11 Free vibrations: one degree of freedom

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1 11 Free vibrations: one deree of freedom 11.1 A uniform riid disk of radius r and mass m rolls without slippin inside a circular track of radius R, as shown in the fiure. The centroidal moment of inertia I C of the disk is iven by I = 1 mr. R O r 1. What is the correct equation of motion for an arbitrary lare anle θ? A B (a) 3 (R r) θ + sin θ = 0 (b) 3 (R + r) θ + cos θ = 0 A (c) 3 (R r) θ sin θ = 0 (d) 1 (R r) θ + sin θ = 0 R (e) 5 (R r) θ sin θ = 0. What is the natural frequency ω n of small oscillations around the equilibrium position θ = 0? (a) ω n = (b) ω n = (c) ω n = (d) ω n = R r 3(R r) 3 3(R r) (e) ω n = 3 3. Given θ(0) = π and θ(0) = 1, what is the value of θ at t = 3? (hint: use the 10 linearized equation of motion) 3(R r) (a) θ(3) = sin 6 + π cos 6 (R r) 10 (R r) (b) θ(3) = π 10 cos 6 (R r) (c) θ(3) = (d) θ(3) = (e) θ(3) = 3(R r) (R r) sin sin 3(R r) sin 6 (R r) (R r) + π 10 cos 3 (R r) 6 (R r) + π 10 cos 6 (R r) 11-1

2 11 Free vibrations: one deree of freedom 11- Solution question 1 Consider a normal and tanential (n-t) coordinate system as shown. On this diaram, the applied forces are shown as well. In order to derive the equation of motion, we could apply anular momentum principle about the center of mass C as Ḣ C + v C P = M ext C, Ḣ C + 0 = M ext C. (11.1) Since the problem is planar, the anular momentum principle has only one component, which can be written as I C α = F f r. (11.) where α is the anular acceleration of the body relative to the inertial frame. The anular acceleration of the body written in terms of the θ and φ is α = φ θ. (11.3) Since the disk rolls without slippin, it holds that r φ = R θ. (11.4) By combinin (11.3) and (11.4) and substitutin into (11.) we obtain mr (R r θ θ) = F f r F f = 1 m(r r) θ. (11.5) We can now write the linear momentum principle in the tanential direction as P t = F t m v C = F t (11.6) where F t denotes the resultant force in the tanential direction. The velocity v C is simply iven by v C = (R r) θ, (11.7) and therefore m(r r) θ = F f m sin θ. (11.8)

3 11 Free vibrations: one deree of freedom 11-3 Then, by substitutin (11.5) into (11.8) we obtain the desired equation of motion 3 m(r r) θ + m sin θ = 0 (11.9) which can be rewritten in the form θ + sin θ = 0. (11.10) 3(R r) Alternative solution: note that the problem could be easily solved also by takin the anular momentum principle w.r.t the contact point. In this way, the effect of the forces at the contact point is nelected. By inspectin all the possible answers for question 1, we can easily see that that option (d) has inconsistent units. Option (c) and (e) feature neative stiffness coefficients, which contradicts the physics of the problem. Option (b) has a cosine term related to the stiffness term, ivin a restorin force for θ = 0, which is not possible. Solution question If we Taylor expand the equation of motion up to linear order in terms of θ and θ around the equilibrium point ( θ, θ) = (0, 0), we obtain the linearized equation of motion f( θ, θ) = θ + 3(R r) sin θ = f(0, 0) + θf(0, 0) θ + θ f(0, 0)θ + O( θ 3 ) (11.11) = θ + 3(R r) θ + O( θ 3 ) = 0. (11.1) Therefore, the equation of motion overnin small oscillations is iven by θ + ω nθ = 0. where the natural frequency ω n is iven by ω n = 3(R r). (11.13) (11.14) Solution question 3 We could immediately exclude options (c), (d) and (e), which have incorrect units. One has to solve the problem to select amon (a) and (b). Lookin at the linearized equation of motion, we know that the solution is of the form: θ(t) = a sin(ω n t) + b cos(ω n t), (11.15) θ(t) = aω n cos(ω n t) bω n sin(ω n t). (11.16) By insertin the initial conditions θ(0) and θ(0), we obtain: a = 1 ω n and b = π 10. (11.17)

