Classical Mechanics Small Oscillations

Transcription

1 Classical Mechanics Sall Oscillations Dipan Kuar Ghosh UM-DAE Centre for Excellence in Basic Sciences, Kalina Mubai Septeber 4, 06 Introduction When a conservative syste is displaced slightly fro its stable equilibriu position, it undergoes oscillation. The cause of oscillation is the restoring forces which are called into play. Restoring forces can do both positive and negative work. When the work done is positive, the restoring forces change the potential energy into kinetic energy and when the work done is negative, they change kinetic energy back into potential energy. For ost echanical systes, when the syste is not too far fro the equilibriu, the restoring force is proportional to the displaceent (F = kx). Such oscillators are called linear oscillators. For linear oscillators, the oscillation frequencies are independent of the aplitude of oscillation. Oscillator otion can be daped in the presence of resistive forces. Resistive forces extract energy fro the oscillator. For low velocities, the resistive forces are proportional to velocity. Oscillators, whether daped or undaped, can be driven by external agencies which continuously supply energy to the oscillator to keep it oscillating. Such oscillators are known as forced or driven oscillators. Driven oscillators can cause aplitude of oscillation to becoe very large when the driving frequency atches the natural frequency of oscillation. This is known as the phenoenon of resonance. Noral Modes. Equilibriu: Consider a syste with {q i } as the generalised coordinates. Since the syste is conservative, the forces acting on the syste are derivable fro a potential energy function

2 c D. K. Ghosh, IIT Bobay V (q, q,..., q N ). Lagrange defined equilibriu as a configuration in which all generalised V forces vanish, i.e. = 0. Clearly, in such a situation, the syste will not change its q i configuration. However, even when Q i = 0, the syste ay not be stable in the sense that if it is slightly disturbed fro a position of equilibriu, it ay not return to the position of equilibriu. If it does, such a configuration is called one of stable equilibriu - otherwise the equilibriu is unstable. Exaple: Siple Pendulu θ The potential energy is given by V (θ) = +gl( cos θ), so that x F (θ) = V θ = gl sin θ = gx V=0 The generalised force corresponding to θ in this case is actually the restring torque. Equilibriu occurs when the restoring torque is zero. There are two such positions, θ = 0 and θ = π. Let us look at the for of the Lagrangian near these two positions. L = l θ gl( cos θ) Near θ = 0, cos θ θ so that L = l θ glθ so that the potential energy is V (θ) = glθ and the corresponding generalised force is glθ which is of restoring nature. On the other hand, near the second position of equilibriu θ = π, cos θ = cos(π + δθ) cos δθ = + δθ. In this situation, L = l θ + gl(δθ) the corresponding force is anti-restoring, aking the equilibriu unstable. For one diensional holonoic systes, equilibriu can be either stable on unstable (leaving out a trivial case of neutral equilibriu where the potential energy function is spatially flat) for which the potential energy has an extreu V q i = 0 ()

3 c D. K. Ghosh, IIT Bobay 3 for every generalised coordinate q i. Let the position of equilibriu be q i0. If the position is one of stable equilibriu, the potential energy has to be iniu. This is because, the syste being conservative, the total energy is constant. If we go away fro the position of iniu potential energy, it leads to an increase in the potential energy and a consequent decrease in the kinetic energy. Thus the syste returns back to the equilibriu position. For stable equilibriu, we, therefore, have The converse would be true for an unstable equilibriu. V q i q j > 0 () Without loss of generality, let us shift the equilibriu position to the origin (q = q =..., = q N = 0). If the syste is disturbed to a configuration {q i }, we can write, V (q, q,...) = V (0, 0,...) + ( ) V q i + ( ) V q i q j + higher order ters q i i 0 q i q j 0 where the partial derivatives are evaluated at the position of equilibriu and all higher order ters which involve third order or higher corrections are neglected. If the potential energy is easured fro its iniu value, we choose V (0, 0,...) = 0. Along with ( ) V = 0 q i The leading ter in the change in potential energy is then ( ) V > 0 q i q j for stable equilibriu. Let us write ( ) V V ij = q i q j so that i,j V = i,j 0 i,j 0 0 V ijq i q j (3) with V ij = V ji. Now, the kinetic energy of a scleronoic syste is, in general, a quadratic in generalised velocities, and can be written as T = T ij q i q j (4) i,j

