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Today: Start chapter 14

•  Oscilla(ons in terms of amplitude, period, frequency and angular frequency

•  Simple harmonic mo(on

•  Simple harmonic mo(on with energy and momentum

T. S(egler 11/24/2014 Texas A&M University

Introduc)on

•  Periodic mo(on is any type of mo(on that repeats itself §  heartbeat §  musical vibra(ons §  rocking chair` §  pendulum swinging §  vibra(ons of molecules in a solid §  AC current

•  We will concentrate on a par(cular type of periodic mo(on, called simple harmonic mo(on (SHM) •  Spring-‐mass systems •  Pendulums


Spring-‐Mass System

k

-A 0 A x

m

•  The force of a spring is given by Hooke’s Law

•  If we apply Newton’s 2nd law we get a 2nd order differen(al equa(on •  Define the angular frequency as

•  Second order differen(al equa(on for simple harmonic mo(on

•  In general simple harmonic mo(on occurs whenever a mechanical system gives rise to a differen(al equa(on of this form.

F = !kx

!kx =m d 2xdt2

! =km

0 = d2xdt2

+! 2x


Variables of periodic mo)on

•  A body undergoing periodic mo(on always has a stable equilibrium posi(on. •  When it is displaced from this posi(on there is a force or torques that will pull it back

towards its equilibrium posi(on. •  This force causes the oscilla(on of the system. When it is directly propor,onal to the

displacement from equilibrium, the resul(ng mo(on is called simple harmonic mo,on.

•  Amplitude = A è magnitude of maximum displacement [m or rad] •  Cycle è a complete “lap” of the mo(on, back to it’s star(ng point •  Period = T è the (me to complete one full cycle [s] •  Frequency = f è the rate of comple(ng a cycle; or # of cycles per unit of (me [Hz = #/s] •  Angular frequency = ω è 2π x frequency [rad/s]


•  Frequency and Period are related (as we have seen before)

•  The units of frequency are cycles per second. A cycle is just a number, i.e. unit-‐less, so the frequency is measured in inverse seconds: Hertz = 1/s

•  Angular frequency isn’t necessarily related to the angular velocity; it is the rate change of the angular quan(ty (whatever “quan(ty” happens to be) measured in radians.

Rela)ons in SHM

f = 1T!T = 1

f

! = 2" f = 2"T


In case you’re not so good with differen)al equa)ons…

Simple harmonic mo(on viewed as mo(on in a circle: •  we want to describe the periodic mo(on mathema(cally but we can’t use the constant

accelera(on equa(ons of mo(on •  instead we can look at simple harmonic mo(on as a projec(on of circular mo(on onto a

diameter

x(t) = Acos(!t)

Since ϕ = ωt is the angle swept out in a (me t, we can say that the posi(on at any (me is:


The period of the mo(on is T = 2π/ω which does not depend on the amplitude.

+A

-‐A

π/(2ω)

2π/ω

π/ω

3π/(2ω)

x(t)

t

when ωt = π rad

when ωt = 2π rad

x(t) = Acos(!t)

Simple harmonic mo)on in a circle


Simple harmonic mo)on w/ arbitrary star)ng posi)on

If we started at ϕ0 instead of at x = A, there’s an offset and the curve looks like this:

ϕ0

+A

-‐A

x(t)

t

x(t) = Acos(!t +"0 )

ϕ0

ϕ0 is called the phase angle, or the posi(on in the cycle when t = 0


This is an x-‐t graph for an object in simple harmonic mo(on.

A. t = T/4

B. t = T/2

C. t = 3T/4

D. t = T

At which of the following (mes does the object have the most nega,ve accelera,on ax?

Clicker Ques)on


Example Mo(on on a spring

A spring agached at one end is stretched by a 6.0N force which causes a displacement of 0.30m. A glider of mass 0.50kg is agached to the end of the spring and pulled to the right a distance 0.020m then released from rest and allowed to oscillate. a) Find the force constant of the ideal spring b) Find the angular frequency, frequency, and period of the resul(ng oscilla(on.

General math:

log (x/y) = log (x)! log (y)

log (xy) = log (x) + log (y)

log (xn) = n log (x)

lnx " loge x

x = 10(log10 x)

x = e(ln x)

ha

hho

!

