UNIVERSITY PHYSICS I CHAPTER 7 Simple Harmonic Oscimation §7.1hook' s force law 1. Ideal model-spring oscillator k 0000000m x A x=0 +A elasticity inertia Equilibrium position 2. Hook's force law Horizontal direction: =-k i 1
1 §7.1 Hook’s force law 1. Ideal model—spring oscillator Equilibrium position − A + A x Horizontal direction: 2. Hook’s force law F kxi ˆ s = − r elasticity inertia
87.2 Simple harmonic oscillation 1. Differential equation of a simple harmonic oscillation and its solution F=lxi =ma -kri=m-i dt d'x +kx=0 cco0000mi/ General Solution: x=0 x x(t=acos at+bsin at A=(a +b or x(=Acos(at+o) tang= 87.2 Simple harmonic oscillation 2. The quantities describing the oscillation Displacemnt at time t Phase x()=Acos(ot+φ) Amplitude Time m Angular frequency constant angle
2 §7.2 Simple harmonic oscillation 1. Differential equation of a simple harmonic oscillation and its solution x 0 d d 2 2 + x = m k t x i t x kxi m ˆ d d ˆ 2 2 Fx kxi max − = r r = − =ˆ General Solution: or ( ) cos( ) ( ) cos sin ω φ ω ω = + = + x t A t x t a t b t a b A a b = − = + tanφ ( ) 2 2 1 2 §7.2 Simple harmonic oscillation A m x 2. The quantities describing the oscillation
87.2 Simple harmonic oscillation Angular frequency, period and frequency dx k k a2x(t)+x()=0 dt x(r)=Acos(ar+小)∴.a2 =—or Angular frequency a is related to the spring constant and the mass. It is decided by the nature of the system. Unit is rad/s The period T: x(t)=Acos(at+o) x(t+T)=Acos[o(t+r)+oI 87.2 Simple harmonic oscillation 2丌 OT=2丌T Unit is s Frequency v: v=2= 0=2元V 2兀 Unit is 1/s or Hertz(Hz). ② Amplitude dx(t) x(t)=Acos(at+o) v(t)=dr -Asin(at+o) Fo=x(0=Acos t=0 vo=v(0)=-A@sin o A=xm=x5+ The range of the oscillation is 2A=2x
3 §7.2 Simple harmonic oscillation ① Angular frequency , period and frequency 0 d d 2 2 + x = m k t x x(t) = Acos(ωt +φ ) m k m k x t m k x t ∴ = = − + = ω ω ω or ( ) ( ) 0 2 2 Q Angular frequency ω is related to the spring constant and the mass. It is decided by the nature of the system. Unit is rad/s. The period T : ( ) cos[ ( ) ] ( ) cos( ) ω φ ω φ = + = + + = + x t T A t T x t A t §7.2 Simple harmonic oscillation ω π ω π 2 T = 2 T = Unit is s. Frequency ν : ω πν π ω ν 2 2 1 = = = T Unit is 1/s or Hertz (Hz). ②Amplitude x(t) = Acos(ωt +φ ) sin( ) d d ( ) ( ) = = −Aω ωt +φ t x t v t t =0 ω φ φ (0) sin (0) cos 0 0 v v A x x A = = − = = 2 2 2 0 0 ω v A x x = m = + The range of the oscillation is 2A=2xm
87.2 Simple harmonic oscillation Initial phase angle and phase t=0 o=(0)=Acos d 如=tanx(-" v=v(0)=- Asinφ o describe the initial state of the spring oscillator It is called initial phase or phase constant A and is related to the initial states or conditions of the system at+o is called the phase of the motion. It describes the states of the oscillation system 87.2 Simple harmonic oscillation x(t)=Acos(at+p) v()> dx(t)_-A@sin(at+o) dt at+o=T/3 x(t)=Ai v(t) oAi aH-m3x()2写训=Y3 3. The graphs ofx(o), v(o) and a(t) of simple harmonic oscillation
4 §7.2 Simple harmonic oscillation ③Initial phase angle and phase t =0 ω φ φ (0) sin (0) cos 0 0 v v A x x A = = − = = tan ( ) 0 1 0 ω φ x v = − − φ describe the initial state of the spring oscillator. It is called initial phase or phase constant. A and φ is related to the initial states or conditions of the system. ωt+φ is called the phase of the motion. It describes the states of the oscillation system. x(t) = Acos(ωt +φ ) sin( ) d d ( ) ( ) = = −Aω ωt +φ t x t v t §7.2 Simple harmonic oscillation ωt+φ=π/3 x t Ai v t Ai ˆ 2 3 ( ) ˆ 2 1 ( ) = = − ω r r ωt+φ= -π/3 x t Ai v t Ai ˆ 2 3 ( ) ˆ 2 1 ( ) = = ω r r 3. The graphs of x(t), v(t) and a(t) of simple harmonic oscillation
87.2 Simple harmonic oscillation x(t)=Acos(@t +o) +4 v(= dx(t)=-Aasin(at+P)E o d d'x(t (a) a()= +a4 dt -A@ cos(at +o) (b) +x.p=0a=-a x=0v=-V.a=0 x=-X.=0a=+a 87.2 Simple harmonic oscillation 4. How to determine if an oscillatory motion is simple harmonic oscillation? (criterion) Criterion 1: F total One force or the sum of several forces Criterion 2: k +“x=0 Criterion 3: x(t)=Acos(at+o)
5 §7.2 Simple harmonic oscillation x(t) = Acos(ωt +φ ) sin( ) d d ( ) ( ) = = −Aω ωt +φ t x t v t cos( ) d d ( ) ( ) 2 2 2 = − ω ω +φ = A t t x t a t m m m m m x x v a a x v v a x x v a a = − = = + = = − = = + = = − 0 0 0 0 +A +A +ωA −ωA A 2 +ω A 2 +ω φ=0 4. How to determine if an oscillatory motion is simple harmonic oscillation?(criterion) F kxi ˆ total = − r Criterion 1: One force or the sum of several forces Criterion 2: 0 d d 2 2 + x = m k t x Criterion 3: x(t) = Acos(ωt +φ ) §7.2 Simple harmonic oscillation
