MECHANICS: MOTION IN A STRAIGHT LINE

Part IIIMECHANICS: MOTION IN ASTRAIGHT LINE1 IntroductionA point mass m is moving along the x-axis under a force F(x, ẋ, t):=⇒ mẍ = F(x, ẋ, t) .0✻✲ xThe particle will be in equilibrium if its acceleration is zero: it may be moving in a straight line withconstant speed. Since Newton’s second law tells us that acceleration is proportional to force, a conditionthat F = 0 is both necessary and sufficient for equilibrium.2 Constant AccelerationIf F is constant then the acceleration is constant: ẍ = a.which is a second-order differential equation.d 2 xdt 2 = aSuppose that at t = 0, x = x 0 and ẋ = u 0 . These are the initial conditions. Integrating, we haveAt t = 0, ẋ = u 0 and so α = u 0 . Hence,dxdt= at + α.v = dxdt = at + u 0.This gives the velocity v. To find x, integrate again:But when t = 0, x = x 0 so β = x 0 :x = 1 2 at2 + u 0 t + β.x = 1 2 at2 + u 0 t + x 0 .So we have found v and then x as a function of t. It is possible to eliminate t and get a relation betweenx and v.We can, however, get this straight away if we write:Acceleration = d2 xdt 2 = dvdt = dv dxdx dt = v dvdx .Thus there are two equations for acceleration: dv/dt or v dv/dx.So in this case we have a = v dv/dx = d/dx( 1 2 v2 ) and, integrating, v 2 = 2ax+c, and the initial conditionsgive u 2 0 = 2ax 0 + c so that v 2 = 2ax + u 2 0 − 2ax 0.Example:A car passes through green traffic lights at 20ms −1 (≈ 40mph). He sees a horse and applies thebrakes which give him a retardation force of 15N per unit mass.How long does it take him to stop and how far away should the horse be for comfort?42

Solution:We must solve mẍ = −15m. Integrating, ∫ ẍdt = ẋ = −15t+20.Thus it takes him 20/15 = 4/3 seconds to stop.✻lighthorseFor the distance, we integrate v dv/dx = −15 to have v 2 /2 = −15x+k. Substituting v = 20 whenx = 0 gives 200 = k so v 2 /2 = 200 − 15x. The car stops when v = 0 so 200 = 15x, i.e. a minimumsafe distance for the horse as: x = 40/3m.✲ x3 Acceleration as a given function of timeWe are solving the systemmẍ = F(t).Example:A particle moves under a force F(t) = mF 0 sinpt where F 0 and p are positive constants.Solution:When t = 0, the particle goes through x = 0 with velocity V 0 î.Note: V 0 could be positive or negative.Show that the particle will eventually move to +∞ if and only if V 0 > −F 0 /p.Integrate : ẍ = F 0 sinptẋ = − F 0p cospt + cWhen t = 0 V 0 = − F 0p + c=⇒ ẋ = − F 0p cospt + V 0 + F 0pIntegrate again : x = − F (0p 2 sinpt + V 0 + F 0pAt t = 0 : 0 = k=⇒ x = − F (0p 2 sinpt + V 0 + F 0p)t + kThe term −(F 0 /p 2 )sinpt has maximum size F 0 /p 2 so for large t it is small compared to (V 0 +F 0 /p)t.Therefore, as t → ∞, x gets very large if and only if (V 0 + F 0 /p) > 0, V 0 > −F 0 /p.)t4 Acceleration as a function of velocitymẍ = F(ẋ).This is a second-order differential equation because the highest order derivative that appears is thesecond. But if we write ẋ = v, it becomes a first order equation for v:m˙v = F(v).We can separate the variables to give (if, at t = 0, v = v 0 ):∫ v ∫dv t1mv 0F(v 1 ) =0dt 1 .43