"

ha = h cos ! = h sin"

ho = h sin ! = h cos"

h2 = h2a + h2

o tan ! =ho

ha

#A = Axi+Ay j +Az k

#A · #B = AxBx +AyBy +AzBz = AB cos ! = A!B = AB!

#A# #B = (AyBz!AzBy )i+ (AzBx!AxBz)j + (AxBy!AyBx)k

= AB sin ! = A"B = AB"

df

dt= natn#1

If f(t) = atn, then

!

"

"

#

"

"

$

% t2t1

f(t)dt = an+1

&

tn+12 ! tn+1

1

'

ax2 + bx+ c = 0 $ x =!b±

%b2 ! 4ac

2a

"

" + ! = 90!" + ! = 180!

"!

!

"

!

"

!

"

!

"!

" !

Equations of motion:translational rotational

constant (linear/angular) acceleration only:

#r(t) = #r$ + #v$t+12#at

2

#v(t) = #v$ + #at

v2f = v2$ + 2a(r ! r$)

#r(t) = #r$ +12 (#vi + #vf )t

!(t) = !$ + $$t+12%t

2

$(t) = $$ + %t

$2f = $2

$ + 2%(! ! !$)

!(t) = !$ +12 ($i + $f )t

always true:

&#v' = !r2#!r1t2#t1

#v = d!rdt

&#a' = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t) dt

#v(t) = #v$ +% t0 #a(t) dt

&$' = "2#"1t2#t1

$ = d"dt

&%' = #2##1

t2#t1%= d#

dt =d2"dt2

!(t) = !$ +% t0 #$(t) dt

$(t) = $$ +% t0 #%(t) dt

Forces, Energy and Momenta:translational rotational

W = #F ·!#r =%

#F · d#r

P = dWdt = #F · #v

#pcm = m1#v1 +m2#v2 + . . .

= M#vcm#J =

%

#Fdt = !#p(

#Fext = M#acm = d!pcm

dt(

#Fint = 0

Ktrans =12Mv2cm

#& = #r # #F and |#& | = F"r

W = & !! =%

&d!

P = dWdt = #& · #$

#L = I1#$1 + I2#$2 + . . .

= Itot#$

= #r # #p(

#&ext = Itot#% = d!Ldt

(

#&int = 0

Krot =12Itot$

2

— Both translational and rotational —

W = !K = Ktrans,f +Krot,f !Ktrans,i !Krot,i

Etot,f = Etot,i +Wother ( Kf + Uf = Ki + Ui +Wother

U = !)

#F · d#r ; Ugrav = Mgycm ; Uelas =12k(r ! requil)

2

Fx(x) = !dU(x)/dx #F = !#)U = !*

$U$x i+

$U$y j +

$U$z k

+

Circular motion: arad =v2

RT =

2'R

v

s = R! vtan = R$ atan = R%

Relative velocity:#vA/C = #vA/B + #vB/C

#vA/B = !#vB/A

Constants/Conversions:

g = 9.80 m/s2 = 32.15 ft/s2 (on Earth’s surface)

G = 6.674# 10#11 N ·m2/kg2

R% = 6.38# 106 m M% = 5.98# 1024 kgR& = 6.96# 108 m M& = 1.99# 1030 kg

1 km = 0.6214 mi 1 mi = 1.609 km1 ft = 0.3048 m 1 m = 3.281 ft1 hr = 3600 s 1 s = 0.0002778 hr

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

1 J = 1 N·m 1 W = 1 J/s1 rev = 360$ = 2' radians 1 hp = 745.7 W

10#9 nano- n10#6 micro- µ10#3 milli- m10#2 centi- c

103 kilo- k106 mega- M109 giga- G

Forces: Newton’s:, #F = m#a, #FB on A = !#FA on B

Hooke’s: #Felas = !k(#r ! #requil)

friction: |#fs| * µs|#n|, |#fk| = µk|#n|

Centre-of-mass:

#rcm =m1#r1 +m2#r2 + . . .+mn#rn

m1 +m2 + . . .+mn

(and similarly for #v and #a)

Gravity:

