Electromagnetism - Oscillation

Question 3-4: refer to a thin, nonconducting ring of radius R, as shown below, which has a charge Q uniformly spread out on it.

A small particle of mass m and charge –q is placed at point P and released. If R ≫ x, the particle will undergo oscillations along the axis of symmetry with an angular frequency that is equal to:

(GR9677 #04)

Solution:

F_electric = kqQ/r²
F_centripetal = mv²/r = mω²r

F_e = F_c
kqQ/r² = mω²r
ω² = kqQ/mr³

with
k = 1/4πɛ₀
r²  = R² + x²
R ≫ x
r² ∼ R²  → r³ ∼ R³

ω = √(^qQ/_4πɛ0mR³)

Answer: A

Notes: see problem GR9277 #65

Electromagnetism - Faraday’s law

A circular wire loop of radius R rotates with an angular speed ω in a uniform magnetic field B, as shown in the figure. If the emf ɛ induced in the loop is ɛ₀ sin ωt, then the angular speed of the loop is

A. ɛ₀ R/B
B. 2πɛ₀/R
C. ɛ₀/BπR²
D. ɛ₀²/BR²
E. tan⁻¹(ɛ₀/Bc)

(GR9677 #46)

Solution:

Faraday's Law: ɛ = − dΦ/dt
with  Φ = NBA  

Given:
N = 1
B uniform (constant)
ɛ =  ɛ₀ sin ωt

→ ɛ = − B dA/dt =  ɛ₀ sin ωt

− B dA =  ɛ₀ sin ωt dt 

 B ∫dA =  ɛ₀ ∫ sin ωt dt 

− BπR² = − ^ɛ₀/_ω  cos ωt

ω = ɛ₀ cos ωt / BπR²

at t = 0,
ω = ɛ₀ / BπR²

Answer: C

Alternative solution:
A = πR²  cos ωt
ɛ = − dΦ/dt = − B dA/dt 
ɛ = − BπR²  d(cos ωt)/dt 
ɛ₀ sin ωt = BπR²ω  sin ωt 
ɛ₀= BπR²ω  

ω = ɛ₀ / (BπR²)

Classical Mechanics - Lagrangian

A bead is constrained to slide on a frictionless rod that is fixed at an angle θ with a vertical axis and is rotating with angular frequency ω about the axis, as shown above. Taking the distance s along the rod as the variable, the Lagrangian for the bead is equal to

A. ½ mṡ ² − mgs cos θ 
B. ½ mṡ ² + ½ m(ωs)² − mgs 
C. ½ mṡ ² + ½ m(ωs cos θ)² + mgs cos θ
D. ½ m(ṡ sin θ)² − mgs cos θ 

E. ½ mṡ ² + ½ m(ωs sin θ)² − mgs cos θ

(GR9677 #68)

Solution:

Lagrangian: L = T − U

Potential energy: U = mgh = mgs cos θ 
Kinetic Energy:  T_kin  =  ½ mṡ ²
Rotational kinetic energy:  T_rot =  ½ Iω²
with moment inertia: I = mr²  = m(s sin θ)²
→ T_rot = ½ m(ωs sin θ)²

L = T_kin + T_rot − U =  ½ mṡ ² + ½ m(ωs sin θ)² − mgs cos θ

Answer: E

Classical Mechanics - Pendulum

Two pendulums are attached to a massless spring, as shown above. The arms of the pendulums are of identical lengths l, but the pendulum balls have unequal masses m₁ and m₂. The initial distance between the masses is the equilibrium length of the spring, which has spring contant K. What is the highest normal mode frequency of this system?

A.

B.

C.

D.

E.

(GR9677 #84)

Solution:

A. FALSE
ω = √(g/l) is angular frequency for single pendulum

B. and C are FALSE
Both answers do not depend on g/l

D. TRUE
If there is no dependence on K → ω = √(g/l)
And if m₂ → ∞,  m₁ will still oscillate with spring constant K but has no dependence on m_2, as if it were connected to a stationary object.

E. FALSE
If there is no dependence on K → ω = √(2g/l), not angular frequency for single pendulum
And if m₂ → ∞,  ω = √(2g/l) has no dependence on K and m₁

Answer: D 

Electromagnetism - Fourier Series

If n is an integer ranging from 1 to infinity, is an angular frequency, and t is time, then the Fourier series for a square wave, as shown above, is given by which of the following?

A.

B. 

C. 

D. 

E.  

(GR9277 #39)

Solution:

The sine function is an odd function.
The cosine function is an even function.
The function in this problem is an odd square wave function.
→ C, D, E are FALSE.

Both A and B are the same, thus the right solutions. However, A is more generalized and B is the simplified answer.

For Answer A, the sine term vanishes when n is even (n = 0, 2, 4, ...) → rewrite n for odd term as 2n +1 with n starts at 0 → the answer can be simplified to B.

Answer: B

Electromagnetism - Coulomb's Law

Two point charges with the same charge +Q are fixed along the x-axis and are a distance 2R apart as shown. A small particle with mass m and charge –q is placed at the midpoint between them. What is the angular frequency ω of small oscillations of this particle along the y-direction? 

A. 

B.

C.

D.

E.

(GR9277 #65)

Solution:

Check the unit, ω → sec⁻¹

with
Q = q → Coulomb
R = meter
m = kg
ϵ₀ = coulomb² / (newton · meter²) = coulomb² · sec²/ (kg · meter³)

A. coulomb² · kg · meter³ /  (coulomb² · sec² · kg · meter²) =  meter/sec²
→ FALSE.

B. Identical unit with A. → FALSE

C. 1/sec² → FALSE

D. √m/sec → FALSE

E. 1/sec → TRUE

Answer: E

Calculation

Coloumb's Law: 

In this case,



(Electric fields from both Qs are in x-and y-direction, but in x-axis, they cancel each other.)