Here, we have written the solution as a definite integral. If you don’t like it, then you can write it as anindefinite integral∫ ∫dvmF(v) = dt + cand then calculate the constant c.Alternatively, we can treat it as a first order differential equation for v(x):so, integrating:or, if v = v 0 where x = x 0 ,mv dvdx = F(v)∫ ∫ v dvmF(v) =dx + c∫ v ∫v 1 dv x1mv 0F(v 1 ) = dx 1 .x 0Example:A particle moving under constant force P per unit mass subject to air resistance proportional tovelocity.✲ P ✲✛ xλẋin which λ > 0 is a constant.ẍ = P − λẋusing F = maSuppose the initial conditions are: at t = 0, x = 0 and v = v 0 > 0. Then the particle is projectedto the right.L19 End(a) Find v as a function of time.(b) Hence find the particle path, x as a function of t.(c) Find a relationship between v and x.Example:Recap: we are solvingwith x = 0 and v = v 0 > 0 at t = 0.If we write ẋ = v, we getẍ = P − λẋ∫ v˙v= P − λv∫ tdv 1=v 0P − λv 1[− 1 v1|]λ ln |P − λv = tv 00dt 11λ ln |P − λv 0| − 1 ln |P − λv|λ = tIf we assume that the sign of P − λv is the same as the sign of P − λv 0 , thenln∣P − λv 0P − λv∣ = ln P − λv 0P − λvP − λv 0P − λv= λt= e λtP − λv 0 = (P − λv)e λtv = P (λ + v 0 − P )e −λtλ44

As t → ∞, v → P/λ, the terminal velocity. It tends from above or below according to themagnitudes of v 0 and P/λ.vPλ✻v 0 > P/λv 0 < P/λ✲ tWe can see that if v 0 > P/λ then v > P/λ for all time, and if v 0 < P/λ then v < P/λ for all time:so our assumption was justified.To find x, we writedxdtx == P (λ + v 0 − P )λPtλ + (v 0 − P λe −λt) (− 1 λ e−λt )+ C.When t = 0, x = 0 and so0 =(v 0 − P ) (− 1 )+ Cλ λ( )As t → ∞, x → ∞.C = 1 λv 0 − P λx = Ptλ + 1 λ(v 0 − P λ) (1− e−λt )Now for the relation between v and x, we return to the original equation:λx =v dvdx = P − λv∫ vλv 1 dv 1=v 0P − λv 1∫ v( ) [P−1 + dv 1 = −v 1 − Pv 0P − λv 1 λ ln |P − λv 1|∫ x0λdx 1] v= v 0 − v + P ( P −v 0λ ln λv0P − λvwhere we used the fact that we had already checked that the sign of P − λv matches P − λv 0 toremove the modulus signs.Example:Now suppose the resistance is proportional to velocity squared. We will use a force of P 2 per unitmass and an air resistance coefficient of λ 2 for convenience.Resistance is always opposite to the direction of motion, so we haveẍ = P 2 − λ 2 ẋ 2 when ẋ > 0ẍ = P 2 + λ 2 ẋ 2 when ẋ < 0.In the previous example it was not necessary to make this distinction because λẋ changes sign withẋ, so the fact that resistance opposes the motion was already built in.At t = 0, x = x 0 and v = v 0 .).45

Case 1: v 0 > 0As before, we put ẋ = v. We haveNow1P 2 − λ 2 v 2 1ẍ = P 2 − λ 2 ẋ 2 as long as ẋ > 0˙v = P 2 − λ 2 v 2 as long as v > 0∫ vdv 1v 0P 2 − λ 2 v12= 1 λ 2 1P 2 /λ 2 − v 2 1== 1 [2Pλ∫ t0dt 11(P/λ) + v 1+]1.(P/λ) − v 1Now since we are assuming that v > 0, we know that (P/λ) + v > 0. If we additionally assumethat (P/λ) − v has the same sign as (P/λ) − v 0 then we can remove the modulus signs when weintegrate: [∣] ln(P/λ) + v 1 ∣∣∣ v∣= 2λPt(P/λ) − v 1 v 0[( ) ( )](P/λ) + v (P/λ) − v0ln= 2Pλt.(P/λ) − v (P/λ) + v 0Taking exponentials on both sides givesand we can solve for v, givingv = P λNow as t → ∞, we can see that v → (P/λ).((P/λ) + v)((P/λ) − v 0 )((P/λ) − v)((P/λ) + v 0 ) = e2Pλt{ e 2Pλt }((P/λ) + v 0 ) − ((P/λ) − v 0 )e 2Pλt .((P/λ) + v 0 ) + ((P/λ) − v 0 )We assumed two things: that v > 0 throughout, and that (P/λ) − v did not change sign.Questions:1. Did v ever become zero?If so,e 2Pλt = ((P/λ) − v 0)((P/λ) + v 0 ) .The LHS is always > 1 and the RHS is always < 1 so this can never happen.2. Was v ever equal to (P/λ)?This means thatand soe 2Pλt ((P/λ) + v 0 ) + ((P/λ) − v 0 ) = e 2Pλt ((P/λ) + v 0 ) − ((P/λ) − v 0 )v 0 = (P/λ)so unless we had that particular set of initial conditions, for which ˙v = 0 and v = v 0 , theassumption was valid. If we did have those initial conditions then our solution does still work.46

