exam2_solutions

exam2_solutions - Exam
2
for
PHYSICS
2210


Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Exam
2
for
PHYSICS
2210
 9:30
to
10:45
am
on
Nov
5.
2009
 
 Name:






























































































Score:
 
 1)
Multiple‐choice
questions
(30
points
total,
and
5
points
each
question).
 
 1a)
For
r,
a,
and
b
real
constants,
the
complex
conjugate
of
 re a +ib 
is:
 
 A)
 re-a +ib ,












 B)
 -re a +ib ,













 € -a +ib C)
 re ,












 a −ib D)
 re ,
 -a −ib E)
 re ,













 
Answer:
D)
 
 1b)
For
2nd
order,
inhomogeneous,
ordinary
differential
equation:

 d2x dx + 3 + 2x = 5e− t .
For
arbitrary
constant
k,
which
of
the
following
form
is
 dt dt 2 correct
for
the
particular
solution,
xp(t)?
 
 A)
 ke-t ,















 B)
 ke-2t ,













 C)
 kte-t ,

















 D)
 kte-2t ,















 E)
 kt 2e-t .
 Answer:
C)
 
 1c)
For
a
driven
oscillator
(spring‐block)
with
damping,
there
is
a
resonant
 frequency
ωR.
If
the
driving
force
is
at
this
resonant
frequency,
the
amplitude
of
the
 response
is
maximized.
Suppose
that
the
spring
constant
is
k,
the
mass
of
the
block
 is
m,
damping
parameter
is
b,
and
the
magnitude
of
driving
force
is
F0.
Which
of
the
 following
statement
is
true
for
ωR?
 
 A)
ωR
depends
on
k,
m,
b,
and
F0.

 B)
ωR
depends
on
k,
m,
and
b.


 C)
ωR
depends
on
k,
m
and
F0.


 D) ωR
depends
on
k
and
m.
 E)
None
of
above.

 
Answer:
B)
 
 
 € € € € € € € € € € € 
 1
 x(t) or G(t,t’) 1d)
The
top
left
panel
shows
the
response
of
an
oscillator
to
an
impulse
force,
i.e.,
 the
Green’s
function,
G(t,t’).
Suppose
that
a
force
function,
f(t),
consists
of
two
 impulses
applied
at
two
distinct
times
(t=2
and
3),
shown
in
the
top
right
panel.

 
 Among
A,
B,
C,
and
D,
what
is
the
correct
response
of
the
oscillator
to
this
force
f(t)?
 
 1 1 0 Force f(t) 0 −1 1 0 −1 1 0 0 1 2 t-t’3 4 5 6 −1 1 0 0 1 2 t 3 4 5 6 A) B) x(t) 0 1 2 t3 4 5 6 x(t) −1 1 0 0 1 2 t 3 4 5 6 C) D) x(t) −1 
 Answer:
C)
 

 
 1e)
How
is
the
block’s
kinetic
energy
KE,
of
an
oscillator
(spring‐block)
dependent
 on
quality
factor
Q
of
the
oscillator?

 
 A)
 KE
decays
with
time
faster
for
larger
Q
 B)
 KE
decays
with
time
slower
for
larger
Q
 C)
 KE
decays
with
time
but
does
not
depend
on
Q.
 D)
 KE
remains
constant
with
time,
independent
of
Q.
 
 Answer:
B)
 

 
 0 1 2 t 3 4 5 6 x(t) −1 0 1 2 t 3 4 5 6 
 2
 1f)
Consider
a
spring‐block
oscillator
with
no
damping.
Suppose
that
the
spring
 k , for x > 0 constant
k
is
given
as:
 k = 0 
,
where
k0
is
a
positive
constant,

Which
 0.5k0 , for x < 0 of
the
following
phase
diagrams
is
correct?
 
 v v A) B) € x x C) v D) v x x 
 

 Answer:
B) 
 
 3
 2)
(50
points
total)
This
problem
has
four
parts.
A
block‐spring
oscillator
(see
figure
 below)
is
placed
in
a
water
tank.
The
spring
constant
is
k,
the
block’s
mass
is
m,
and
 the
resisting/damping
force
on
the
block
is
given
as
 Fr = −bv ,
where
b>0
and
v
is
 velocity.
Suppose
that
at
the
beginning
(i.e.,
t=0),
the
block
is
at
the
equilibrium
 position
but
moves
to
the
right
side
at
a
speed
of
v0.
Suppose
that
the
oscillation
is
a
 under­damping
system.

 € 
 2a)
(15
points)
(i)
Setup
a
coordinate
system,
by
 k specifying
and
drawing
the
origin
and
positive
 m direction
of
the
coordinate
system.
(ii)
Write
down
 the
Newton’s
second
law
as
a
second
order
 differential
equation.
(iii)
What
is
the
condition
for
 
 under‐damping?
 
