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
m 0
-A
x
A
• The force of a spring is given by Hooke’s Law F = !kx • If we apply Newton’s 2nd law we get a 2nd order differen(al equa(on k • Define the angular frequency as !=
d2x !kx = m 2 dt
m
• Second order differen(al equa(on for simple harmonic mo(on
d2x 0 = 2 +!2x dt • In general simple harmonic mo(on occurs whenever a mechanical system gives rise to a differen(al equa(on of this form. 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.
• • • • • T. S(egler
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]
11/24/2014
Texas A&M University
Rela)ons in SHM • Frequency and Period are related (as we have seen before)
f=
1 1 !T = T f
• 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.
2" ! = 2" f = 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
Since ϕ = ωt is the angle swept out in a (me t, we can say that the posi(on at any (me is: x(t) = A cos(! t) T. S(egler
11/24/2014
Texas A&M University
Simple harmonic mo)on in a circle The period of the mo(on is T = 2π/ω which does not depend on the amplitude.
x(t) = A cos(! t) when ωt = 2π rad
x(t) +A
π/ω 2π/ω
t
-‐A π/(2ω)
3π/(2ω)
when ωt = π rad
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:
x(t) = A cos(! t + "0 ) x(t) +A ϕ0 t
-‐A 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
Clicker Ques)on This is an x-‐t graph for an object in simple harmonic mo(on.
At which of the following (mes does the object have the most nega,ve accelera,on ax? A. t = T/4 B. t = T/2 C. t = 3T/4 D. t = T
T. S(egler
11/24/2014
Texas A&M University
Example Mo(on on a spring
Rθ
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 eleased nd 1 1scillate. I =2 r12 M L2 from Irest = 13aM L2allowed Ito =o12 M (a2 + b2 ) I = 31 M a2 2πR a) =Fvind tThe arad = force constant of the ideal spring R v b) Find the angular frequency, frequency, and period of the resul(ng oscilla(on. vtan = Rω
atan = Rα
L
#vA/B = −#vB/A
a
sions:
slender rod, axis 2
slender rod, axis
32.15 ft/s (on Earth’s through centre surface) through one end N · m2 /kg2 M⊕ = 5.98 × 1024 kg 30 2 M I =&1= M1.99 (R×2 10 + Rkg ) I = 1 M R2 1
2
214 mi 1 mi = 048 m 1m= 0s 1s= 248 lb 1 lb = ·m R1 1 W = R=2 radians 1 hp
b
L
#vA/C = #vA/B + #vB/C
2
2
1.609 km 3.281 ft 0.0002778 hr 4.448 N 1 J/s 745.7 W
hollow cylinder
R solid cylinder
b a
rectangular plate, axis through centre I = M R2
R thin-walled hollow cylinder
thin rectangular plate, axis along edge I = 52 M R2
R
I = 23 M R2
R
solid sphere thin-walled n 103 kilok hollow sphere µ 106 mega- M m 109 gigaG c ! For a point-like particle of mass M a distance R from the axis of rotation: I = M R2 ,# # # F = m#a! , FParallel −Ftheorem: B on A = A on B axis Ip = Icm + M d2 F#elas = −k(#r − #requil ) Damped/forced: SHM: |f#s | ≤ µs |#n|, |f#k | = µk |#n| ω = 2πf = 2π/T x(t) = A cos (ω " t)e−(b/2m)t ! ! x(t) = A cos (ωt + φ◦ ) ! I/mgd pend m1#r1T+ m2= #r22π + . . L/g . + m=n#r2π n v(t) = −ωA sin (ωt + φ◦ ) ω " = k/m − b2 /4m2 ! " m + mn Tspring 2π. . . m/k 1 + m= 2+ a(t) = −ω 2 A cos (ωt + φ◦ ) ! A = Fmax / (k − mωd2 )2 + b2 ωd2 T = 2π I/κ torsion similarly for #v and T. S(egler #a) 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. ! ! ! ! ! dx dv d 2 x = = and a(t) = • Star(ng with x(t) = A cos(! t) we already know v(t) dt dt dt 2