87.3 a vertically oriented spring 1. Is the motion of this system simple harmonic? Fares x=0 m New equilibrium position: l= mg total =-k(xe+x)i+ mgi 87. 3 a vertically oriented spring Ftotal=-k(x '+r)i+mgi=-kxi Kxi= ma,I=m-t dt dx k +x=0 dt m The origin ofx is the new equilibrium position. 6
6 §7.3 A vertically oriented spring 1. Is the motion of this system simple harmonic? New equilibrium position: kx mg e ′ = F k x x i mgi e ˆ ˆ ( ) total = − ′ + + r §7.3 A vertically oriented spring i t x kxi max i m ˆ d d ˆ ˆ 2 2 − = = F k x x i mgi kxi x e ˆ ˆ ˆ ( ) ,total = − ′ + + = − r m k x m k t x = + = ω 0 d d 2 2 The origin of x is the new equilibrium position
87.3 A vertically oriented spring Example 1: as shown in Fig. 1, n when the block of mass m falls freely and make a completely h unelastic collision with the plate of mass m, the system will oscillate up and down. Find the k T,Aandφ of the motion. 1g. Solution: The system is composed of 2 m and k. The angular frequency and the period are respectively 2m T=2丌 2m k 7.3 A vertically oriented spring From the initial condition ≈、mg∠0 k h m2gh=2m→V >0 t=0 2 We can obtain the amplitude k A=1x6+2 42 mg 1+ kh mg
7 §7.3 A vertically oriented spring Solution: The system is composed of 2 m and k. , 2m k ω = k m T 2 = 2π The angular frequency and the period are respectively m m h k Example 1: as shown in Fig. 1, when the block of mass m falls freely and make a completely unelastic collision with the plate of mass m, the system will oscillate up and down. Find the T, A and φ of the motion. Fig. 1 §7.3 A vertically oriented spring 0 2 0 = > gh v 0 0 2 2 0 m gh mv k mg x = =− < m m h k x o t =0 0 v r x0 From the initial condition mg kh k mg k mgh k v m g A x = + = + = + 1 2 2 2 2 2 2 0 0 ω We can obtain the amplitude
87.3 A vertically oriented spring From the initial condition x 小==兀 vn=- Dosing>0→sinφ<0 B=arct(--0)+r=arct 十兀 g 87.4 simple harmonic motion and the uniform circular motion I. Gelileo's observation of the moons of Jupiter 5 an.15202530Feb.510152025Mar.1 What can you imagine from the results? 8
8 π π ω φ = − + = + mg kh x v arctg( ) arctg 0 0 sin 0 cos 0 0 0 = − > = π From the initial condition §7.3 A vertically oriented spring §7.4 simple harmonic motion and the uniform circular motion 1. Gelileo’s observation of the moons of Jupiter What can you imagine from the results?
87.4 simple harmonic motion and the uniform circular motion >0 vKO KO 0 Simple harmonic motion can be described as the projection of a uniform circular motion along a diameter of the circle. A is called rotating vector. 7. 4 simple harmonic motion and the uniform circular motion v(1) x(t)=Acos(@t+o) v(t)=- Ao sin(ot+φ)
9 §7.4 simple harmonic motion and the uniform circular motion Simple harmonic motion can be described as the projection of a uniform circular motion along a diameter of the circle. A is called rotating vector. r A r ϕ π 2 3 = ϕ =π ϕ = 0 ϕ = π 2 0 0 v x 0 0 > > v x 0 0 > < v x 0 0 < < v x ω A r §7.4 simple harmonic motion and the uniform circular motion A ωA x(t) = Acos(ωt +φ ) v(t) = −Aω sin(ωt +φ )
87.4 simple harmonic motion and the uniform circular motion The virtues of describing the simple harmonic motion by using the uniform circular motion +中 Express the A, Tand a(1)P Q汁+ of simple harmonic motion; determine the initial phase of oscillation easily make the superposition a(t=-A cos(ar +o) of several oscillations conveniently. 7. 4 simple harmonic motion and the uniform circular motion Example 1: There is a simple harmonic oscillation of amplitude 0.24 m and period 3s. At initial time, t=0, xo=0. 12m, vo<0. Find the initial phase and the shortest time interval in which the oscillator arrive at position x=-0 12m Solution Draw the rotating vectors at t=0 and t A(t=0) p=丌/3 0.120012024x(m 4t!=-T=0.5s 6 10
10 §7.4 simple harmonic motion and the uniform circular motion A 2 ω ( ) cos( ) 2 a t = −Aω ωt +φ The virtues of describing the simple harmonic motion by using the uniform circular motion: 1express the A, T and ω t+φ of simple harmonic motion; 2determine the initial phase of oscillation easily; 3make the superposition of several oscillations conveniently. §7.4 simple harmonic motion and the uniform circular motion Example 1: There is a simple harmonic oscillation of amplitude 0.24 m and period 3s. At initial time, t=0, x0=0.12m, v0<0. Find the initial phase and the shortest time interval in which the oscillator arrive at position x= −0.12m. x(m) o 0.24 Solution: -0.12 A(t) r 0.5 s 6 1 ∆tmin = T = Draw the rotating vectors at t =0 and t . φ = π 3 0.12 A(t = 0) r φ