Fgrav = GM1M2

R212

Ugrav = !GM1M2

R12T =

2'a3/2%GM

Phys 218 — Final Exam Formulae

rectangular plate,axis through centre

thin rectangular plate,axis along edge

hollow spheresolid spherethin-walled hollow

cylindersolid cylinder

I = 112ML2 I = 1

3ML2 I = 112M(a2 + b2) I = 1

3Ma2

a

L

through centreslender rod, axis

through one endslender rod, axis

thin-walled

RRR2

R1

hollow cylinderR R

b

a

bL

I = 12M(R2

1 + R22) I = 1

2MR2 I = MR2I = 2

5MR2 I = 23MR2

! For a point-like particle of mass M a distance R from the axis of rotation: I = MR2

! Parallel axis theorem: Ip = Icm +Md2

SHM:

! = 2"f = 2"/T

Tpend = 2"!

L/g = 2"!

I/mgd

Tspring = 2"!

m/k

Ttorsion = 2"!

I/#

x(t) = A cos (!t+ $!)

v(t) = !!A sin (!t+ $!)

a(t) = !!2A cos (!t+ $!)

Damped/forced:

x(t) = A cos (!"t)e#(b/2m)t

!" =!

k/m! b2/4m2

A = Fmax/"

(k !m!2d)

2 + b2!2d

Waves:k = 2"/%

v = ±%f

v2string = FT /µ

µ = M/L

y(x, t) = ASW sin (kx) sin (!t)standingwave

#

$

%%n =2L

n" fn = n

v

2L, n=1, 2, 3, . . .

travellingwave : y(x, t) = A cos

&

2"

'

x

%#

t

T

()

= A cos (kx# !t)

P =!

µFT !2A2 sin2 (kx# !t), $P % =1

2

!

µFT !2A2

I(r) =power

surf area at r

'

=P

4"r2for 3D sphere

(


Velocity and accelera)on in SHM

•  We can use calculus to derive the equa(ons for velocity and accelera(on as a func(on of (me for simple harmonic mo(on.

•  Star(ng with

•  The deriva(ves of the trig func(ons that we need are and

•  Therefore:

x(t) = Acos(!t)!v(t) = d

!xdt

!a(t) = d!vdt=d 2 !xdt2

we already know and

ddt(cos! ) = !sin! d!

dt"

#$

%

&'

ddt(sin! ) = cos! d!

dt!

"#

$

%&

x(t) = Acos(!t)v(t) = !A! sin(!t)a(t) = !A! 2 cos(!t)


Example SHM mass and spring system

For simple-‐harmonic-‐mo(on due to a mass agached to a spring the period T = 0.20 s. Released from rest, 2.4 cm from the equilibrium posi(on. a) What are the frequency f and angular frequency ω for this system? b) What is the amplitude A and what are vmax and amax? c) What are x(t), v(t) and a(t) at t = 0.22 s aker release?





I = 112ML2 I = 1

3ML2 I = 112M(a2 + b2) I = 1

3Ma2

a

L



thin-walled

RRR2

R1

hollow cylinderR R

b

a

bL

I = 12M(R2

1 + R22) I = 1

2MR2 I = MR2I = 2

5MR2 I = 23MR2



SHM:

! = 2"f = 2"/T

Tpend = 2"!

L/g = 2"!

I/mgd

Tspring = 2"!

m/k

Ttorsion = 2"!

I/#

x(t) = A cos (!t+ $!)

v(t) = !!A sin (!t+ $!)

a(t) = !!2A cos (!t+ $!)

Damped/forced:


!" =!

k/m! b2/4m2

A = Fmax/"

(k !m!2d)

2 + b2!2d

Waves:k = 2"/%

v = ±%f

v2string = FT /µ

µ = M/L


#

$

%%n =2L

n" fn = n

v

2L, n=1, 2, 3, . . .


&

2"

'

x

%#

t

T

()

= A cos (kx# !t)

P =!

µFT !2A2 sin2 (kx# !t), $P % =1

2

!

µFT !2A2

I(r) =power

surf area at r

'

=P

4"r2for 3D sphere

(





I = 112ML2 I = 1

3ML2 I = 112M(a2 + b2) I = 1

3Ma2

a

L



thin-walled

RRR2

R1

hollow cylinderR R

b

a

bL

I = 12M(R2

1 + R22) I = 1

2MR2 I = MR2I = 2

5MR2 I = 23MR2



SHM:

! = 2"f = 2"/T

Tpend = 2"!