" d! % d • The deriva(ves of the trig func(ons that we need are (cos! ) = !sin ! $ ' # dt & dt ! d! $ d (sin ! ) = cos! # & and " dt % dt • Therefore:
x(t) = A cos(! 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 1
1
1
I =simple-‐harmonic-‐mo(on ML I = 3M L I t=o 12 +abgached ) M a the period T = 0.20 s. 12 3 For due aM m(aass to aI s=pring Released from rest, 2.4 cm from the equilibrium posi(on. 1 1 2 = 3 Ma) L2What aIre = t12 Mf(a + b2 ) = 31 M af2requency ω for this system? he requency f and aIngular L L A and what are v b and a b b) What is the amplitude max max? a c) What are x(t), v(t) and a(t) at at = 0.22 s aker release? 2
L slender rod, axis through centre a
2
b slender rod, axis through one end
a
2
b rectangular plate, axis through centre
thin rectangular plate, rod, axis 1 2 I = 2 M (Rrectangular + R22 ) Iplate, = 12 M R2 =M R2 1 axis Ialong edge one end axis through centre
=
1 M R2 2
R1
I = MR R2
hollow cylinder
R ylinder
2
R solid cylinder
I = 52 M R2
1
2
2
thin rectangular plate, axis along edge I = 52 M R2
I = 23 M R2
I = 23 M R2
R thin-walled hollow cylinder R
R
R solid sphere
thin-walled hollow sphere
R R hollow solidMsphere !thin-walled For a point-like particle of mass a distance R from the axis of rotation: I = M R2 thin-walled cylinder 2 hollow sphere ! Parallel axis theorem: Ip = Icm + M d
Damped/forced: SHM: icle of mass a distance R from the axis of rotation: I = M R2 ω =M2πf = 2π/T m: Ip = Icm + M! d2 x(t) = A cos (ω " t)e−(b/2m)t ! x(t) = A cos (ωt + φ◦ ) ! Tpend = 2π L/g = 2π I/mgd v(t) = −ωA sin (ωt + φ◦ ) ω " = k/m − b2 /4m2 ! Damped/forced: " Tspring = 2π m/k a(t) = −ω 2 A cos (ωt + φ◦ ) ! A = F / (k − mωd2 )2 + b2 ωd2 max Tx(t) = cos 2π (ωt I/κ x(t) = A cos (ω " t)e−(b/2m)t torsion =A + φ◦ ) ! " & ' () v(t) = −ωA sin (ωt + φ◦ ) ω = k/m − b2 /4m2 Waves: travelling : y(x, t) = A cos 2π x ∓ t " 2 2 vstring = FT /µ wave 2π/λ a(t) = −ω kA=cos (ωt + φ◦ ) λ T A = Fmax / (k − mωd2 )2 + b2 ωd2 T. S(egler v = ±λf 11/24/2014 µ = M/L = A cos (kx ∓ ωt)
Texas A&M University
Prelecture: SHM ques(on 1
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)
T. S(egler
11/24/2014
Texas A&M University
Clicker Ques)on and Prelecture: SHM Problem 2
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.? Compare the period of oscilla(on in the two cases: k a) T2 = T1 != m b) T2 = 2T1 2" m T= = 2" c) T2 = 4T1 ! k T. S(egler
11/24/2014
Texas A&M University
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? x(t) a) x = -‐A +A b) x = 0 t c) x = +A -‐A amax = d2x/dt2 max value T. S(egler
11/24/2014
Texas A&M University
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.