4 11 Free vibrations: one deree of freedom 11-4 The motion is therefore iven by: θ(t) = 1 ω n sin(ω n t) + π 10 cos(ω nt). (11.18) And therefore, at t = 3, 3(R r) 6 θ(3) = sin (R r) + π 10 cos 6 (R r). (11.19)

5 11 Free vibrations: one deree of freedom A block of mass m is suspended by two sprins of stiffness k 1 and k and zero unstretched lenth l 1 and l. (a) (b) 1. What is the equivalent sprin stiffness k eq of a sinle sprin that yields the same period of oscillation T of the system shown in Fiure (a), where the two sprins are arraned in parallel? (a) k eq = k 1 + k (b) k eq = k 1+k k 1 k (c) k eq = k 1k k 1 +k (d) k eq = k 1 + k (e) k eq = 1 (k 1 + k ) k 1 k m Parallel k 1 k m Series. What is the equivalent sprin stiffness k eq of a sinle sprin that yields the same period of oscillation T of the system shown in Fiure (b), where the two sprins are arraned in series? (a) k eq = k 1k k 1 +k (b) k eq = k 1 k k1 +k (c) k eq = k 1 k (d) k eq = k 1 k + k k (e) k eq = k 1 + k Solution question 1 The free-body diaraam is sketched here. k 1 k F 1 F F 1 F x m m

6 11 Free vibrations: one deree of freedom 11-6 The forces of the two sprins are iven by F 1 = k 1 x (11.0) F = k x (11.1) (11.) The linear momentum principle for the mass m is simply iven mẍ = m F 1 F (11.3) = m (k 1 + k )x. (11.4) Introducin the equivalent stiffness k eq = k 1 + k, (11.5) we have mẍ + k eq x = m. (11.6) Introducin y := x m k eq ( ÿ = ẍ), (11.7) we can shift the oriin to the equilibrium position 1. We are now back to the familiar equation of motion for undamped, unforced vibrations. ÿ + k eq m y = 0. (11.8) With ω0 := keq, the period of oscillation T becomes m T = π ω 0 = π m k 1 + k. (11.9) 1 You can picture this by thinkin of an equivalent sprin that has stiffness k eq. Once we attach the mass m to its end, due to the ravitational force m, the sprin will extend to the new lenth m k eq. This is the term that we subtracted in (11.7).

7 11 Free vibrations: one deree of freedom 11-7 Analysis of the multiple choice answers the correct units. Clearly, options (b), (c) and (d) do not have Solution question As done previously, the free-body diaram in shown. k 1 F F x 1 k F F m m x Since the spins are arraned in series, the force F must be equal, The elonation x 1 of the first sprin should be equal to F, and F = k 1 (x 1 l 1 ) (11.30) F = k (x l ). (11.31) The elonation of the equivalent sprin, x eq = x 1 +x, multiplied by its stiffness k eq should also be equal to F : F = k eq (x 1 + x ) = k eq x eq = k eq (x 1 + x ). (11.3) We have 3 equations to solve for k eq. Equalin (11.30) and (11.31), we et x = k 1 k x 1, (11.33)

8 11 Free vibrations: one deree of freedom 11-8 and pluin into (11.3), F = k eq (x 1 + k 1 k x 1 ). (11.34) Comparin this to (11.30), we need k eq (x 1 + k 1 k x 1 ) = k 1 x 1. (11.35) The result for k eq is k eq = k k 1 k = k 1k k 1 + k. (11.36) The period of oscillation T is therefore oin to be m(k 1 + k ) T = π. (11.37) k 1 k

