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Phys 3610, Fall 2008
Exam #
- Give definitions (explanations) of:
a) Stable equilibrium
Configuration (position) of a system where the force on a particle is zero but for which a small
displacement from that position will send the mass moving far away, not returning.
b) Decay time of a damped oscillator
Time required for the amplitude (or position itself, if there are no oscillations to decay to a small
fraction (like 1 /e) of its maximum value.
c) Resonance frequency of a driven damped system.
The frequency of the driving force (or possibly the natural frequency of the oscillator if we are
looking changing that quantity) for the amplitude of the long-term solution is a maximum.
- Consider the isotropic harmonic oscillator in 2 dimensions, for a particle of mass m, where
we have
F = −kr
a) Find the general solution for the coordinates of the particle, x(t) and y(t).
The force law decomposes as
Fx = −kx and Fy = −ky
So they x and y coordinates separately oscillate with general solutions:
x(t) = Ax cos(ωt − δx) y(t) = Ay cos(ωt − δy)
b) Suppose at t = 0 the particle has coordinates and velocity given by
r 0 = Rxˆ and v 0 = v 0 yˆ
Solve for x(t) and y(t).
Here we have
x(0) = R y(0) = 0 x˙(0) = 0 y˙(0) = v 0
so, considering that
x ˙(t) = −ωAx sin(ωt − δx) y˙(t) = −ωAy sin(ωt − δy)
a little cogitation gives
x(t) = R cos(ωt) y(t) =
v 0
ω
sin(ωt)
as the correct solutions for x(t) and y(t).
- The potential energy for a mass m moving in one dimension is given by
U(x) = −
A
cosh
2 (αx)
where A and α are positive constants.
a) Sketch this function; what is the point of equilibrium? Is the equilibrium stable?
y
x
Since cosh(x) has its minimum at x = 0 then (^) cosh(^1 x) must have
its maximum at x = 0 and go to zero at x = ±∞. Then the given
function ought to behave as shown in the graph here.
Clearly the minimum value is at x = 0, as seen from the deriva-
tive of U:
dU
dx
= +2Aα
sinh(αx)
cosh
3 (αx)
which is zero at x = 0.
b) Well, it is stable! For oscillations about this point find the equivalent harmonic potential
and identify the force constant k.
We need to find the behavior of the function for small x. To get, find the Taylor series for the
given U(x) up to second order. We need the second derivtive, and we find:
U
′′ (x) =
2 Aα^2
cosh
2 (αx)
6 Aα^2 sinh
2 (αx)
cosh
4 (αx)
so that
U
′′ (0) = 2Aα
2
Then the Taylor expansion for U(x) is
U(x) = U(0) +
U
′ (0)x +
U
′′ (x)x
2
= −A + 0 + Aα
2
which is equivalent to a constant term plus a potential
1 2 kx
2 if we indentify
1 2
k = Aα
2 =⇒ k = 2Aα
2
- Sketch (reasonably accurately) a graph of the motion x(t) of a damped harmonic oscillator
if it has displacement x = A and v = 0 at t = 0 and
a) β = 15 ω 0 (Hint: What is decay time?)
so if we point the y axis downward, then the time spent on each path element is the length divided
by the speed:
dt =
√ x′(y)^2 + 1dy √ 2 gy
and the integral we want to minimize is
S =
2 g
∫ (^2)
1
√ x′(y)^2 + 1dy √ y
Of course we can just minimize the integral without the 1 /
2 g factor in front.
- Find the equation of the path joining the origin O to the point P (1, 1) that makes the
integral
∫ (^) P O (y
′ 2
2 )dx stationary.
Here we have
f(y
′ , y, x) = (y
′ 2
2 )
and we calculate ∂f
∂y
= 2y
∂f
∂y′^
= 2y
′
then the E-L equation gives
2 y =
d
dt
(2y′) = 2y′′^ =⇒ y′′^ = y
which has the general solution
y(x) = A cosh(x) + B sinh(x)
Since (0, 0) is on the curve, this gives 0 = A and since (1, 1) is on the curve, we get
1 = B sinh(1) =⇒ B =
sinh(1)
so the solution is
y(x) =
sinh(x)
sinh(1)
- Find and describe the path y = y(x) for which the integral
∫ (^) x 2
x 1
x
√ 1 − y′^2 dx is station-
ary.
(On this one, you can leave any constants of integration undetermined, since I didn’t give
values for x 1 and x 2 .)
Here we have
f(y
′ , y, x) =
x
√ 1 − y′^2
which we note does not depend on y so as we've seen, the E-L equation implies
∂f
∂y′^
= const =⇒
xy
′
1 − y′^2
xy
′ √ 1 − y′^2
= C
This gives
xy
′ 2 = C
2 (1 − y
′ 2 ) =⇒ (C
2
′ 2 = C
2 =⇒ y
′
C
C^2 + x
This integrates pretty easily to give
y =
C 12
√ C 12 + x + C 2 =⇒ y − C 2 =
C 12
√ C 12 + x
Then squaring both sides gives
(y − C 2 )
2
C
4 1
4
(C
2 1 +^ x)
Anyway, this is a parabola.