L20 EndIf you want to find position, it is often easier to find a relation between x and v, using ẍ = v dv/dx.Our equation becomes∫ vv dvdx = P 2 − λ 2 v 2∫ xv 1 dv 1v 0P 2 − λ 2 v12 = dx 1x 0− 1 [ln |P 22λ 2 − λ 2 v1 2 |] v= x − xv 00∣ ∣1 ∣∣∣2λ 2 ln P 2 − λ 2 v02 ∣∣∣P 2 − λ 2 v 2 = x − x 0P 2 − λ 2 v 2P 2 − λ 2 v 2 0= e −2λ2 (x−x 0)This method is easier than integrating the expression for the velocity.As t → ∞, v → P/λ (from the earlier calculation) so x → ∞.Case 2: v 0 < 0In this case we expect the particle to stop at some point, after which we will have Case 1 again.Let’s just ask: at what time t 0 does the particle come to a halt?For this, we must solve ẍ = P 2 + λ 2 ẋ 2 :∫dv0 ∫dt = P 2 + λ 2 v 2 dvt0P 2 + λ 2 v 2 =If we make the substitution v = (P/λ)tanθ this becomest 0 =∫ 0arctan [λv 0/P]v 0(P/λ)sec 2 ∫θ dθ 0P 2 + λ 2 (P 2 /λ 2 )tan 2 θ =arctan [λv 0/P]Once the particle comes to rest, it will turn round and give case 1 again.0dtdθPλ = − 1 [ ]Pλ arctan λv0.P5 Acceleration as a function of displacement only5.1 EnergyWe are solving the systemmẍ = F(x)with initial conditions: at t = 0, x = x 0 and v = v 0 .As before, we write ẋ = v, ẍ = v dv/dx. This gives the equation asmv dvdx = F(x), i.e. m d ( ) 1dx 2 v2 = F(x).∫ x1i.e.2 mv2 − 1 2 mv2 0 = F(x 1 )dx 1 . (35)x 0Now 1 2 mv2 is called the kinetic energy of the particle.∫ xx 0F(x 1 )dx 1is called the work done by the force F(x) as the particlemoves from x 0 to x.So (35) says that “Change in Kinetic Energy = Work done by force”.47

In order that we can integrate the RHS of (35), we define a potential function V (x) so thatwhere the minus sign is by convention.F(x) = −m dVdxThen (35) becomes 1 2 mv2 − 1 2 mv2 0 = −mV (x) + mV (x 0), i.e.12 v2 + V (x) = 1 2 v2 0 + V (x 0 ). (36)V (x) is the potential energy per unit mass and so mV (x) is the potential energy. We havekinetic energy + potential energy = constantthroughout the motion. (36) is the energy equation.Denote the total energy per unit mass on the RHS of (36) by E (a constant). Then we can rewrite (36)as12 v2 = E − V (x).Since v 2 ≥ 0, we must have E ≥ V (x) for possible motion.Suppose V (x) looks like:EV (x)✻x 1 x 2✲ xL21 EndThe particle will never traverse to x 1 because in that region V (x) > E: i.e. the particle cannot use thepiece of the x-axis between x 1 and x 2 since we must have E ≥ V (x).Suppose the particle starts in x > x 2 with ẋ < 0. Where will it stop? It stops where v = 0, i.e.E = V (x): in this case at x = x 2 .What happens next? The original equation gave mẍ = F(x) = −mdV/dx. We see from the diagramthat V ′ (x 2 ) < 0 i.e. ẍ > 0 hence, at x 2 , ẍ > 0 and the particle will move off to the right.The particle will never stop since V (x) does not cross E again. Since V (x) ≠ E after x 2 , the particlewill travel to ∞.Recap: Energy equationFor a systemthe energy equation isẍ = − dVdx12 v2 + V (x) = E.This means the particle cannot be in any region where V (x) > E.48