9 11 Free vibrations: one deree of freedom Consider the followin pendulum consistin of a homoeneous thin rod with center of mass C, and of a point mass m riidly attached at the tip B of the rod. The rod has mass M and lenth l. It is suspended at point O, and may rotate without friction about it. We denote the anular displacement with respect to the vertical axis by ϕ. A massless sprin of stiffness k connects points B and A, as shown. The x and y components of the distance between point A and O are iven by a and b = l/, respectively. Let the restorin force of the sprin be zero when r BA = 0. Gravity is actin downwards, as shown. the centroidal moment of inertia of the rod is iven by I C = 1 1 Ml. a O e z M,l ' C e x m B e y k A b 1. What is the correct equation of motion of the system? (a) ϕ = ka cos ϕ + ( 3 3 ml 8 k + 3 ) m l cos ϕ (b) ϕ = ka cos ϕ ( 1 3 ml 4 k + 3 ) m l sin ϕ (c) ϕ = 3 k cos ϕ ( 3 4 mal 8 k + 3 ) m l sin ϕ (d) ϕ = 3 ka sin ϕ + ( 3 4 ml 8 k + 3 ) m l sin ϕ (e) ϕ = 3 ka cos ϕ ( 3 4 ml 8 k + 3 ) m l sin ϕ. What is the value of a that results in an equilibrium at ϕ 0 = π? Assume M = m. 4 (a) a = l + 5 m kl (b) a = l + 3 m k (c) a = l 3 m k (d) a = 0 (e) a = l 3. Introduce the new variable x(t) = ϕ(t) ϕ 0. What is the correct linearized equation of motion? ( ) (a) ẍ + 3 k m x = 0 l ( ) (b) ẍ k m x = 0 l ( ) (c) ẍ + 3 k x = 0 ( ) (d) ẍ + 3 k 4 m + 3 x = 0 l ( (e) ẍ 3 k + ) 3 4 m l x = 0

10 11 Free vibrations: one deree of freedom Solution question 1 We will need the followin vectors: r OC = l cos ϕe x + l sin ϕe y r CB, (11.38) r OB = l cos ϕe x + l sin ϕe y, (11.39) r OA = be x + ae y, r BA = r OA r OB. (11.40) (11.41) We could separate the rod from the point mass and hihliht the reaction forces R = R x e x + R y e y and T = T x e x + T y e y : e z e x e y O ' M R C T B T m We can now write the linear momentum principle for mass m as kr BA m r OB = T + m + kr BA. (11.4) Differentiatin (11.39) twice, usin the other expressions from Step I to write out kr BA, and then solvin for T, we et T x = kb + m kl cos ϕ + ml cos(ϕ) ϕ + ml sin(ϕ) ϕ, (11.43) T y = ka kl sin ϕ + ml sin(ϕ) ϕ ml cos(ϕ) ϕ. (11.44) The linear momentum principle for the rod writes M r OC = R + M + T, (11.45) which yields the followin components of the reeaction force R R x = 1 [ M + Tx + Ml cos(ϕ) ϕ + Ml sin(ϕ) ϕ ], (11.46) R y = 1 [ Ty Ml sin(ϕ) ϕ + Ml cos(ϕ) ϕ ]. (11.47) The anular momentum principle for the rod, with respect to its center of mass C, ives Ḣ C = M ext C. (11.48) By introducin the centroidal moment of inertia I C, the anular momentum H C can be written as H C = 1 1 Ml ϕ e z. (11.49)

11 11 Free vibrations: one deree of freedom The external torque M ext C is iven by M ext C = r CO R + r CB T (11.50) = l [( R y + T y ) cos ϕ + (R x T x ) sin ϕ] e z. (11.51) By usin all the derived relations, the e z -component of the anular momentum principle (11.48) is obtained as 1 1 Ml ϕ = 1 l ( 4ka cos ϕ + (kb + m + M) sin ϕ + l(4m + M) ϕ). (11.5) 4 Solvin the last equation for ϕ yields ϕ = 3 ka cos ϕ (kb + (m + M)) sin ϕ. (11.53) (3m + M)l Usin b = l/ and M = m, this simplifies to ϕ = 3 ka 4 ml cos ϕ 3 ( ) k 8 m + 3 sin ϕ. (11.54) l Alternative solution: The problem can be easily solved also by considerin the mass and the bar as a unique body, and then by takin the anular momentum principle w.r.t. the hine O. In this case, only the ravity force and the sprin force enerate a torque. note that, in this case, one has to compute the moment of inertia of the assembly w.r.t. O. This is obtained as I O = 1 1 Ml Ml + ml = ( 1 3 M + m)l. Analysis of the multiple choice answers Option (c) has inconsistent units. All the other options are in principle plausible, as it is a priori not possible to infer about the sins of the various terms by inspection. One has to solve the problem fully to provide the correct answer. Solution question At equilibrium, we have ϕ = 0 and ϕ = 0. All that needs to be done is to insert the desired equilibrium value ϕ = ϕ 0 into Eq. (11.54) and set the left hand side to zero: 0 = 3 ka 4 ml cos ϕ 0 1 ( 3 k ) 8 m + 9 sin ϕ 0. (11.55) l With ϕ 0 = π 4, and thus sin π 4 = cos π 4 = 1, we et a = l + 3 m k. (11.56) With this choice of parameter, the ODE (11.54) becomes ϕ + 3 [ ] k 8 m + 3 (sin ϕ cos ϕ) = 0. (11.57) l

