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MODULE-II --- SINGLE DOF FREE S VTU-NPTEL-NMEICT Project Progress Report The Project on Development of Remaining Three Quadrants to NPTEL Phase-I under grant in aid NMEICT, MHRD, New Delhi SME Name : Course Name: Type of the Course Module Subject Matter Expert Details Dr.MOHAMED HANEEF PRINCIPAL, VTU SENATE MEMBER Vibration engineering web II DEPARTMENT OF MECHANICAL, GHOUSIA COLLEGE OF, RAMANARAM -562159 Page 1 of 40

MODULE-II --- SINGLE DOF FREE S Sl. No. 1. CONTENTS DISCRETION Lecture Notes (Single-DOF free Vibration). 2. Quadrant -2 a. Animations. b. Videos. c. Illustrations. 3. a. Wikis. Quadrant -3 b. Open Contents 4. Quadrant -4 a. Problems. b. Assignments c. Self Assigned Q & A. Page 2 of 40

MODULE-II --- SINGLE DOF FREE S MODULE-II SINGLE DEGREE OF FREEDOM, FREE 1.LECTURE NOTES SINGLE DEGREE OF FREE S Un-damped Free Vibrations of Single Degree of Freedom Systems When the elastic system vibrate because of inherent forces and no external forces is included, it is called free vibration. If during vibrations there is no loss of energy due to friction or resistance it is known as undamped vibration, free vibrations which occur in absence of external force are easy to analyse for single degree of freedom systems. A vibratory system having mass and elasticity with single degree of freedom in the simplest case to analyse. The determination of natural frequency to avoid resonance is essential in machine elements. Module-II is define the following i. Vibration model, Equation of motion-natural frequency. ii. Energy method, Rayleigh method. iii. Principle of virtual work, damping models. 2.1. MODEL, EQUATION OF MOTION-NATURAL FREQUENCY 2.1. i. Spring mass system displaced vertically. a) MODEL: Consider a spring mass system as shown in fig constrained to move in a collinear manner along with the axis of spring. The spring having stiffness is fixed at one end and carries a mass m at its free end. The body is displaced from its equilibrium position vertically downwards. This equilibrium position is called static equilibrium. The free body diagram of the system is shown in fig: 2.1 (a). Page 3 of 40

MODULE-II --- SINGLE DOF FREE S Fig (a): Spring mass system Fig (b): Free body diagram of spring mass system Figure: 2.1 (a) b) EQUATION OF MOTION: In equilibrium position, the gravitational pull mg is balanced by a spring force such that mg = kδ [From fig: 2.1 (b)] Where δ is the static deflection of the spring. Since the mass is displaced from its equilibrium position by a distance x and then released, so after time t as per Newton s II law. Net Force = mass acceleration mg k (δ + x) = m x [equation of motion ] m x = mg kδ kx (: mg = k δ) m x = kx m x + kx = 0 x + ( k /m) x = 0 (1) Eq. (1) is differential Equation of motion for free vibration c) NATURAL FREQUENCY Equation (1) is a differential equation. The solution of which is x = A sin K/m t + B cos K/m t. Where A & B are constant which can be found from initial conditions. The circular frequency ω n = k/m Page 4 of 40

MODULE-II --- SINGLE DOF FREE S The natural frequency of vibration ƒ n = ω n 2π R ƒ n = 1 2 π k /m = 1 2 π k (kδ/g) = 1 2 π g δ Where δ = static deflection 2.1. ii. Spring mass system displaced horizontally. a) MODEL: In the system shown in fig: 2.2, a body of mass m is free to move on a fixed horizontal surface. The mass is supported on frictionless rollers. The spring of stiffness is attached to a fixed frame at one side and to mass at other side. Figure: 2.1 (b) b) EQUATION OF MOTION: As per Newtons II law Mass x acceleration = resultant force on mass m x = kx m x + kx = 0 [equation of motion ] x + (k /m) x = 0 (1) Page 5 of 40