Another potential:y = V (x)✻E 2 x 1 x 2 x 3 x 4 x 5✲ xThe particle could move along the projections of the thick lines onto the x-axis because in these regionsE 2 > V (x).The particle can move between x 1 and x 2 , or x 3 and x 4 , or all of x > x 5 . But, for example, x 2 to x 3 isout of bounds.If the particle starts between x 1 and x 2 , it will remain between these two points for ever. At the endpoints it will stop (E = V (x) so v = 0) and the acceleration pulls it back towards the low in the potentialfunction. So the particle oscillates between x 1 and x 2 .Time for oscillation from x 1 to x 2 and back to x 1 .We haveso( ) 2 dx= 2(E − V (x))dt( ) dx= √ 2(E − V (x)) as long as the particle moves to the rightdt∫ x∫dx t1√x 1 2(E − V (x1 )) = dt 10Thus the time, T 1 , for the particle to get to x 2 isT 1 =∫ x2dx 1√x 1 2(E − V (x1 )) .From x 2 back to x 1 , the velocity is negative so( ) dx= − √ 2(E − V (x))dt∫ xSuppose we are back at x 1 at time T. Thenso T = 2T 1 .∫ x1∫dx t√ 1x 2 2(E − V (x1 )) = −T 1dt 1dx√ 1= −T + T 1x 2 2(E − V (x1 ))} {{ }=−T 1The motion is periodic with period T (this is the time taken to get back to the starting position). Theparticle turns round every time it gets to an end point. It always has the same speed at the same place.5.2 Simple Harmonic Motion (SHM).Simple Harmonic Motion is given by the special caseẍ = −ω 2 x.49

✻0 ✛x✲✲ xThe particle is always attracted towards 0, with a force proportional to the displacement.v dvdx = −ω2 x where ω is a constant.Let us suppose that when t = 0, x = x 0 and v = v 0 . Then integrating gives12 v2 = − 1 2 ω2 x 2 + E.where E = 1 2 (v2 0 + ω2 x 2 0 ); now choose a so that E = 1 2 ω2 a 2 and we haveHere the potential V (x) = 1 2 ω2 x 2 .−a✻E✲a xv 2 = ω 2 (a 2 − x 2 ).The particle oscillates between −a and a. The period of this oscillation isT = 2= 4∫ a−a∫ a0∫dx a√ 12(E − V (x1 )) = 2dx 1ω √ (a 2 − x 2 ) = 4 ω−adx 1ω √ (a 2 − x 2 )[ ( sin −1 x a=a)] 4 0 ω × π 2 = 2π ω .Can we find x(t)?To find x as a function of t, we writev = dxdt = √ 2(E − V (x)) = ω √ a 2 − x 2as long as the particle is moving in the positive direction.We separate variables to give∫i.e., rearranging,∫dx√a2 − x = 2ωdtx = a sin (ωt − δ).(sin −1 x)= ωt − δaWe should now take the negative square root for the velocity and return from x = a. Integrating asbefore gives the same formula as above.[Exercise: try this!]• a is called the amplitude. It is the furthest displacement from the origin of the oscillating particle.• 2π/ω is the period since sin(ω(t + 2π/ω) − δ) = sin (ωt − δ).• ω is called the frequency of the oscillation.• δ is called the phase angle.Easier approachBecause the equation ẍ = −ω 2 x is linear, this process can be done more easily.Now ẍ = −ω 2 x is linear and the coefficients are constant. We can try solutions of the form x = e λt :λ 2 e λt = −ω 2 e λt50

