Physics 3221 Mechanics I
Fall Term 2010
Quiz 3
This is a 10 min. quiz (closed book). There are two multiple choice problems.
Problem 1. [2.5 pts] 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 w of small oscillations
about the minimum of the potential is equal to
(A)
q
4b
m
(B)
q
a
2b
(C) 2π
q
a
m
(D) 2π
q
m
a
(E)
q
2a
m
Solution. E. The potential has three extrema, but only the one at x = 0 is a minimum.
Then the equation of motion is
mẍ = Fx = −
dU
= −2ax + 4bx3 ≈ −2ax
dx
ẍ +
from where ω 2 =
2a
x=0
m
2a
.
m
Notice that answers A, B and D can be ruled out simply on dimensional grounds (they have
the wrong units).
Problem 2. [2.5 pts.] A particle of mass m undergoes harmonic oscillation with period T0 .
A retarding force f proportional to the speed v of the particle, f = −bv is introduced. If the
particle continues to oscillate, the period with f acting is now
(A) larger than T0
(B) smaller than T0
(C) independent of b
(D) gradually increasing with time
(E) gradually decreasing with time
Solution. A. This is anqunderdamped oscillator (it continues to oscillate) whose angular
frequency is given by ω1 = ω02 − β 2 (see formula sheet). Therefore ω1 < ω0 and consequently,
T1 > T0 . (Recall that ω ∼ T1 .)
Formula sheet
Simple harmonic oscillator:
mẍ + kx = 0
x(t) = A sin(ω0 t − δ)
x(t) = A cos(ω0 t − φ)
s
2π
ω0 = 2πν0 =
=
τ0
k
m
Damped oscillator:
b
ẍ + 2β ẋ + ω02 x = 0, 2β =
m
√ 2 2
√ 2 2 x(t) = e−βt A1 e β −ω0 t + A2 e− β −ω0 t
Underdamped motion
x(t) = Ae−βt cos(ω1 t − δ),
ω1 =
q
ω02 − β 2
Critically damped motion
x(t) = (A + Bt)e−βt
Overdamped motion
i
h
x(t) = e−βt A1 eω2 t + A2 e−ω2 t ,
Driven oscillator
ω2 =
q
β 2 − ω02
F0
ẍ + 2β ẋ + ω02 x = A cos ωt, A =
m
√ 2 2 √ 2 2
xc (t) = e−βt A1 e β −ω0 t + A2 e− β −ω0 t
A
xp (t) = q
cos(ωt − δ)
2
(ω0 − ω 2 )2 + 4ω 2β 2
δ = tan
−1
q
2ωβ
2
ω0 − ω 2
!
ω02 − 2β 2
ωR
Q=
2β
ωR =
RLC circuit
VL = L
Gauss’s law
Z
S
dI
dt
VR = RI
~n · ~g da = −4πG
VC =
Z
V
ρ dv
q
C