For small θ, sin θ ≈ tan θ = y/R

Hooke's Law in y-axis: 

Angular frequency,  

Optics - Dispersion Curve

The dispersion curve shown above relates the angular frequency to the wave number k. For waves with wave numbers lying in the range k₁ < k< k₂ which of the following is true of the phase velocity and the group velocity?

A. They are in opposite directions.
B. They are in the same direction and the phase velocity is larger.
C. They are in the same direction and the group velocity is larger.
D. The phase velocity is infinite and the group velocity is finite.
E. They are the same in direction and magnitude.

(GR9277 #79)

Solution:

Phase velocity: v_p = ω/k
Group velocity: v_g = dω/dk

In the range k₁ < k< k₂:
v_p → positive quantity
v_g → negative quantity (negative slope)
→ v_p and v_g are in opposite directions.

Answer: A

Electromagnetism - RLC Circuit

In the RLC circuit shown, the applied voltage is ε(t) = ε_m cos ωt For a constant ε_m, at what angular frequency ω does the current have its maximum steady-state amplitude after the transients have died out?

A. 1/RC
B. 2L/R
C. 1/√(LC)
D. √[(1/LC) − (R/2L)²]
E.  √[(1/RC)² − (L/R)²]

(GR9277 #81)

Solution:

I_max when X_L = X_C

→ ωL = 1 / ωC

ω = 1/√(LC)

Answer: C

Classical Mechanics - Angular Speed

A thin plate of mass M, length L, and width 2d is mounted vertically on a frictionless axle along the z-axis. Initially the object is at rest. It is then tapped with a hammer to provide a torque τ, which produces an angular impulse H about the z-axis of magnitude H = ∫ τ dt. What is the angular speed ω of the plate about the z-axis after the tap? 

A. H/2Md²
B. H/Md²
C. 2H/Md²
D. 3H/Md²
E. 4H/Md²

(GR9277 #82)

Solution:

H = ∫ τ dt

Torque: τ = Iα
Angular acceleration: α = ω/t → ω = αt

H = ∫ τ dt = Iα ∫dt = Iαt = Iω

Moment inertia for the plate about the z-axis: I = ¹/₃Md²

H = ¹/₃Md²ω
ω = 3H/Md²

Answer: D

Classical Mechanics - Pendulum

The figure above represents a point mass m attached to the ceiling by a cord of fixed length l. If the point mass moves in a horizontal circle of radius r with uniform angular velocity ω, the tension in the cord is

A. mgr / l
B. mg cos(θ/2)
C. mωr sin(θ/2)
D. m(ω²r² +g²)^½
E. m(ω⁴r² +g² )^½

(GR8677 #37)

Solution:

Horizontal components:

T sin (θ/2) = F_a = mv² / r = mω²r

Vertical components:
T cos(θ/2) = mg

The magnitude of T:
T² = [T sin(θ/2)]² + [T cos(θ/2)]² = (mω²r)² + (mg)²
T = m(ω⁴r² +g² )^½

Answer: E

Lab Methods - Graph

The gain of an amplifier is plotted versus angular frequency ω in the diagram above. If K and a are positive constants, the frequency dependence of the gain near ω = 3 × 10⁵ second⁻¹ is most accurately expressed by

A. Ke^−aω
B. Kω²
C. Kω
D. Kω⁻¹
E. Kω⁻²

(GR8677 #39)

Solution:

(A) g = Ke^−aω → decreasing (exponential decay). FALSE.


(B) g = Kω² → increasing (quadratic function/parabola). FALSE.


(C) g = Kω → increasing (linear function). FALSE.
(D) g = K/ω → decreasing (linear function). FALSE.

(E) g = K/ω² → decreasing (quadratic function). TRUE.

Check:
At ω = 10⁶, g ≈ 10
g = K/ω² 
10 = K/10¹²
K = 10¹³

At ω = 3 × 10⁵, K = 10¹³
g = 10¹³/(9 × 10¹⁰) ≈ 0.11 × 10³
10²  g  10³

Answer: E

Sound and Wave - Group Velocity

The dispersion law for a certain type of wave motion is ω = (c²k²+m²)^½, where ω is the angular frequency, k is the magnitude of the propagation vector, c and m are constants. The group velocity of these waves approaches 

A. Infinity as k → 0 and zero as k → ∞
B. Infinity as k → 0 and c as k → ∞
C. c as k → 0 and zero as k → ∞
D. Zero as k → 0 and infinity as k → ∞
E. Zero as k → 0 and c as k → ∞

(GR8677 #59)

Solution:





As k → 0, v_g = 0
As k → ∞, v_g = c
since m ≪ →

Answer: E

Classical Mechanics - Potential Energy

A Particle of mass m moves in a one-dimensional potential V(x) = −ax² + bx⁴, where a and b are positive constants. The angular frequency of small oscillations about the minima of the potential is equal to

A. π(a/2b)^1/2
B. π(a/m)^1/2
C. (a/mb)^1/2
D. 2(a/m)^1/2
E. (a/2m)^1/2

(GR9677 #92)

Solution:

The minima of the potential (most probable value x or the equilibrium position of the mass):





Angular Frequency:

Conservative force:




Thus, the angular frequency about the minima of the potential:



Answer: D