4 c D. K. Ghosh, IIT Bobay 4 where the coefficients t ij are, in general, functions of generalised coordinates. One can expand t ij in a Taylor series about the equilibriu position T ij (q, q,...) = t ij (0, 0,...) + ( ) tij q k +... q k k 0 ( ) tij It turns out that the quantities and the higher order derivatives are negligibly q k 0 sall so that the coefficients t ij s can be essentially treated as constants having the sae values as they would have in the equilibriu position. Thus around the equilibriu position, the Lagrangian has the following structure: L = T V = (t ij q i q j V ij q i q j ) The Lagrangian equations of otion ( ) d L dt q k can then be written as i,j L q k = 0 d dt [t ijδ ik q j + t ij q i δ kj ] V ij (δ ik q j + q i δ kj ) = 0 i,j i,j [ t kj q j + ] [ t ik q i V ik q j + ] V ik q i = 0 (5) j i j i Changing the duy suation index j to i in the first and the third ters of the above and using the syetry of V ij and of t ij, we get t ik q i + V ik q i = 0 i i for each k. We seek a solution to the above equations of the for q i = A i e iωt which gives ( ) Vik ω t ik Ai = 0 (6) i The equation is a hoogeneous equation in A i s and the condition for existence of the solution is det(v ik ω t ik ) = 0 which is a single algebraic equation of n th degree in ω. This equation has n roots soe of which are real and soe coplex (soe of the roots ay be degenerate). We are

5 c D. K. Ghosh, IIT Bobay 5 only interested in real roots of the above equation. ω k s deterined fro this equation are known as characteristic feequencies or eigenfrequencioes. Fro physical arguents it is clear that for real physical situations, the roots are real and positive. This is because the existence of an iaginary part in ω would ean tie dependence of q k and q k such that the total energy would not be conserved in tie and such solutions are unacceptable. We can arrive at the sae conclusion atheatically as well. Multiplying (6) with A k and suing over k we get (V ik ω t ik )A ka i = 0 i,k so that ω i,k = V ika k A i i,k t ika k A i Both the nuerator and the denoinator are real because V ik = V ki and t ik = t ki. It is seen that the ters are positive as well because expressing A i = a i + ib i, we have t ik A i A k = t ik (a i ib i )(a k + ib k ) i,k i,k = i,k t ik (a i a k + b i b k ) where the iaginary ters cancel because of syetry of t ik. Thus we have been able to express i,k t ika i A k as a su of two positive sei-definite ters ( t ik a i a k = a T ta is positive definite).. Matrix Forulation Let us rewrite (6) (using syetry properties of V and T ) as (V ki λt ki )A i = 0 where λ = ω. Let us define a colun vector i A A A =... A N The atrices V and t are given by V V... V N V V = V... V N V N V N... V NN t t... t N t T = t... t N t N t N... t NN

6 c D. K. Ghosh, IIT Bobay 6 Using these, the equation (6) can be expressed as a atrix equation V A = ω T A (7) Note that this equation is not in the for of an eigenvalue equation as V A is not equal to a constant ties A but a constant ties T A. (If T is invertible, one can get an eigenvalue equation T V A = ω IA. Since we have N hoogeneous equations, we have N odes, i.e. N solutions for ω. Let us denote the k-th ode frequency by ωk = λ k. Let the vector A corresponding to this ode be written as We then have A k A A k = k... A kn V A k = λ k T A k (8) Taking conjugate of this equation and changing the index k to i, we get à i V = λ i à i T (9) where we have used à to denote the transpose of the atrix A.. Fro (8) we get by ultiplying with Ãk Fro (8) and (9) it follows that λ k = ÃkV A k à k T A k (0) so that à i V A k = λ k à i T A k à i V A k = λ i à i T A k (λ k λ i )ÃiT A k = 0 Thus, if the eigenvalues are non-degenerate, i.e. if λ i λ k, we get the orthogonality condition à i T A k = 0 () Note that this is different fro the orthogonality condition on eigen vectors for a regular eigenvalue equation. Since (7) does not uniquely deterine A, we define noralization condition as à i T A i = () Exaple : Consider two asses = and = connected by three springs, as shown.