Hence λ 2 = −ω 2 and so λ = ±iω, which works because differentiating doesn’t alter the form of thefunction.We now have a solution e iωt and another e −iωt . Indeed, we can multiply either of them by a constant,say A or B, to have another solution: Ae iωt or Be −iωt . This works by linearity.We can also use their sum to give us another solution. This also works because the equation is linear.So now we have solutions: Ae iωt , Be −iωt , Ae iωt + Be −iωt .It often pays to write the solution asAe iωt + Be −iωtNow let us define= A(cos ωt + i sinωt) + B(cosωt − i sinωt) = (A + B)cosωt + i(A − B)sinωt= α cosωt + β sinωtαsinδ = −√ α2 + β , cosδ = β√ 2 α2 + β , a = √ α 2 + β 22in order to transform our solution:L22 Endx = a [− sinδ cosωt + cosδ sinωt] = a sin (ωt − δ) as before.Summary of Simple Harmonic Motion• mẍ = mF(x) = −mω 2 x• ẋ = v and ẍ = v dv/dx• v dv/dx = −ω 2 x so integrating gives 1 2 v2 = − 1 2 ω2 x 2 + E = E − V (x) where V (x) = 1 2 ω2 x 2 .• E depends on the initial conditions: if at t = 0, x = x 0 and v = v 0 then E = 1 2 v2 0 + 1 2 ω2 x 2 0 and wecan set a such that E = 1 2 ω2 a 2 .• Where E = V (x), ẋ = 0 and ẍ acts to turn the particle around and it returns to the region whereV (x) < E.• Motion is described byx = a sin (ωt − δ)v = ωa cos(ωt − δ).Elastic strings✛ l ✲ ✛ e ✲✛A B T CWe have an elastic string of natural length l, fixed at A. We stretch the string so that end B moves toC.Once the string is stretched, there is a tension in the string given by Hooke’s Law, which saysT = λe , where λ is the modulus of elasticity.lIf we have a string, the tension is zero if the extension is negative. If we have a spring, it is OK to usethis formula with negative extension – we get a thrust.Example:A particle of mass m on the end, B, of an elastic string is projected at speed v 0 away from A.AB✲ v051

Find the greatest length of the string in the subsequent motion.Solution:Measure x from B. The governing equation ismẍ = −T = − λxlwhich is the equation of SHM with ω = √ (λ/ml).v dvdx = −λx ml=⇒ ẍ = − λxml=⇒ 1 2 v2 = − λ x 2ml 2 + 1 2 v2 0 .The greatest string length occurs when v = 0, so we haveλ x 2ml 2 = 1 ( ) v2 1/22 v2 0 =⇒ x = 0 ml.λ( ) v2 1/2This is the extension, so the greatest length is l+ 0 ml.λBackground material for the next sectionTaylor seriesYou will see Taylor series elsewhere in due course: but you will need them here so if you have not seenthem before, this is a quick reference.For a well-behaved function f(x) near a point x = a, we can writeO notationf(x) ≈ f(a) + (x − a)f ′ (a) +(x − a)2f ′′ (a) +2!(x − a)3f ′′′ (a) + · · ·3!L23 EndIf f(x) = O(x n ) this means that as x → 0, f(x)/x n remains bounded, in other words f(x) → 0 at leastas fast as x n .f(x) = o(x n ) means f(x)/x n → 0 as x → 0.f(x) ∼ x n means f(x)/x n → a finite limit as x → 0.5.3 Equilibrium and StabilityWe have a particle moving in a potential:ẍ = − dVdx12 v2 = E − V (x) where E is a constant depending on the initial conditions.Equilibrium points occur where particles could be at rest and stay at rest.These are where ẍ = 0, i.e. V ′ (x) = 0.y = V (x)✻✲ x52

Let x = c be one of these points. Then V ′ (c) = 0.Near x = c,V ′ (x) = V ′ (c) +(x − c)V ′′ (c) +} {{ }=0(x − c)2V ′′′ (c) + O ( (x − c) 3)2and the governing equation becomes, neglecting terms as small as (x − c) 2 ,Let x − c = X, then this gives Ẍ = −XV ′′ (c).If V ′′ (c) > 0 this is SHM with period 2π/ √ V ′′ (c).ẍ = −(x − c)V ′′ (c).Stable means the particle stays near x = c: clearly true in this case.If V ′′ (c) < 0 then Ẍ has the same sign as X and the particle will accelerate away from the equilibriumpoint if it is not exactly on the point. In this case the equilibrium is unstable.A maximum of V (x)A minimum of V (x)is an unstable equilibrium pointis a stable equilibrium pointExample:V (x) =kxx 2 + b 2 with k > 0.Find the period of small oscillations about the position of stable equilibrium. Give also an expressionfor the period of oscillation for a particle which starts from rest at the point x = −b/2.Solution: Let’s look first at the shape of the potential function:V ′ (x)k=1x 2 + b 2 − 2x 2(x 2 + b 2 ) 2 = b2 − x 2(x 2 + b 2 ) 2 ,so there are equilibrium points at x = ±b.V (x)✻−bb✲ x✻stable equilibrium pointWe could differentiate again and use the formula 2π/ √ V ′′ (−b). Alternatively we could investigatedirectly.To examine the behaviour near x = −b, write x = −b + X where |X| is small.This equation becomes:ẍ = − dVdx = −k(b2 − x 2 ) −k(b − x)(b + x)(x 2 + b 2 ) 2 =(x 2 + b 2 ) 2Ẍ =which is SHM with period 2π √ (2b 3 /k).−k(2b − X)X((−b + X) 2 + b 2 ) 2 ≈ −2kbX4b 4= −kX2b 3 ,53