L/g = 2"!

I/mgd

Tspring = 2"!

m/k

Ttorsion = 2"!

I/#

x(t) = A cos (!t+ $!)

v(t) = !!A sin (!t+ $!)

a(t) = !!2A cos (!t+ $!)

Damped/forced:


!" =!

k/m! b2/4m2

A = Fmax/"

(k !m!2d)

2 + b2!2d

Waves:k = 2"/%

v = ±%f

v2string = FT /µ

µ = M/L


#

$

%%n =2L

n" fn = n

v

2L, n=1, 2, 3, . . .


&

2"

'

x

%#

t

T

()

= A cos (kx# !t)

P =!

µFT !2A2 sin2 (kx# !t), $P % =1

2

!

µFT !2A2

I(r) =power

surf area at r

'

=P

4"r2for 3D sphere

(


The above curve represents the displacement versus (me of a mass oscilla(ng on a spring. The axes cross ωt = 0. Which of the following func(ons best describes this curve? a) A sin(ωt) b) A sin(ωt+π/2) c) A sin(ωt-‐π/2)

Prelecture: SHM ques(on 1


and Prelecture: SHM Problem 2

Compare the period of oscilla(on in the two cases: a) T2 = T1 b) T2 = 2T1 c) T2 = 4T1

A block having mass m is agached to a spring having a spring constant k and oscillates with simple harmonic mo(on. In Case 2 the amplitude of the oscilla(on is twice as big as it is in Case 1.?


! =km

T = 2"!

= 2" mk

Clicker Ques)on

Checkpoint: SHM Problem 1

A mass on a spring moves with simple harmonic mo(on as shown. Where is the accelera(on of the mass most posi(ve? a) x = -‐A b) x = 0 c) x = +A


+A

-‐A

x(t)

t

amax = d2x/dt2 max value


Suppose the two sinusoidal curves shown above are added together. Which of the plots shown below best represents the result?


Energy in SHM

Use mass on a spring as an example •  As it has been the total mechanical energy is conserved throughout the mo(on.

•  At the maximum displacement, x = A and v = 0

•  So at any other (me:

! E = 12mv2 + 1

2kx2

! E = 12m(0)2 + 1

2k(A)2 = 1

2kA2

12kA2 = 1

2mv2 + 1

2kx2 =

mv2 = k(A2 ! x2 )

" v = ± km(A2 ! x2 )

constant




In the two cases shown the mass and the spring are iden(cal but the amplitude of the simple harmonic mo(on is twice as big in Case 2 as in Case 1. How are the maximum veloci(es in the two cases related? a) Vmax,2 = Vmax,1 b) Vmax,2 = 2 Vmax, c) Vmax,2 = 4 Vmax,1

12kA2 = 1

2mv2 + 1

2kx2

! v = ± km(A2 " x2 )

vmax (x = 0) =kmA2

Example Energy in SHM

General math:

log (x/y) = log (x)! log (y)

log (xy) = log (x) + log (y)

log (xn) = n log (x)

lnx " loge x

x = 10(log10 x)

x = e(ln x)

ha

hho

!

"

ha = h cos ! = h sin"

ho = h sin ! = h cos"

h2 = h2a + h2

o tan ! =ho

ha

#A = Axi+Ay j +Az k

#A · #B = AxBx +AyBy +AzBz = AB cos ! = A!B = AB!

#A# #B = (AyBz!AzBy )i+ (AzBx!AxBz)j + (AxBy!AyBx)k

= AB sin ! = A"B = AB"

df

dt= natn#1

If f(t) = atn, then

!

"

"

#

"

"

$

% t2t1

f(t)dt = an+1

&

tn+12 ! tn+1

1

'

ax2 + bx+ c = 0 $ x =!b±

%b2 ! 4ac

2a

"

" + ! = 90!" + ! = 180!

"!

!

"

!

"

!

"

!

"!

" !