1 1 ! E = mv 2 + kx 2 2 2 • At the maximum displacement, x = A and v = 0
1 1 1 2 2 2 ! E = m(0) + k(A) = kA 2 2 2 • So at any other (me:
1 2 1 2 1 2 kA = mv + kx = constant 2 2 2
mv 2 = k(A 2 ! x 2 ) k 2 2 "v=± (A ! x ) m T. S(egler
11/24/2014
Texas A&M University
Checkpoint: SHM Problem 3
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 T. S(egler
11/24/2014
1 2 1 2 1 2 kA = mv + kx 2 2 2 k 2 2 !v=± (A " x ) m vmax (x = 0) =
k 2 A m
Texas A&M University
Example Energy in SHM
If f (t) = atn , then
% t2 t1 f (t)dt =
a n+1
&
' 1 tn+1 −Itn+1 2 1=
12
I = 13 M L2
M L2
Equations of motion: translational rotational
Circular motion:
arad =
I= v2 R
T =
1 M 12
2πR v
A glider with mass 0.500kg is agached to constant the end of a spring with konly: = 450N/m undergoes simple vtan = Rω atan (linear/angular) acceleration s = Rθ L harmonic mo(on with an amplitude #ro(t)f 0=.040m. #r◦ + #v◦ t + 12 #at2 θ(t) = θ◦ + ω◦ t + 12 αt2 L Relative velocity: #vA/C = #vA/B + #v (t) = #v◦ + #at ω(t) = ω◦ + αt a) what is the maximum speed of the glider? a #vB/C 1 1 2 2 2 2 2 I = 1 M L2 I = 3 M L #vA/B = −#vIB/A = 12 M( vf = v◦ + 2a(r − r◦ ) ωf = ω◦ + 2α(θ12 − θ◦ ) 1 1 slender rod, axis #r(t) = #r◦ + 2 (#vi + #vf )t θ(t) = θ◦ + 2 (ω i + ωf )t slender rod, axis rectangu Constants/Conversions: through centre always true: through one end axis throu 2 2 g = 9.80 m/s = 32.15 ft/s (on Earth’s su θ −θ r d! r dθ &ω' = &#v ' = !rt −! # v = ω = −t dt t −t dt G = 6.674 × 10−11 N · m2 /kg2 ! v −! v ω −ω d! v d ! r dω d θ 2 Ldt α = dt 1= &α' = t −t b) what is the speed at x = -‐.015m? &#a' = t −t #a = dt = dt 2 R⊕ =2 6.38 × 106L × 10 R2 ) I =8 12mM R2 M⊕ = 5.98 I= M3 %t % t I = 2 M (R1 + R = 6.96 × 10 m M = 1.99 × 10 & & #r(t) = #r◦ + 0 #v (t) dt θ(t) = θ◦ + 0 ω # (t) dt a %t %t #v (t) = #v◦ + 0 #a(t) dt ω(t) = ω◦ + 0 α # (t) dt 1 km = 0.6214 mi 1 mi = 1.609 1 ft =axis 0.3048 m 1 m = 3.281p slender rod, axis slender rod, rectangular Forces, Energy and Momenta: through centre 1 hr = 3600 s 1 sthrough = 0.0002 through one end axis translational rotational m 1 kg = 1 N = 0.2248 lb 1 lb = 4.448 c) magnitude of the maximum accelera(on? s #τ = #r × F# and R |#τ |1= F⊥ r R 1 J = 1 N·m 1 W = 1 J/s % % 2 R 2 R 1 τ dθ 2 1 2 2 W =I τ ∆θ = W = F# · ∆#r = F# · d#r ◦ = 2 Mcylinder (R1 + R12 )rev = I 360 = 2M Rradians ==M R = 2π 1Ihp 745.7 hollow thin-walled h solid cylinder # · #v P = dW = #τ · ω # P = dW = F dt dt cylinder 10−9 nanon 3 # = I1 ω p # = m # v + m # v + . . . 10 kilo L # + I ω # + . . . cm 1 1 2 2 1 2 2 10−6 micro- µ 6 10 meg = M#vcm = Itot ω # 10−3 millim 9 % 10 ! For a point-like particle of mass Rgiga fro 10−2 centic M a distance = #r × p# J# = F# dt = ∆# p 2 ! Parallel axis theorem: I = I + M d ( , p cm ( d) accelera(on at x = -‐.015m? ! L Rα F# = m#a, F#B on A = Forces: Newton’s: #τext = Itot #1 = ddt F#ext = M#acm = d!pdt R 2 ( R Hooke’s: F#elas = −k(#r R ( SHM: − #requil ) #τint hollow =0 F#int = 0 cylinder thin-walled solid cylinder ω = 2πf = 2π/T # friction: |fs | ≤ µs |#n|, |f#khollo |=µ ! ! 