Special Relativity - Doppler Effect

The Lyman alpha spectral line of Hydrogen (λ = 122 nanometers) differs by 1.8 × 10⁻¹² meter in spectra taken at opposite ends of the Sun’s equator. What is the speed of a particle on the equator due to the Sun’s rotation, in kilometers per second?

A. 0.22
B. 2.2
C. 22
D. 220
E. 2200

(GR9677 #60)

Solution:

The problem deals with v ≪ c since the  answer is in km/s.

For v ≪ c, Redshift parameter, z = Δλ/λ ≈ v/c

v =  cΔλ/λ

Given:
Δλ = 1.8 × 10⁻¹² m 
λ = 122 nm = 1.22 × 10⁻⁷ m
c = 3 × 10⁸ m/s

v = (3 × 10⁸)(1.8 × 10⁻¹²)/(1.22 × 10⁻⁷) 

= (5.4/1.22) × 10³ m/s  
= (5.4/1.22) km/s ≈ 2.2 km/s

Answer: B

Electromagnetism - Cyclotron

An electron in a metal has an effective mass m* = 0.1 m_e. If this metal is placed in a magnetic field of magnitude 1 tesla, the cyclotron resonance frequency ω_c, is most nearly

A. 930 rad/s
B. 8.5 × 10⁶ rad/s
C. 2.8 × 10¹¹ rad/s
D. 1.8 × 10¹² rad/s
E. 7.7 × 10²⁰ rad/s

(GR9277 #60)

Solution:





Angular velocity: 



Answer: D

Classical Mechanics - Harmonic Oscillator

A particle of mass m that moves along the x-axis has potential energy V(x) = a + bx² , where a and b are positive constants. Its initial velocity is v₀ at x = 0. It will execute simple harmonic motion with a frequency determined by the value of

A. b alone
B. b and a alone
C. b and m alone
D. b, a and m alone
E. b, a, m and v₀

(GR8677 #60)

Solution:



Answer: C

Electromagnetism - Gauss' Law

An infinite, uniformly charged sheet with surface charged density σ cuts through a spherical Gaussian surface of radius R at a distance x from its center, as shown in figure. The electric flux Φ through the Gaussian surface is:

A.  πR²σ / ε₀
B.  2πR²σ / ε₀
C.  π(R − x)²σ / ε₀
D.  π(R² −  x²)σ / ε₀
E.  2π(R² −  x²)σ / ε₀

(GR0177 #60)

Solution:

The Gaussian surface marks a circle around the plane with radius:  

r² = R² −  x²

The net charge within the Gaussian surface:

q_enc = σA = σπr² = σπ(R² −  x²)

Electric Flux:

Φ = q_enc / ε₀ = σπ(R² −  x²) / ε₀ 

Answer: D