Equations of motion:translational rotational

constant (linear/angular) acceleration only:

#r(t) = #r$ + #v$t+12#at

2

#v(t) = #v$ + #at

v2f = v2$ + 2a(r ! r$)

#r(t) = #r$ +12 (#vi + #vf )t

!(t) = !$ + $$t+12%t

2

$(t) = $$ + %t

$2f = $2

$ + 2%(! ! !$)

!(t) = !$ +12 ($i + $f )t

always true:

&#v' = !r2#!r1t2#t1

#v = d!rdt

&#a' = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t) dt

#v(t) = #v$ +% t0 #a(t) dt

&$' = "2#"1t2#t1

$ = d"dt

&%' = #2##1

t2#t1%= d#

dt =d2"dt2

!(t) = !$ +% t0 #$(t) dt

$(t) = $$ +% t0 #%(t) dt

Forces, Energy and Momenta:translational rotational

W = #F ·!#r =%

#F · d#r

P = dWdt = #F · #v

#pcm = m1#v1 +m2#v2 + . . .

= M#vcm#J =

%

#Fdt = !#p(

#Fext = M#acm = d!pcm

dt(

#Fint = 0

Ktrans =12Mv2cm

#& = #r # #F and |#& | = F"r

W = & !! =%

&d!

P = dWdt = #& · #$

#L = I1#$1 + I2#$2 + . . .

= Itot#$

= #r # #p(

#&ext = Itot#% = d!Ldt

(

#&int = 0

Krot =12Itot$

2

— Both translational and rotational —

W = !K = Ktrans,f +Krot,f !Ktrans,i !Krot,i

Etot,f = Etot,i +Wother ( Kf + Uf = Ki + Ui +Wother

U = !)

#F · d#r ; Ugrav = Mgycm ; Uelas =12k(r ! requil)

2

Fx(x) = !dU(x)/dx #F = !#)U = !*

$U$x i+

$U$y j +

$U$z k

+

Circular motion: arad =v2

RT =

2'R

v

s = R! vtan = R$ atan = R%

Relative velocity:#vA/C = #vA/B + #vB/C

#vA/B = !#vB/A

Constants/Conversions:

g = 9.80 m/s2 = 32.15 ft/s2 (on Earth’s surface)

G = 6.674# 10#11 N ·m2/kg2

R% = 6.38# 106 m M% = 5.98# 1024 kgR& = 6.96# 108 m M& = 1.99# 1030 kg

1 km = 0.6214 mi 1 mi = 1.609 km1 ft = 0.3048 m 1 m = 3.281 ft1 hr = 3600 s 1 s = 0.0002778 hr

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

1 J = 1 N·m 1 W = 1 J/s1 rev = 360$ = 2' radians 1 hp = 745.7 W

10#9 nano- n10#6 micro- µ10#3 milli- m10#2 centi- c

103 kilo- k106 mega- M109 giga- G

Forces: Newton’s:, #F = m#a, #FB on A = !#FA on B

Hooke’s: #Felas = !k(#r ! #requil)

friction: |#fs| * µs|#n|, |#fk| = µk|#n|

Centre-of-mass:

#rcm =m1#r1 +m2#r2 + . . .+mn#rn

m1 +m2 + . . .+mn

(and similarly for #v and #a)

Gravity:

Fgrav = GM1M2

R212

Ugrav = !GM1M2

R12T =

2'a3/2%GM

Phys 218 — Final Exam Formulae





I = 112ML2 I = 1

3ML2 I = 112M(a2 + b2) I = 1

3Ma2

a

L



thin-walled

RRR2

R1

hollow cylinderR R

b

a

bL

I = 12M(R2

1 + R22) I = 1

2MR2 I = MR2I = 2

5MR2 I = 23MR2



SHM:

! = 2"f = 2"/T

Tpend = 2"!

L/g = 2"!

I/mgd

Tspring = 2"!

m/k

Ttorsion = 2"!

I/#

x(t) = A cos (!t+ $!)

v(t) = !!A sin (!t+ $!)

a(t) = !!2A cos (!t+ $!)

Damped/forced:


!" =!

k/m! b2/4m2

A = Fmax/"

(k !m!2d)

2 + b2!2d

Waves:k = 2"/%

v = ±%f

v2string = FT /µ

µ = M/L


#

$

%%n =2L

n" fn = n

v

2L, n=1, 2, 3, . . .


&

2"

'

x

%#

t

T

()

= A cos (kx# !t)

P =!

