Physics 3221 Mechanics I Fall Term 2010 Quiz 3

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 m¨ x = Fx = −

dU = −2ax + 4bx3 ≈ −2ax dx x¨ +

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: m¨ x + 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 x¨ + 2β x˙ + 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 x¨ + 2β x˙ + 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