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
T. S(egler 11/24/2014 Texas A&M University
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
T. S(egler 11/24/2014 Texas A&M University
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]
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• 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
T. S(egler 11/24/2014 Texas A&M University
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:
T. S(egler 11/24/2014 Texas A&M University
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
T. S(egler 11/24/2014 Texas A&M University
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
T. S(egler 11/24/2014 Texas A&M University
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
T. S(egler 11/24/2014 Texas A&M University
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
(
T. S(egler 11/24/2014 Texas A&M University
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)
T. S(egler 11/24/2014 Texas A&M University
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?
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
(
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
(
T. S(egler 11/24/2014 Texas A&M University
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
T. S(egler 11/24/2014 Texas A&M University
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.?
T. S(egler 11/24/2014 Texas A&M University
! =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
T. S(egler 11/24/2014 Texas A&M University
+A
-‐A
x(t)
t
amax = d2x/dt2 max value
Checkpoint: SHM Problem 2
Suppose the two sinusoidal curves shown above are added together. Which of the plots shown below best represents the result?
T. S(egler 11/24/2014 Texas A&M University
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
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Checkpoint: SHM Problem 3
T. S(egler 11/24/2014 Texas A&M University
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
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
(
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
(
T. S(egler 11/24/2014 Texas A&M University
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.
T. S(egler 11/24/2014 Texas A&M University
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
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
(
T. S(egler 11/24/2014 Texas A&M University
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
T. S(egler 11/24/2014 Texas A&M University
v(t)! dxdt> 0 (moving in +x-dir)
a(t)! d 2xdt 2 < 0 (slowing down)