7 c D. K. Ghosh, IIT Bobay 7 k = 4k k = k k = k 3 = = We know that a single ass spring syste with ass and spring constant k has a natural frequency of oscillation k/. Let us consider the syste in the figure. Let the generalised coordinates be displaceent of the asses fro their equilibriu positions, the ass being displaced by x while the ass by an aount x. The central spring is then copressed or stretched by an aount x x. Let us attept to solve the proble using the force ethod. The equations of otion for and are ẍ = 4kx k(x x ) ẍ = kx k(x x ) (3a) (3b) The sign of the last ter is fixed by taking x to be large positive so that the force on is in the positive direction, as it ought to be. We will attept to solve this pair of coupled equations (3a) and (3b) by using a bit of guesswork and a bit of luck. This will not work except in cases which show sufficient syetry. The idea is to find a linear cobination of x and x so that the coupled equations becoe uncoupled. Let us rewrite (3a) and (3b) as ẍ = 5kx + kx ẍ = kx 3kx (4a) (4b) It can be easily seen that (ẍ ẍ ) = 7 k(x x ) (ẍ + ẍ ) = k(x + x ) (5a) (5b) Thus, if we define new noral coordinates the equations for y and y are uncoupled, y = x x y = x + x (6) ÿ = 7 ky ÿ = ky (7) The noral coordinates oscillate with independent frequencies k/ and 7k/, which are known as the noral odes. These new generalised coordinates execute

8 c D. K. Ghosh, IIT Bobay 8 siple periodic oscillations known as noral oscillations. The new noral coordinates satisfy ÿ α + ω αy α = 0 (8) With this intuitive background, let us look at the proble ore forally. The Lagrangian of the syste is L = ẋ + ẋ k x k (x x ) k 3x (9) Define t and V atrices t = V = ( ) T 0 = ẋ i ẋ j 0 ( ) V 5k k = x i x j k 3k (0) () Condition for existence of non-trivial solution is which gives det(v ik ω t ik ) = 0 5k ω k k 3k ω = 0 which has the solution ω = 7k/ and k/. The noral odes are now found fro (6) ω T A = V A Writing A = ( A A ), we get, for ω = 7k 7k ( 0 0 ) ( ) ( ) ( ) A 5k k A = k 3k For this noral ode we have A = A so that (noralizing) A = ( ) 6 A A For ω = k, a siilar calculation gives A = ( ) 3 The general solution can then be written as ( ) ( ) ( ) ( ) ( ) x = A cos ω t + B sin ω t + C cos ω + t + D sin ω + t x

9 c D. K. Ghosh, IIT Bobay 9 We can fix the constants A, B, C and D by using initial conditions. Suppose we we pull the two asses aside and release, we then have x (0) = x 0, x (0) = x 0, ẋ (0) = ẋ (0) = 0 A + C = x 0 which gives A C = x 0 Bω + Dω + = 0 Bω Dω + = 0 so that It can be checked that, A = 3 x x 0 B = D = 0 C = 3 x 0 3 x 0 ( x = 3 x 0 + ) ( 3 x 0 cos ω t + 3 x 0 ) 3 x 0 cos ω + t () ( x = 3 x 0 + ) ( 3 x 0 cos ω t 3 x 0 ) 3 x 0 cos ω + t (3) x x = (x 0 x 0 ) cos ω + t x + x = (x 0 + x 0 ) cos ω t as expected. Exaple : As a second exaple, consider a pair of identical pendulus consisting of a pair of assless rigid rods of length l each at the end of each of which a ass is attached. The id-points of the rods are connected by a spring of force constant k. In the absence of coupling, each oscillator has a frequency ω = proble by force ethod. g l. As before, let us try to solve the A B k

10 c D. K. Ghosh, IIT Bobay 0 Let the unstretched lengths of the spring be equal to the horizontal separation between the rods when the rods are vertical. When the rods ake θ and θ with the vertical (easured in the sae sense) the extension of the spring is l (θ θ ). Since we are interested in the angular accelerations of the rods, let us take torques about the points A and B. Tension and reaction at the support both pass through A and B and do not contribute to the torque. The torque about A (and B) is gl sin θ + k(θ θ ) l l cos θ = I θ gl sin θ k(θ θ ) l l cos θ = I θ Since the rods are assless I = l. For sall angles, θ = g l θ + k 4 (θ θ ) θ = g l θ k 4 (θ θ ) These equations can be decoupled by adding and subtracting (4a) and (4b). Let (4a) (4b) We have, Θ ± = θ ± θ Θ + = ( θ + θ ) = g l (θ + θ ) = g l Θ + Θ = ( θ θ ) = g l (θ θ ) k (θ θ ) [ g = l + k ] Θ (4c) (4d) Thus the noral odes are ω + = g l and ω = g l + k We will derives the sae equations using the Lagrangian foralis. Lagrangian for a siple pendulu, when the rod akes an angle θ with the vertical is Thus for two uncoupled pendulu, L = l θ gl( cos θ) l θ glθ L = l ( θ + θ ) gl(θ + θ )