MODULE-II --- SINGLE DOF FREE S Eq. (1) is differential Equation of motion for free vibration c) NATURAL FREQUENCY: Equation (1) is a differential equation. The solution of which is x = A sin K/m t + B cos K/m t. Where A & B are constant which can be found from initial conditions. The circular frequency ω n = k/m The natural frequency of vibration, ƒ n = ω n 2π R 1 ƒ n = 2 π k /m = 1 2 π Where δ = static deflection k (kδ/g) = 1 2 π g δ 2.1. iii. System having a rotor of mass (Torsional Vibrations) a) MODEL: Consider a system having a rotor of mass moment of inertia I connected to a shaft at its end of torsional stiffness K t, let the rotor be twisted by an angle θ as shown in fig: 2.2(c). The body is rotated through an angle θ and released, the torsional vibration will result, the mass moment of inertia of the shaft about the axis of rotation is usually negligible compressed to I. Fig: 2.2 (c) The free body diagram of general angular displacement is shown Fig: 2.2 (c) b) EQUATION OF MOTION: As per Newtons II law Mass x acceleration = resultant force on mass The equation of motion is. I θ = K t θ Page 6 of 40

MODULE-II --- SINGLE DOF FREE S I θ + K t θ = 0 c) NATURAL FREQUENCY: θ + K t θ = 0 I Equation (1) is a differential equation. The solution of which is x = A sin K t I initial conditions. The circular frequency ω n = K t I The natural frequency of vibration, Where K t = GJ L t + B cos K t t. Where A & B are constant which can be found from I ; J = π d3 ω n = K t I SPRINGS IN ARBITRARY DIRECTION 32 ƒ n = ω n 2π & f n = 1 2π K t I Fig shows a spring K making an angle α with the direction of motion of the mass m. Page 7 of 40

MODULE-II --- SINGLE DOF FREE S If the mass is displaced by x, the spring is deformed by an amount xcosα along its axis (spring Axis). The force along the spring axis is kx cos α. The component of this force along the direction of motion of the mass is kx cos 2 α. The equation of motion of the mass m is mx + (kcos 2 α) x = 0. From the above equation it may be noted that the equivalent stiffness K e of a spring making angle-α with the axis of motion is K e = K cos 2 α. EQUIVALENT SHIFTNESS OF SPRING COMBINATIONS Certain systems have more than one spring. The springs are joined in series or parallel or both. They can be replaced by a single spring of the same shiftness as they all show the same shiftness jointly. SPRINGS IN PARALLEL The deflection of individual spring is equal to the deflection of the system. i. e K 1 X + K 2 X = K e X K 1 + K 2 = K e The equivalent spring shiftness is equal to the sum of individual spring shiftiness. SPRINGS IN SERIES Page 8 of 40

MODULE-II --- SINGLE DOF FREE S The total deflection of the system is equal to the sum of deflection of individual springs. X = X 1 + X 2 force K e 1 = force K 1 K e = 1 K 1 = 1 K 2 = force K 2 Thus when springs are connected in series, the reciprocal of equivalent spring shiftiness is equal to the sum of the reciprocal of individual spring shiftiness. 2.2. ENERGY METHOD, RAYLEIGH METHOD Other Methods of Finding Natural Frequency The above method is called Newton s method. The other methods which are commonly used in vibration for determination of frequency are, (i) Energy Method (ii) Rayleigh s Method 2.2. i. Energy Method: Consider a spring mass system as shown in fig: 2.2 (a), assume the system to be conservative. In a conservative system the total sum of the energy is constant in a vibrating system the energy is partly potential and partly kinetic. The Page 9 of 40

MODULE-II --- SINGLE DOF FREE S K.E, T is because of velocity of the mass and Potential energy V is stored in the spring because of its elastic deformation As per conservation law of energy. T + V = constant Differentiating the above equation w.r.t. t d (T + V) = 0 dt For a spring mass system shown K. E = T = ½ mx 2 P. E = V = ½ kx 2 Hence the natural frequency is d dt (½ mx 2 + ½ kx 2 ) = 0 d dt ( mx 2 + kx 2 ) = 0 m x x + kx x = 0 x + (k/m) x = 0 ωn = k/m ƒ n = 1 /2π k/m = 1/ 2π g / δ Fig:2.2 (c) 2.2. ii. Rayleigh s Method: Page 10 of 40

= MODULE-II --- SINGLE DOF FREE S Consider the spring mass system as shown. In deriving the expression, it is assumed that the maximum K.E at mean position is equal to the maximum P.E at the extreme position. The motion is assumed to be SH Then x = A sin ω n t X= displacement of the body from mean position after time t A = Maximum. displacement from mean position to extreme position. Differentiating w. r. t x = Aω n cos ω n t Maximum Velocity at mean position x = ω n A Maximum kinetic energy at mean position = 1 2 (mx 2 ) = 1 2 (mω n 2 A 2 ) And maximum potential energy at Extreme position W.K.T, P. E = 1 2 (ka2 ) Hence the natural frequency is K.E = P.E = 1 2 (mω n 2 A 2 )P ω 2 n = (k/m) ω n = (k/m) 1 2 (ka2 ) Fig:2.2 (b) ƒ n = 1 /2π k/m = 1/ 2π g / δ Page 11 of 40