Now what about the larger oscillations? If, initially, we have x = −b/2 and v = 0, then the energyequation is12 v2 + V (x) = V (−b/2).In our case V (−b/2) = −2k/5b so we have12 v2 + kxx 2 + b 2 = −2k 5b .The particle will start at x = −b/2, move left to the next point where V (x) crosses the lineE = −2k/5b, and then turn around and move right again.What is the position of the turnaround point? Say it is X; thenV (X) =kXX 2 + b 2 = −2k 5b2X 2 + 5bX + 2b 2 = 0 (2X + b)(X + 2b) = 0so the next point where the particle stops is at x = −2b. Then we return to the energy equation:v 2 = − 4k5b − 2kxx 2 + b 2 = + 5bx + 2b 2 )−2k(2x2 5b(x 2 + b 2 )√ ( )dxdt = ± − 2k(2x2 + 5bx + 2b 2 )5b(x 2 + b 2 )and find the period by integrating the positive velocity from −2b to −b/2 and then the negativevelocity back again; remember these two integrals give the same answer soT = 2∫ −b/2−2b(−5b(x 2 + b 2 )) 1/2dx.2k(2x 2 + 5bx + 2b 2 )We can simplify this integral by making a substitution x = by:T = 2bNumerically we can calculate( 5b2k∫ −1/2y=−2) 1/2 ∫ −1/2y=−2( −(y 2 ) 1/2+ 1)(2y 2 dy.+ 5y + 2)( −(y 2 ) 1/2+ 1)(2y 2 dy ≈ 3.64+ 5y + 2)L24 Endso we end up with T ≈ 11.5 √ b 3 /k which takes longer than the small SHM oscillations, whoseperiod is T SHM = 2π √ (2b 3 /k) ≈ 8.89 √ b 3 /k.5.4 Phase PlaneWe are looking at systems of the formv dvdx = −dV dxNow, since v = ẋ, 1 2ẋ2 = E − V (x).mẍ = F(x) = −m dVdx=⇒ 1 2 v2 = E − V (x) where E is a constant.Sometimes it is helpful to draw the phase plane, i.e. set ẋ = y and plot y as a function of x for variousvalues of E.54

Example: ẍ = a where a is a constant and a > 0.y = ẋ✻ẍ = av dvdx = a12 v2 = ax + Eso if we put y = v then✒−E/a✲ x✻ E = 0x = 12a y2 − E a .These curves are all parabolas on theirsides. They are all similar but shifted tothe right by different amounts.Suppose at t = 0, x = x 0 and ẋ = y = 0.Then E = −ax 0 .Since the force is positive, ẍ > 0 so ẋ increasesand we draw the arrow of motionupwards in the direction of increasing t.Time t is not shown explicitly on the curve. Note that in the upper half plane, ẋ > 0 so x increasesand the arrow points to the right; in the lower half plane the arrow points to the left.Simple Harmonic Motion✻ V (x) 12 ω2 x 2E 2 > 0✲ xE 1 < 0ẍ = −ω 2 x = − dVdx12 v2 = − 1 2 ω2 x 2 + E = −V (x) + Ey 2 = −ω 2 x 2 + 2ERecall that for E 1 < 0 there is no motion; for E 2 > 0 the particle will oscillate over the thick-linedregion; and for E = 0 the particle stays at x = 0.For the phase plane, we draw y 2 + ω 2 x 2 = 2E for various E:y✻✲ x✙This is a CENTRE and is STABLE. The behaviour of a system close to a stable equilibrium alwayslooks like this.The point in the middle of the diagram (x = 0, ẋ = 0) is a stable equilibrium point given by E = 0.Suppose a particle starts at x = a with ẋ = 0. Then 2E = ω 2 a 2 and ẋ starts by decreasing: hence thearrow, in the direction of increasing t.One orbit takes a period of 2π/ω.The general solution isx = a sin (ωt − δ)ẋ = aω cos(ωt − δ).55