 
 
Answer:
Let
x
be
the
displacement
from
the
 equilibrium
position
of
the
block,
with
positive
x
 pointing
to
the
right
side
(I
ignore
the
drawing
 here).
The
Newton’s
second
law
in
this
coodinate
system
is

 b k b k ˙ ˙ ˙ m˙˙ = −kx − bx ,
or

 ˙˙ + x + x = 0 ,
or
 ˙˙ + 2βx + ω 2 x = 0 ,
where
 β = x x , ω 2 = .
 x 0 0 m m 2m m 2/(4m).
or
 β 2 < ω 2 .
 For
under‐damping,
k>b € 
 € € € 
 2b)
(15
points)
Solve
the
equation
for
the
solution
of
oscillation
with
time,
using
the
 € initial
conditions
(Hint:
Use
the
cosine
or
sine
function
form
for
the
general
solution.
 If
you
introduce
new
variables
such
as
β
or
ω,
you
need
to
state
how
they
are
related
 to
k,
m,
and
b).
 
 Answer:
The
solution
before
considering
initial
condition
is
 0 x(t) = Ae- βt cos(ωt - δ ) ,
where
 ω = ω 2 − β 2 ,
and
A
and
δ
are
constants
to
be
 0 € € € determined.
 At
t=0,
x(0)=0,
and
v(0)=v0.
 ˙ v(t) = x(t) = -Ae- βt [ βcos(ωt - δ ) + ωsin(ωt - δ )] .
 € From
x(0)=0,
δ=π/2.
And
from
v(0)=v0,
A=v0/ω.
 v π v Therefore,
 x(t) = 0 e- βt cos(ωt - ) = 0 e- βt sin(ωt) ,
and
 ω 2 ω v0 - βt ˙ v(t) = x(t) = - e [ βsin(ωt) - ωcos(ωt)] .
 ω 
 € 
 
 
 
 
 4
 c)
(10
points)
Sketch
the
phase
diagram.

 v v0 x € € 
 
 
 
 
 
 
 
 
 d)
(10
points)
The
total
mechanical
energy,
E,
of
the
oscillation
system
decays
with
 time,
due
to
damping.
Derive
an
expression
for
the
rate
of
the
energy
decay
with
 time:
dE/dt,
using
the
results
from
part
b).
 
 1 1 Answer:

 E = T + U = mv 2 + kx 2 .
 2 2 dE ˙ ˙ ˙x ˙˙x ˙ = mvv + kxx = mx˙˙ + kxx = x(m˙˙ + kx) = -bx 2 < 0 ,

 dt ˙ x where
we
considered
 m˙˙ = −kx − bx .
 € 2 bv 0 dE dE ˙ = - 2 e-2 βt [ βsin(ωt) - ωcos(ωt)] 2 .
 = -bx 2 


 dt dt ω 
 € 
 
 
€ 
 
 5
 3)
(20
points)
A
wooden
block
with
uniform
cross
section
area,
A,
and
a
height
H,
 floats
in
a
water
pool
at
the
equilibrium
position.
Suppose
that
the
densities
for
the
 block
and
water
are
ρb
and
ρ0,
respectively.
The
block
is
perturbed
slightly
by
 pulling
the
block
up
a
little
and
then
releasing
it.
Assume
that
the
viscous
resisting
 force
and
frictional
force
can
be
ignored.
i)
Set
up
a
coordinate
system
(the
origin
 and
the
positive
direction),
and
write
down
the
Newton’s
2nd
law
for
the
block,
ii)
 demonstrate
that
the
block
undergoes
oscillatory
motion
after
the
perturbation,
and
 iii)
determine
the
period
of
the
oscillation
and
express
the
solution
in
terms
of
the
 variables
given
above
(g
is
the
gravitational
acceleration).

 
 air H ρb ρ0 water € € € 
 
 
 Answer:

Set
up
a
coordinate
system
with
the
origin
at
the
center
of
mass
of
the
 block
at
the
equilibrium,
and
positive
direction
pointing
upward
(drawing
is
ignored
 here).

 First,
at
the
equilibrium,
the
weight
of
the
block
is
balanced
by
buoyancy.

 mg = Fb_e ,

 where
m=ρbHA,
is
the
mass
of
the
block,
and
Fb_e=ρ0heAg,
is
the
buoyancy
at
the
 equilibrium,
where
he
is
the
height
of
the
block
under
the
water
at
equilibrium.
 he=Hρb/ρ0.

 After
the
perturbation
is
introduced,
the
Newton’s
2nd
law
is
 m˙˙ = −mg + Fb ,

 x where
the
buoyancy
force
Fb=ρ0(he‐x)Ag=ρ0heAg‐ρ0xAg=mg‐ρ0xAg.
Therefore,

 ρg x ρ b HA˙˙ = − ρ 0 gAx 


or
 ˙˙ + 0 x = 0 .
 x ρb H This
equation
is
the
standard
form
for
oscillation
with
an
angular
frequency

 ρ0g 2π ρH 

and
period
 T = ω= = 2π b .
 ρb H € ω ρ0g 
 € € 
 6
 ...
View Full Document

Ask a homework question - tutors are online