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

Classical Mechanics - Pendulum

A cylindrical tube of mass M can slide on a horizontal wire. Two identical pendulum, each of mass m and length l, hang from the ends of the tube, as shown above. For small oscillations of the pendulums in the plane of the paper, the eigenfrequencies of the normal modes of oscillation of the system are 0,,   and …

A.

B.

C.

D.

E.

(GR9277 #07)

Solution:

2 modes in which the system can oscillate:

1. The 2 pendulum oscillate out of phase so there is torsional effect on the tube M → there is M in the equation.

2. In phase → no torsional effect on M → No M in the equation.

Answer: A

Classical Mechanics - Physical Pendulum

A long, straight, and massless rod pivots about one end in a vertical plane. In configuration I, shown above, two small identical masses are attached to the free end; in configuration II, one mass is moved to the center of the rod. What is the ratio of the frequency of small oscillations of configuration II to that of  configuration I? 

A. (6/5)^½
B. (3/2)^½
C. 6/5
D. 3/2
E. 5/3

(GR9277 #61)

Solution:

Angular frequency for physical pendulum: 

ω = (MgL/I)^1/2

Pendulum I
M = 2m
I = mr² +  mr² = 2mr²
L = r + r = 2r

ω_I = (2mg2r/2mr²)^1/2 = (2g/r)^1/2

Pendulum II
M = 2m
I = mr² +  m(r/2)² = (⁵/₄)mr²

L = ½ r + r = (³/₂)r

ω_II =[ 2mg (³/₂)r / (⁵/₄)mr²]^1/2 = (12g/5r)^1/2

The ratio:

ω_II / ω_I = (12g/5r)^1/2 / (2g/r)^1/2 = 6/5

Answer: A  

Classical Mechanics - Small Oscillation

Two circular hoops, X and Y, are hanging on nails in a wall. The mass of X is four times that of Y, and the diameter of X is also four times that of Y. If the period of small oscillations of X is T, the period of small oscillations of Y is

A. T
B. T/2
C. T/4
D. T/8
E. T/16

(GR9277 #74)

Solution:

Small oscillation → simple pendulum.

Period of simple pendulum:

→

In this case, = radius of the hoop.



Answer: B