Hyperbolic Motion.ẍ = ω 2 x = −dV/dx.V (x)✻E 1unstable equilibrium point✠ ✲ xE 212 v2 = 1 2 ω2 x 2 + E = −V (x) + EThere are no small oscillations in this diagram. A smalldisplacement from the origin will make the particle fall.At E 2 the particle is restricted to one or other of the thicklinedregions but will still travel to ∞.E = 0 ✒y✻E > 0[1]❘✒E < 0✲ xE < 0✠■E > 0[2]This is a SADDLE POINT and is UNSTABLE. The behaviour of a system close to an unstableequilibrium always looks like this.The point where the lines cross is an unstable equilibrium point, where E = 0, x = 0 and ẋ = 0.Suppose E = 0, the possible trajectories for the system are the straight lines y = ±ωx, so ẋ = ±ωx.Taking the +ve root x = Ae ωt ẋ = Aωe ωtTaking the −ve root x = Be −ωtẋ = −Bωe −ωtL25 EndThe first of these, with A > 0 (we assume ω > 0) corresponds to the line [1], on which as t increases, ẋincreases and x increases.Line [2] corresponds to x = Be −ωt with B > 0. x is always positive and ẋ is always negative: and xdoes not reach zero until time is infinite.The general solution isx = a sinhωt + b coshωtẋ = aω cosh ωt + bω sinhωt.5.5 The Pendulum✟❆❑❆❆❆✟❆❆❆❆ T❆❑l❆❆❑ a 2❆ ❆❑ ❆❝✟ ✟✯ ✟✯a 1❆❆❆❯✟✟❄mgConsider a mass m at the end of a light inextensible string of length l.Resolve:T − mg cosθ = ma 2−mg sinθ = ma 1To find T, we need a 2 [see MATH1302].along stringperpendicular to stringTo get the position θ, we need a 1 . In fact, we might suspect the correctform, a 1 = l¨θ (to see this shown, see MATH1302).56

Our equation isl¨θ = −g sinθ. (37)Suppose θ is very small, i.e. there will be small oscillations about the equilibrium position. Since θ issmall, sinθ ≈ θ.So ¨θ = (−g/l)θ, which is SHM of period 2π √ (g/l).This is called a simple pendulum: θ = Asin ( √ (g/l)t − δ).However, we can (to a certain extent) do better. Let us write ˙θ = v. ThenNow equation (37) becomes¨θ = dvdt = dv dθdθ dt = vdv dθ .v dvdθ= −(g/l)sinθ12 v2 = (g/l)cosθ + E = −V (θ) + EE is dependent on the initial conditions. Suppose at t = 0, θ = 0 and the velocity perpendicular to thestring, i.e. tangential velocity is l ˙θ = l ˙θ 0 . Then E = 1 2 ˙θ 2 0 − g/l.V (θ)✻E 1✲ θE 2E 3E 4E 4 : This is no good since nothing can happen: this value of E never occursE 3 : Our simple pendulum oscillates about the stable equilibrium point θ = 0 (or θ = 2π, 4π etc.if we choose the origin as being at one of these points)E 2 : This is the same as E 3 except that the oscillations are no longer small.✟ θIf θ becomes larger than π/2 we should check that the tension does not becomezero: for this we need a 2 from M13B.E 1 : This corresponds to complete revolutions.Note that at E 2 we do not have true Simple Harmonic Motion; even at E 3 SHM is only a good approximation.Phase planeThe energy equation is12 ( ˙θ) 2 = E + (g/l)cosθ = E − V (θ)with potential function is V (x) = −(g/l)cosx.For the phase plane,Taking c = −1 gives12 y2 = E + g l cosx = g (c + cosx).l12 y2 = g (cosx − 1), i.e. the points x = 0, ±2π, ±4π, . . . and y = 0.l57