2 x(t) = A cos (ωt + φ◦ ) cylinder Krot = 21 Itot ω 2 Ktrans = 12 M vcm Tpend = 2π L/g = 2π I/mgd Centre-of-mass: v(t) = −ωA sin (ωt + φ◦ ) ! m1#r1 + m2#r2 + . . . + mn#rn — Both translational and rotational —= 2π m/k Tspring #rcm =a(t) = −ω 2 A cos (ωt + φ ) e) the total energy at any point? ! m1a+distance m2 + . . R . +from mn◦ th ! For a point-like particle of mass M 2
1
2
1
2
1
2
1
2
2
1
2
1
2
2
2
2
1
1
2
2
cm
T. S(egler
W = ∆K = Ktrans,f + Krot,f − Ktrans,i − Krot,i Ttorsion = 2π I/κ ! Parallel axis theorem: Ip = Icm + M d2 Etot,f = Etot,i + Wother ⇔ Kf + Uf = Ki + Ui + Wother 1 (and 1similarly1for #v and #a) ) kA 2 = mv 2 + kx 2 = const.tra Waves: Gravity: 2v 2 SHM: 2 2 w = 2π/λ )2 U = − F# · d#r ; Ugrav = M gycm ; Uelas = 21 k(r − rkequil string = FT /µ ω = 2πf = 2π/T M M M M 1µ 2 1 2 v = ±λf + !F * = M/L AU cos (ωt=+−G φ◦ ) grav = G x(t) grav # = − ∂U ˆi + ∂U ˆj!+ ∂U kˆ 2 =k Fx (x) = −dU (x)/dx F# = −∇U 2 2 R12 R ∂x ∂z = 2π I/mgd Tpend = ∂y 2π L/g 12 ! =(kx) ± =sin−ωA (A ! x ) ! y(x, t) = ASW v sinv(t) (ωt)sin (ωt + φ◦ ) P = m Tspring standing = 2π m/k a(t) = −ω 2 A cos (ωt + φ◦ ) ! vexas 2L 11/24/2014 T =wave T &M λ = ⇔ f = n , n =A1, 2, 3,U.niversity .. n 2π I/κ n torsion
Clicker Ques)on 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?
A. t = T/8
1 2 1 2 1 2 K = mv = kA ! kx 2 2 2 K max when x = 0
B. t = T/4
C. t = 3T/8
D. t = T/2
E. more than one of the above
1 K max = kA 2 2
1 2 1 2 1 2 kA = mv + kx = const. 2 2 2 T. S(egler
11/24/2014
Texas A&M University
Example Energy in SHM I =
1
I = 1 M L2
M L2
1
I =
I = 1 M a2
M (a2 + b2 )
3 12 3 A 1.6 12 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? b
L
L
b
a slender rod, axis through centre
slender rod, axis through one end
I = 21 M (R12 + R22 )
R1
R2
rectangular plate, axis through centre
I = 12 M R2
thin rectangular plate, axis along edge I = 52 M R2
I = M R2
I = 23 M R2
R
R
R thin-walled hollow cylinder
R solid cylinder
hollow cylinder
a
solid sphere
thin-walled hollow sphere
! For a point-like particle of mass M a distance R from the axis of rotation: I = M R2 ! Parallel axis theorem: Ip = Icm + M d2 Damped/forced:
SHM: ω = 2πf = 2π/T ! ! Tpend = 2π L/g = 2π I/mgd ! Tspring = 2π m/k ! Ttorsion = 2π I/κ
1 2
2
kA = Waves:
1 2 1 2 mv + kx = const. 2k = 2π/λ 2 v2
T. S(egler v = ±λf
a(t) = −ω 2 A cos (ωt + φ◦ )
string
x(t) = A cos (ω " t)e−(b/2m)t ! ω " = k/m − b2 /4m2 " A = Fmax / (k − mωd2 )2 + b2 ωd2
x(t) = A cos (ωt + φ◦ ) v(t) = −ωA sin (ωt + φ◦ )
travelling : wave
= FT /µ
µ = M/L
& ' () x t y(x, t) = A cos 2π ∓ λ T
11/24/2014 = A cos (kx ∓ ωt)
Texas A&M University
Clicker Ques)on 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? dx > 0 (moving in +x-dir) dt d2x a(t) ! 2 < 0 (slowing down) dt
v(t) !
y(t) (A) (C)
t (B)
T. S(egler
11/24/2014
Texas A&M University