µFT !2A2 sin2 (kx# !t), $P % =1

2

!

µFT !2A2

I(r) =power

surf area at r

'

=P

4"r2for 3D sphere

(





I = 112ML2 I = 1

3ML2 I = 112M(a2 + b2) I = 1

3Ma2

a

L



thin-walled

RRR2

R1

hollow cylinderR R

b

a

bL

I = 12M(R2

1 + R22) I = 1

2MR2 I = MR2I = 2

5MR2 I = 23MR2



SHM:

! = 2"f = 2"/T

Tpend = 2"!

L/g = 2"!

I/mgd

Tspring = 2"!

m/k

Ttorsion = 2"!

I/#

x(t) = A cos (!t+ $!)

v(t) = !!A sin (!t+ $!)

a(t) = !!2A cos (!t+ $!)

Damped/forced:


!" =!

k/m! b2/4m2

A = Fmax/"

(k !m!2d)

2 + b2!2d

Waves:k = 2"/%

v = ±%f

v2string = FT /µ

µ = M/L


#

$

%%n =2L

n" fn = n

v

2L, n=1, 2, 3, . . .


&

2"

'

x

%#

t

T

()

= A cos (kx# !t)

P =!

µFT !2A2 sin2 (kx# !t), $P % =1

2

!

µFT !2A2

I(r) =power

surf area at r

'

=P

4"r2for 3D sphere

(


A glider with mass 0.500kg is agached to the end of a spring with k = 450N/m undergoes simple harmonic mo(on with an amplitude of 0.040m. a) what is the maximum speed of the glider? b) what is the speed at x = -‐.015m? c) magnitude of the maximum accelera(on? d) accelera(on at x = -‐.015m? e) the total energy at any point?

v = ± km(A2 ! x2 )

12kA2 = 1

2mv2 + 1

2kx2 = const.


A.  t = T/8

B.  t = T/4

C. t = 3T/8

D. t = T/2

E. more than one of the above

This is an x-‐t graph for an object connected to a spring and moving in simple harmonic mo(on.

At which of the following (mes is the kine,c energy of the object the greatest?

Clicker Ques)on

K =12mv2 =

12kA2 !

12kx2

Kmax when x = 0

Kmax =12kA2

12kA2 = 1

2mv2 + 1

2kx2 = const.

A 1.6 kg mass agached to a spring oscillates with an amplitude of 7.3 cm and a frequency of 2.6 Hz. What is the total energy of the system? How fast is the mass moving when it is 3.0 cm from the equilibrium posi(on?

Example Energy in SHM





I = 112ML2 I = 1

3ML2 I = 112M(a2 + b2) I = 1

3Ma2

a

L



thin-walled

RRR2

R1

hollow cylinderR R

b

a

bL

I = 12M(R2

1 + R22) I = 1

2MR2 I = MR2I = 2

5MR2 I = 23MR2



SHM:

! = 2"f = 2"/T

Tpend = 2"!

L/g = 2"!

I/mgd

Tspring = 2"!

m/k

Ttorsion = 2"!

I/#

x(t) = A cos (!t+ $!)

v(t) = !!A sin (!t+ $!)

a(t) = !!2A cos (!t+ $!)

Damped/forced:


!" =!

k/m! b2/4m2

A = Fmax/"

(k !m!2d)

2 + b2!2d

Waves:k = 2"/%

v = ±%f

v2string = FT /µ

µ = M/L


#

$

%%n =2L

n" fn = n

v

2L, n=1, 2, 3, . . .


&

2"

'

x

%#

t

T

()

= A cos (kx# !t)

P =!

µFT !2A2 sin2 (kx# !t), $P % =1

2

!

µFT !2A2

I(r) =power

surf area at r

'

=P

4"r2for 3D sphere

(


12kA2 = 1

2mv2 + 1

2kx2 = const.

A mass oscillates up & down on a spring. Its posi(on as a func(on of (me is shown below. At which of the points shown does the mass have posi(ve velocity and nega(ve accelera(on?

t

y(t)

(A)

(B)

(C)

Clicker Ques)on


v(t)! dxdt> 0 (moving in +x-dir)

a(t)! d 2xdt 2 < 0 (slowing down)

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