Taking c = 1 gives12 y2 = g 2g(cosx + 1) =l√lg( xy = ±2l 2)cos .( cos 2 x).2If c > 1, y does not vanish.ystable equilibrium pointsc > 1✁unstable equilibrium pointsc = 1✁❅ ❅❘ ✲ ✁ ✁✁ ✁−1 < c < 1✁ ✁−4π −2π ✁☛✁☛2π 4π x✻c = −1✛c > 1The points where V ′ (x) = 0 and y = 0, i.e. the equilibrium points, are called the critical points in thephase plane.L26 End[Handout 12 Here]5.6 Further Simple Harmonic MotionFor SHM we hadwith solutionswhich oscillate with period 2π/ω.ẍ + ω 2 x = 0x = Asin (ωt − δ),Damped SHM or The Harmonic OscillatorNow the particle oscillating on the end of an elastic string will be submerged in a bath of oil.✻T + DXT✻❄❄❄mgmg + DParticle moving downwards Particle moving upwardsẊ > 0 Ẋ < 0D is the drag or resistance due to the oil.Stokes drag is given to be proportional to Ẋ.The equation is the same whether the particle is going up or down: if we take X = 0 at the equilibriumposition, thenẌ = − λ X − kẊ with k > 0,ml58

which we can write asTry the solution e µt :Ẍ + kẊ + ω2 X = 0 where ω 2 = λ/ml.The general solution isµ 2 + kµ + ω 2 = 0 =⇒ µ = −k ± √ k 2 − 4ω 2αe µ1t + βe µ2t .2= µ 1 , µ 2 .Case 1: Weak damping ω 2 > k 2 /4.Since ω 2 − k 2 /4 is positive let us set ω 2 − k 2 /4 = r 2 . Then µ = − 1 2k ± ir. Hence the solution isX(t) = e −1 2 kt (αe irt + βe −irt ) = e −1 2 kt (Acos rt + B sinrt) = A ′ e −1 2 kt sin (rt − δ)in which A ′ = √ A 2 + B 2 .X(t)✻A ′ e −1 2 kt✲t−A ′ e −1 2 ktCase 2: Strong damping ω 2 < k 2 /4.Since k 2 /4 − ω 2 is positive let us set k 2 /4 − ω 2 = s 2 . Then µ = − 1 2k ± s and these values of µ are bothnegative. The solution isand in this case there is no oscillation:X(t) = αe −(1 2 k+s)t + βe −(1 2 k−s)tX(t)✻✲tCase 3: Critical damping ω = k/2.Here there is only one (repeated) root of the quadratic equation, µ = − 1 2k. The solution isX(t) = e −1 2 kt (A + Bt).This tends to zero, but not as fast as the strongly-damped case because of the Bt term.59

X(t)✻✲tForced SHM and ResonanceLet us look at simple harmonic motion with a periodic external force:ẍ + ω 2 x = F cospt.E.g. with our particle on a string, wobbling the “fixed” point at the top of the string up and down. Thesolution isx = Asin (ωt − δ) + F cospt(ω 2 − p 2 ) .This is fine for ω ≠ p, i.e. if the natural frequency is not equal to the forcing frequency.We can see thati.e. x remains bounded for all time.X(t)✻|x| < |A| +|F ||ω 2 − p 2 |✲tIf ω = p then this solution won’t do.We need to solve ẍ+ω 2 x = F cosωt. We have a problem, because cosωt is part of the solution. Insteadwe can solve:ẍ + ω 2 x = Fe iωt . (38)Substituting into (38) givesWe try xẋ= γte iωt= γe iωt + iωγte iωtẍ = 2iωγe iωt − ω 2 γte iωt .2iωγe iωt = Fe iωt =⇒ γ = F2iω = −Fi 2ω .60

Now since the true forcing, F cosωt = Re(Fe iωt ) we have the solution given byThis solution is unbounded: resonance.x = α cosωt + β sinωt + Re(γte iωt )(= α cosωt + β sinωt + Re − Fi)2ω t(cosωt + i sinωt)= α cosωt + β sinωt + Ft2ω sinωtX(t)✻✲tL27 End61