12 11 Free vibrations: one deree of freedom 11-1 Analysis of the multiple choice answers Option (a) has inconsistent units. Also, by inspectin the system, it is clear that a = 0 (option (d)) is not the correct answer. Now we linearize about ϕ 0 = π 4 by introducin the new variable x = ϕ ϕ 0. The expressions we need to plu into (11.57) are therefore ϕ = ϕ 0 + x = π 4 + x ϕ = ẋ ϕ = ẍ. (11.58) We also need the followin Taylor expansions about the equilibrium point: ( π ) sin 4 + x = x + O(x ), (11.59) ( π ) cos 4 + x = 1 1 x + O(x ). (11.60) With these equations, Eq. (11.57) can be expanded as ẍ + 3 [ ] ( k 1 8 m x ) x + O(x ) l = 0. (11.61) The linear ODE is therefore ẍ + 3 ( k 4 m + 3 ) x = 0, (11.6) l which overns free undamped vibrations of natural frequency ( 3 k ω 0 = 4 m + 3 ). (11.63) l Analysis of the multiple choice answers Option (c) and (d) feature inconsistent units.

13 11 Free vibrations: one deree of freedom The four diarams A, B, C, D shown below are time responses of damped vibrations. The vertical axis measures the displacement y of the oscillator, the horizontal axis shows the time t. Denote with M the mass of the oscillator, with δ the characteristic dampin, and with k the sprin constant. 1. Which plot corresponds to free linear vibrations? (a) A (b) B (c) C (f) D. Let us introduce the loarithmic decrement Λ = lo y(t), with T bein the oscillation period, and lo the natural loarithm. What is the correct formula for the y(t+t ) Lehr s dampin D? (a) D = (b) D = (c) D = (d) D = Λ Λ+4π Λ Λ +4π Λ Λ +4π Λ Λ π

14 11 Free vibrations: one deree of freedom (e) D = Λ Λ+4π 3. Consider the response below. What is an approximate value of the Lehr s dampin D? (a) D 0.03 (b) D 0.3 (c) D 0.9 (d) D 10 (e) D Solution question 1 We see oscillations with decayin amplitudes. We are therefore lookin for free, underdamped vibrations. in this case, the eneral solution can then be written as y(t) = e δt (C 1 cos ω δ t + C sin ω δ t), (11.64) where ω δ = ω 0 δ, (11.65) with ω 0 = k/m bein the natural frequency of the undamped system. As discussed in the lecture, eq. (11.64) may also be written as y(t) = C 0 e δt sin(ω δ t + ϕ 0 ), (11.66) where ϕ 0 is a constant. Clearly, we must have an oscillation with constant period T δ = π ω δ and exponentially decayin amplitude C 0 e δt. By inspectin the iven answer options, we observe that Curve A decays linearly, and not as C 0 e δt. Curve C and D shows oscillations with variable period. Curve B is the only choice consistent with (11.66).

