of a vector field around a closed curve is equal to the surface integral of the curl over that vector field.

∮ F · dl = ∫ (∇ × F) · dA

Given: u = yi − xj + zk

∇ × u =
$\left| \begin{array}{ccc} \hat{i} & \hat{j}& \hat{k}\\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ y & -x & i \end{array} \right| \begin{array}{cc} \hat{i} & \hat{j} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} \\ y & -x \end{array}$
= 0 + 0 − k̂ − k̂ − 0 − 0 = − 2k̂

∫ (∇ × u) · dA = − ∫ 2k̂ dA = − 2k̂ ∫ dA = − 2k̂ A = − 2πR²k̂

Answer: C

Classical Mechanics - Lagrangian

If ∂L/∂q_n = 0, where L is the Lagrangian for a conservative system without constraints and q_n is a generalized coordinate, then the generalized momentum p_n is

A. An ignorable coordinate
B. Constant
C. Undefined
D. Equal to $\frac{d}{dt} \left (\frac{\partial L}{\partial q_n} \right )$
E. Equal to the Hamiltonian for the system

(GR9277 #43)

Solution:

Lagrangian Equation of Motion for the generalized coordinate q:

$\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q}$

Momentum,

$\begin{aligned} p&=mv=m\dot{x} \\L&=T-U=\frac{1}{2}m\dot{x}^2-mgy\\&\rightarrow \frac{\partial L}{\partial \dot{x}}=m\dot{x}=p \end{aligned}$

For the generalized coordinate q:

If $\frac{\partial L}{\partial q_n}=0 \rightarrow \frac{\partial L}{\partial \dot{q_n}}=p_n= \text{constant}$

Answer: B

Classical Mechanics - Simple Harmonic Oscillation

Three masses are connected by two springs as above. A longitudinal normal mode with frequency ¹/₂_π√^k/_m is exhibited by

A, B, C all moving in the same direction with equal amplitude.
A and C moving in opposite, and B at rest.
A and C moving in the same direction with equal amplitude, and B moving in the opposite direction with the same amplitude.
A and C moving in the same direction with equal amplitude, and B moving in the opposite direction with twice the amplitude.
None of the above.

(GR8677 #43)

Solution:

Angular frequency, ω = 2πf → f = ^ω/₂_π
For Simple Harmonic Oscillation (SHO), ω = √^k/_m
Thus, f = ¹/₂_π√^k/_m is frequency of SHO.

SHO is a periodic motion of a particle about a fixed point (the centre of motion).

Only choice (B) gives a fixed point (mass B at rest).

Answer: B

Quantum Mechanics - Commutation Relations

The components of the orbital angular momentum operator $\vec{L}=(L_x,L_y,L_z )$ satisfy the following commutation relations.

$\\\left[L_x,L_y \right ]=i \hbar L_z \\\left[L_y,L_z \right ]=i \hbar L_x \\\left[L_z,L_x \right ]=i \hbar L_y$

What is the value of the commutator $\left[L_x L_y,L_z \right ]$ ?

A. $2i\hbar L_x L_y$
B. $i\hbar (L_x^2+L_y^2 )$
C. $-i\hbar (L_x^2+L_y^2 )$
D. $i\hbar (L_x^2-L_y^2 )$
E. $-i\hbar (L_x^2-L_y^2 )$

(GR0177 #43)

Solution:

Commutator identities: [A, B] = − [B, A]

and, [B, AC] = A[B, C] + [B, A]C

Therefore,
$\left[\underset{A}{\underbrace{L_x L_y}},\underset{B}{\underbrace{L_z}}\right] = - \left[\underset{B}{\underbrace{L_z}}, \underset{A}{\underbrace{L_x L_y}}\right]$

$\left[\underset{B}{\underbrace{L_z}}, \underset{A}{\underbrace{L_x}}\underset{C}{\underbrace{L_y}}\right]=L_x [L_z,L_y ]+[L_z,L_x ] L_y$

$\begin{aligned} -[L_z,L_x L_y ]&=-L_x [L_z,L_y ]-[L_z,L_x ] L_y \\&=-L_x (-i\hbar L_x)-i\hbar L_y L_y \\&=i\hbar L_x^2-i\hbar L_y^2 \\&=i\hbar (L_x^2-L_y^2 ) \end{aligned}$

Answer: D