11 c D. K. Ghosh, IIT Bobay Additional potential energy due to the spring is k ( l ) (θ θ ). Thus the Lagrangian for the coupled pendulu is L = l ( θ + θ ) gl(θ + θ ) 8 kl (θ θ ) (5) [We can derive (4a) and (4b) using the Euler-Lagrange equations corresponding to θ and θ.] The t and V atrices are ( ) l 0 t = 0 l (6) gl + kl kl V = 4 4 (7) kl gl + kl 4 4 The secular equation is det V ω t = 0 which gives gl + kl 4 ω l kl 4 = 0 kl gl + kl 4 4 ω l which has solutions ω = g l ( ω +) and g l + k ( ω ). Let us look at the coordinates when the frequency equals one of the noral ode frequencies. Let us rewrite (4a) and (4b) which can be written in atrix for as ω + g l + k 4 k 4 ω θ = g l θ + k 4 (θ θ ) ω θ = g l θ k 4 (θ θ ) k 4 ω + g l + k 4 ( θ θ ) = 0 (8a) (8b) For ω = ω +, this gives θ = θ, so that the ode is a syetric ode having the following solution for the ode coordinate. θ + = A + cos(ω + t + δ + ) (9) For ω = ω, the ode is antisyetric with θ = θ and the noral coordinate is θ = A cos(ω t + δ ) (30)

12 c D. K. Ghosh, IIT Bobay The constants A +, A, δ + and δ are fixed fro initial conditions. When the syste is in the syetric ode, the two pendulus ove in phase while when they are in the antisyetric ode they are out of phase by π. A B k Anti syetric ode Syetric Mode We can cobine the two odes to get solutions for θ and θ, θ = A + cos(ω +t + δ + ) + A cos(ω t + δ ) (3a) θ = A + cos(ω +t + δ + ) A cos(ω t + δ ) (3b) Suppose the two pendula are both pulled aside by θ 0 and released. We then have A = 0. We then have θ = θ = A + cos(ω +t + δ + ) which shows that the syste always reains in the syetric ode. Except when the syste is in a single ode, there is transfer of energy between the two odes at a frequency known as the beat frequency. When the two odes are present with the sae strength, for instances when only one of the pendula is drawn aside and released initially, we can take A = A + = A and δ + = δ = 0. In this case, θ = A and θ = 0 at t = 0. The expressionds for θ and θ are given by so that We can rewrite the aboves as θ = A cos(ω + t) + A cos(ω t) θ = A cos(ω + t) A cos(ω t) θ + = θ + θ = A cos(ω + t) θ = θ θ = A cos(ω t) θ = A cos( ω + ω θ = A sin( ω + ω t) cos( ω + + ω t) t) sin( ω + + ω t)

13 c D. K. Ghosh, IIT Bobay 3 Define beat frequency ω b and the average frequency ω as in ters of which ω b = ω + ω and ω = ω + + ω θ = A cos(ω b t) cos(ωt) θ = A sin(ω b t) sin(ωt) For weak couplings ω b ω. After a tie t = π, we have θ = 0 and θ = A sin ωt, ω b i.e. the first pendulu has oentarily stopped and all the energy has been transferred to the second. Exaple 3: Noral Modes of a Triatoic Molecule Molecules like CO have linear structures (O C O) with two equal asses at either ends and a central ass M. These are odelled through the three asses connected by two identical springs, as shown in the figure. Clearly, there are three noral odes one when all the three asses ove in unison, a second where the end asses ove in opposite directions while the central ass reains at rest and a third where the two end asses ove in opposite directions while the central ass oves towards the ass oving towards it. k k M The Lagrangian is written as L = (ẋ + ẋ 3) + Mẋ k [ (x x ) + (x 3 x ) ] The atrices T and V are as follows k k V = k k k T = 0 M 0 0 k k 0 0

14 c D. K. Ghosh, IIT Bobay 4 The secular equation is det(v ω T ) = 0, which gives k ω k 0 k k ω M k 0 k k ω = 0 which has the solutions ω = ω, ω, ω 3, where ω = 0 ω = k ω 3 = k + k M. For ω = 0, the equation (8) V A ω T A = 0 can be written as k k 0 A k k k A = 0 0 k k A 3 which gives A = A = A 3. Taking each of the to be equal to (in soe unit) we get A = N where N is to be fixed by noralization. Noralization condition gives N 0 0 ( ) 0 M 0 = 0 0 i.e. M + = N. Thus the noralized vector is (3) + M As expected, the displaceent of each ass is the sae and the asses ove in unision.. Consider ω = k. For this case, we have 0 k 0 A k k k A = 0 0 k 0 A 3