15 11 Free vibrations: one deree of freedom Solution question We list the followin definitions and identities: δ = ω 0 D, T δ = π, ω δ ω δ = ω 0 δ = ω 0 ω 0 (11.67) (11.68) ( ) δ 1 = ω 0 1 D. (11.69) The expression for the loarithmic decrement Λ is iven by: Λ = y(t) lo y(t + T δ ) = e δt sin(ω δ t + ϕ 0 ) lo e δ(t+t δ) sin(ω δ (t + T δ ) + ϕ 0 ) (11.70) (11.71) = lo e δt δ (11.7) = δt δ (11.67) (11.69) = πd 1 D By solvin the last equation for (positive) D, we et (11.73) (11.74) D = Λ Λ + 4π. (11.75) (c) In order to obtain Λ, we need to measure y(t) and y(t + T δ ) from the raph of y: On the printout, y(t) cm, y(t + T δ ) 1.65 cm, and therefore Λ = lo y(t) y(t + T δ ) Usin Eq. (11.75), we et (11.76) D (11.77)

16 11 Free vibrations: one deree of freedom Analysis of the multiple choice questions Options (c) and (e) are neative: this is in contrast with the decayin behavior depicted. Options (a), (b) and (d) are orders of manitudes apart to cancel the effect of small measurin errors.

17 11 Free vibrations: one deree of freedom Two uniform cylinders of mass m 1 and m and radius R 1 and R are welded toether. This composite object rotates without friction about a fixed point O. Two inextensible massless strins are wrapped without slippin around the larer cylinder. The two ends of the strins are connected to the round throuh a sprin of constant k and a damper of dampin coefficient b. The smaller cylinder is connected to a block of mass m 0 via an inextensible massless strin wrapped without slippin around the smaller cylinder. The block is constrained to move vertically; assume the strins are also able to sustain compressive forces. (Hint: the centroidal moment of inertia of a disk of mass m and radius r is iven by I = 1 mr.) b R m 1 R 1 m m 0 k 1. What is the equation of motion of the system in terms of the rotation anle θ? (a) (b) [ (m0 ) ] + m 1 R R 1 + m θ + br θ + kr θ = m 0 R 1 [ ] m 0 R1 R + m θ + br θ + kr θ = m 0 R 1 (c) [( ) ] m 0 + m 1 R R 1 + m θ + br θ + kr θ = m 0 R 1 [ (m0 ) ] (d) + m 1 R R 1 m θ br θ + kr θ = m 0 R 1 (e) m R θ + br θ + kr θ = m 0 R 1. Which are the correct values of the natural frequency and characteristic dampin of the system? (a) ω 0 = ( m0 + m 1 kr ), δ = R R 1 + m br [ (m0 ) + m 1 R R 1 + m ] (b) ω 0 = kr ( ), δ = m0 + m 1 R R 1 + m br [ (m0 ) ] + m 1 R R 1 + m kr br (c) ω 0 =, δ = (m 0 + m 1 + m ) R1 (m 0 + m 1 + m ) R1 (d) ω k 0 = (m 0 + m 1 + m ), δ = b (m 0 + m 1 + m ) (e) ω 0 = ( m0 + m 1 kr ) R 1 + m R, δ = 0 Solution question 1 The reference system and the free-body diaram are shown below. By evaluatin the

18 11 Free vibrations: one deree of freedom e y br B e z R R 1 m 1 O N m m 0 m 0 kr e x anular momentum of the entire system with respect to point O it is possible to avoid havin to find the tension in the cord connected to the hanin mass. The cord exerts only an internal force in the system and does not need to be evaluated. The anular momentum of the system is H O = H rotor + r OB (P m0 ). (11.78) The anular momentum of the rotor is H rotor = I O θez, (11.79) where the moment of inertia I O = m 1 R1/ + m R/ therefore [ ] R1 H rotor = m 1 + m R θe z. (11.80) The remainin cross product can be calculated as r OB (P m0 ) = m 0 R 1 θe z. (11.81) Usin (11.78), (11.80) and (11.81), we et [ ( H O = m 0 + m ) ] 1 R1 R + m θe z. (11.8) We can now write the anular momentum principle as Ḣ O + v O P = M ext O. (11.83) Since v O = 0, we et d dt H O = M ext O d [[ ( m 0 + m ) ] ] 1 R1 R + m θe dt z = M ext O, (11.84) [ ( m 0 + m ) ] 1 R1 R + m θe z = (m 0 R 1 br θ krθ)e z, (11.85) and, rearranin, the followin equation of motion is obtained [ ( m 0 + m ) ] 1 R1 R + m θ + br θ + krθ = m 0 R 1. (11.86)

19 11 Free vibrations: one deree of freedom Solution question By definition, the natural frequency of the system is iven by ω 0 = ( m0 + m 1 kr while the characteristic dampin is ) R 1 + m R (11.87) δ = ( ( m 0 + m 1 br ) R 1 + m R ). (11.88)

20 11 Free vibrations: one deree of freedom Homework problem A wheel of mass m is suspended by three cords of equal lenth L. When a small anular displacement of θ about the z-axis is iven to the wheel, and then released, small oscillations of period τ are observed. The center of mass of the wheel is on the z-axis. Its centroidal moment of inertia about the z-axis is iven by I z = mρ z, where ρ z is the so-called radius of yration. What is the period τ of small oscillations? /3 A L C L z /3 B L (a) τ = π ρz r (b) τ = π ρz r (c) τ = π ρ z L L L r r /3 (d) τ = π ρz (e) τ = π ρz r 3L Remarks: Assume that the cords remain taut, and that L and r are of the same order of manitude. Treat the cords as straiht lines, enclosin a small anle with the z-axis when the wheel is rotated by a small anle θ. Solution This system can be fully described by a sinle deree of freedom.we choose θ as the eneralized coordinate. The fiure below shows the relevant eometric quantities, with φ bein the anle between each cord and the vertical axis. L z r clearly, this eometric layout is the same for the other two ropes. Since we are interested in derivin a model that holds for small anles θ and φ, we can write L tan φ r tan θ. (11.89) The Taylor expansion of tan x about x = 0 is tan x = x + O(x 3 ), (11.90)

21 11 Free vibrations: one deree of freedom 11-1 which is a ood approximation (only) for x 1. We use this to approximate Eq. (11.89) as φ = r L θ. (11.91) The free-body diaram is here shown, where the tension of each strin is iven by T = T x e x + T y e y + T z e z with T = T. C A T m B T T First, we apply the linear momentum principle for the wheel, to determine the strin tension T. By takin the z-component, we obtain m z = 3T cos φ m. (11.9) Since we are dealin with small rotations, we can consistently assume that z const., i.e. z = 0. Therefore, we et that T = m 3 cos φ. (11.93) Subsequently, we apply the anular momentum principle with respect to the center of mass of the wheel The z-component ives I z θ = 3M str O,z, (11.94) where MO,z str is the z-component of the torque that each cord exerts on the wheel, with respect to the oriin O at x = y = z = 0. Considerin the free-body diaram, we obtain M str O,z = rt sin φ. (11.95) E.. use the vector indicated on the left of the free-body diaram: M str O,z = (re x T ) e z = rt y = r ( T cos(90 φ))

22 11 Free vibrations: one deree of freedom 11- Toether with I z = mρ z, the anular momentum principle yields mρ z θ = 3rT sin φ. (11.96) We eliminate T usin (11.93): mρ z θ = rm sin φ cos φ. (11.97) Now we replace φ by θ usin (11.91), to obtain a non-linear equation of motion for θ: θ + r ( r ) tan ρ z L θ = 0. (11.98) Within the small-anle approximation, we may use tan x = x + O(x 3 ) aain and obtain the linear ODE θ + r ρ zl θ = 0. (11.99) This describes free vibrations with natural circular frequency ω 0 = r ρ z L, (11.100) and period of oscillation ( ) τ = π ρ z L = π. (11.101) ω 0 r Analysis of multiple choices Option (a) is not unit-consistent. Option (d) does not show any dependency on the lenth of the strins L, which is in contradiction with the physics of the problem.

Chapter K. Oscillatory Motion. Blinn College - Physics Terry Honan. Interactive Figure

Chapter K. Oscillatory Motion. Blinn College - Physics Terry Honan. Interactive Figure K. - Simple Harmonic Motion Chapter K Oscillatory Motion Blinn Collee - Physics 2425 - Terry Honan The Mass-Sprin System Interactive Fiure Consider a mass slidin without friction on a horizontal surface.