15 c D. K. Ghosh, IIT Bobay 5 which gives A = 0 and A = A 3. Using the orthogonality with the ode ω, we take this ode to be 0 N Noralization condition fixes N = and the displaceent vector is 0 which shows the central ass to be at rest while the end asses displace by the sae aount in the opposite direction. 3. For ω = k + k, the equation is M k M k 0 k km k 0 k k M A A = 0 A 3 which gives A = A 3 and A = M A. The noralization can be done as in the previous cases. The displaceent vector is given by M /M M + 4 Exaple 4 Three asses each, initially located equidistant fro one another on a horizontal circle of radius R. They are connected in pairs by three springs of force constant k each and of unstretched length πr/3. The spring threads the circuloar tract so that the ass is constrained to ove on the circle. Find the noral odes.

16 c D. K. Ghosh, IIT Bobay 6 Let θ, θ and θ 3 be the angular displaceents of the three asses fro their equilibriu positions. The Lagrangian is given by L = T V, where, T = R ( θ + θ + θ 3) and V = kr [ (θ θ ) + (θ θ 3 ) + (θ 3 θ ) ] The V and T atrices are V = kr 0 0 T = R One can write the secular equation as before and easily show that the noral odes are 3k given by ω = 0 and ω = (doubly degenerate). For ω = 0, the atrix equation is k k k A R k k k A = 0 k k k A 3 which gives A = A = A 3. The noralized noral coordinate is given by 3R For ω = 3k/, we have

17 c D. K. Ghosh, IIT Bobay 7 k k k A k k k A = 0 k k k A 3 which gives A + A + A 3 = 0. The eigenvectors cannot be uniquely fixed. However, if we use the orthogonality with the zero ode, we can have the following (non-unique) choices: 6R and 0 R 3 Daped and Forced Oscillations 3. Rayleigh Disspipation Function We have seen that for systes where forces are derivable fro a potential, Euler-Lagrange equation is given by ( ) d L L = 0 dt q i q i This was seen to be true even when a velocity dependent potential exists, as was in the case of Lorentz force. If on the otrher hand, there are soe non-potential forces Q i, we get, using the sae procedure as was used to derive the Euler-Lagrange equation fro d Alebert s principle, d dt ( ) L q i L q i = Q i where Q i = j F j r j q i. Exaple of a non-potential force is frictional force which, in ost cases of interest, is proportional to the velocity of the particle, and, is given by F = k v. More generally, the for of the frictional force is F = k x v x î k y v y ĵ k z v zˆk where k x, k y and k z are all positive constants. One can include such non-potential forces in the Lagrangian foralis by defining Rayleigh Dissipation Function R, R = (k x vx + k y vy + k z v z) (33) i

18 c D. K. Ghosh, IIT Bobay 8 The force F i is then given by i.e. F ix = R v ix F i = vi R where the gradient is taken with respect to velocity field and the index i is the particle index. Work done by the syste in overcoing resistive force is so that dw = F d r = F vdt = Rdt dw dt The generalized force can be written as Q i = j = j = R (34) F j r j q i F j ṙ j q i = j vj R ṙ j q i = R q i (35) where in the second step of the above equation we have used the dot cancellation. Thus we can use this ter in the Euler Lagrange equation by incorporating an additional ter ( ) d L L + R = Q i (36) dt q i q i q i where Q i are non-potential forces which are not derivable either fro a potential or fro a dissipation function. 3. Daped Oscillator Consider a linear oscillator with dissipative force proportional to the velocity. Taking Rayleigh function to be R = α q and using the Lagrangian L = q kq, we get fro the odified Euler Lagrange equation (36) q + α q + kq = 0

19 c D. K. Ghosh, IIT Bobay 9 Defining ω 0 = given by k and γ = α, the get the equation for the daped oscillator to be q + γ q + ω 0 = 0 (37) We seek solution of the for q = q 0 e λt which gives dq dt = q 0λe λt and d q dt = q 0λ e λt. Substituting these in (37), we get λ + γλ + ω 0 = 0 (38) which has the solution Thus the solution of (37) is λ = γ ± γ ω 0 ] γ q = e [Ae γt ω0 γ t + Be ω0 t (39) Let us consider soe special cases.. Underdaped Motion: If γ < ω 0, we can write (39) in ters of trigonoetric function, q = q 0 e γt cos(ωt + δ) (40) where the frequency of oscillation Ω = ω 0 γ is of a lower value than the undaped frequency and the aplitude of oscillation decreases expopnentially with tie. x= e - t cos( t) e - t t

20 c D. K. Ghosh, IIT Bobay 0 The energy of the oscillator decreases in the presence of dissipative forces. Using E = L q L we get q de dt = d ( ) L dt q q dl dt = d ( ) L q + L ( ) L dt q q q L q + q q q = R q q = R where we have used (36) and (34) in the last step. Since R > 0, there is always a dissipation of energy.. Overdaped Motion : If γ > ω 0, the solution is of the for q(t) = Ae γ t + Be γ t where γ, = γ γ ω 0 are both positive. and the oscillation dies off fast. 3. Critical Daping : Critical daping occurs when the roots of the characterisatic equation becoes equal, i.e., when ω 0 = γ = λ. In such a case the two constants A and B in (39) becoes a single constant C. However, as we have a second order differential equation, we ust have two constants. Let us try a solution of the for q(t) = u(t)e λt. We then have dq dt = du dt eλt + uλe λt ( ) d q d dt = u eλt dt + λdu dt + λ Substituting this in the differential equation q + γ q + ω 0q = 0, we get d u + (λ + γ)du dt dt + (λ + γλ + ω0)u = 0 However, λ being a root of the characteristic equation, we have λ +γλ+ω 0 = 0. In addition, we are looking for the situation where λ = γ. Thus we get d u/dt = 0, which has the solution u(t) = A + Bt. The solution for q is q = (A + Bt)e γt The oscillation is said to be critically daped. The following figure shows how the equilibriu is reached for the three cases discussed above. It can be seen that for the case of critical daping the approach to equilibriu is the fastest.

21 c D. K. Ghosh, IIT Bobay Underdaped Critically Daped Overdaped x A C B O t 3.3 Driven Oscillations If, in addition to the haronic potential, there is a tie dependent potential V (q, t), we can, to first order express it as V (q, t) V (q, t) = V (0, t) + q q q=0 = qf (t) where we have chosen V (0, t) = 0 as the reference to the potential. The Lagrangian in such a case is given by The Euler Lagrange equation gives L = q kq + qf (t) q + kq = F (t) Let the driving force be sinusoidal with a frequency ω, i.e. let F = F 0 cos ωt. In addition, let us introduce a velocity dependent daping. The equation of otion is then written as q + γ q + ω 0q = F 0 eiωt (4) where ω 0 = k/ where we have taken an exponential for instead of the cosine as the driving ter and at the end of the calculation, we would take the real part. The solution of this proble has two parts, a solution of the hoogeneous equation q +γ q +ω 0q = 0 and a particular solution. If γ > 0, the hoogeneous equation gives transients, the oscillations die down in one of the three ways discussed above. We therefore look at the particular

22 c D. K. Ghosh, IIT Bobay solution. The equation is solved by an ansatz q = Ae iωt which gives q = iωq and q = ω q. Substituting these in the differential equation (4), we get We then have so that We then have ( ω iωγ + ω 0)Ae iωt = F 0 e iωt A = F 0 ω0 ω iωγ q = R( F 0 ω0 ω iωγ e iωt ) A = F 0 / (ω 0 ω ) + 4ω γ e iϕ where the phase ϕ is given by ϕ = tan ( ) ωγ ω ω 0 Thus we can write q as Using the expression for tan ϕ, we get cos ϕ = sin ϕ = q = A cos(ωt + ϕ) ω ω 0 (ω 0 ω ) + 4ω γ ωγ (ω 0 ω ) + 4ω γ The otion is oscillatory with the frequency sae as that of the driving frequency. As the strength of daping decreases, the aplitude A gets peaked when the driving frequency becoes equal to the natural frequency ω 0. For values of driving frequencies well below that of natural frequency, the response of the oscillator is in phase with the driving frequency (sin ϕ 0 = ϕ 0). Near resonant frequency, sin ϕ, i.e. ϕ = π/. For ω ω 0, sin ϕ 0 but cos ϕ, i.e. it becoes in anti-phase. The energy loss of a weakly daped oscillator is characterized by Q-factor which is defined as π ties the energy stored in the oscillator divided by the energy loss in one period. It can be shown that near resonance Q = ω 0. Thus a sharp resonance results in γ a high Q value.