MODULE-II --- SINGLE DOF FREE S 2.3. PRINCIPLE OF VIRTUAL WORK, DAMPING MODELS 2.3. i. PRINCIPLE OF VIRTUAL WORK: The virtual work method is another scalar method besides the work and energy method. It is useful especially for systems of interconnected bodies of higher DOF. The principle of virtual work states that If a system in equilibrium under the action of a set of forces is given a virtual displacement, the virtual work done by the forces will be zero. In other words (1) δw = 0 For static equilibrium. δw = δw + δw inertia for dynamic equilibrium. (2) Virtual displacements should satisfy the displacement boundary conditions. It will be shown that these conditions are not crucial. Virtual displacement: imaginary (not real) displacement Example: Use the virtual work method; determine the equation of motion for the system below. Draw the system in the displaced position x and place the forces acting on it, including inertia and gravity forces. Give the system a small virtual displacement δ x and determine the work done by each force. Using the fact that virtual work done by external forces equals virtual work done by inertia forces, we then obtain the equation of motion for the system. δw = m. x. δx The virtual work done by inertia forces is δw = kx. δx Page 12 of 40

MODULE-II --- SINGLE DOF FREE S Equating the two quantities above and canceling δ x, we have the equation of motion m. x + kx = 0 Example: Simple Pendulum Use the virtual work method; determine the equation of motion for the system below. δw = δw inertia + δw c + δw nc δw nc = 0: Neoconservative force (external force or damping force) δw inertia = mlθ (lδθ) δw c = (mg sinθ) (lδθ) δw = δw inertia + δw c + δw nc δw = mlθ (lδθ) + (mg sinθ) (lδθ) + 0 δw = mlθ + mg sinθ (lδθ) δw = 0; δθ(t): arbitrary mlθ + mg sinθ = 0 Nonlinear equation If θ is small, sinθ = θ mlθ + mg θ = 0 lθ + g θ = 0 θ + g θ = 0 Linear equation l 2.3. DAMPING MODELS DAMPED FREE OF SINGLE DEGREE OF FREEDOM SYSTEMS In general, all physical systems are associated with one or the other type of damping. In certain cases the amount of damping may be small and in other cases large. When damped free vibrations takes place, the amplitude of vibration gradually becomes small and finally is Page 13 of 40

MODULE-II --- SINGLE DOF FREE S completely lost. The rate, at which the amplitude decays, depends upon the type and amount of damping in the system. The aspects we are primary interested in damped free vibrations are 1) the frequency of damped oscillations 2) the rate of decay Different Models of Damping Damping is associated with energy dissipation. There are several types of damping. Four of which are important types which are discussed here. 1) Viscous damping 2) Coulomb damping 3) Structural damping or solid damping 4) Slip or Interfacial damping Viscous Damping: Viscous damping is encountered by bodies moving at moderate speed through a liquid. This type of damping leads to a resisting force proportional to the velocity. The damping force. F d α dx dt F d = cx When c is the constant of proportionality and is called viscous damping Co-efficient with the dimension of N-s/m. Coulomb Damping: - This type of damping arises from sliding of dry surfaces. The friction force is nearly constant and depends upon the nature of sliding surface and normal pressure between them as expressed by the equation of kinetic friction. F = µ N When µ = co- efficient of friction N = normal force Page 14 of 40

MODULE-II --- SINGLE DOF FREE S Now coming towards damping force, if we analyze above expression, we can deduce result that it only depends upon normal force irrespective of displacement, velocity of the body. Therefore in order to findout mathematical solution of single degree of freedom with coulomb damping, we consider this reciprocatory motion into two cases. In this case, when the displacement x of the body is positive and dx/dt will be positive or displacement x is negative, dx/dt will still be positive. Such kind of condition can only be fulfilled if the body moves from left side to the right side. Therefore Newton s second law of motion will be mẍ = kx μn mẍ + kx = μn The above equation is second order nonhomogeneous differential equation. In order to verify and to make calculations easier, we will assume that the system exhibit harmonic motion. Therefore x(t) = A 1 cos ω n t + A 2 sin ω n t μn / k where in the above expression ω n = (k / m) 1/2 and A 1 and A 2 are constant and there values can be find out by using initial conditions. Solid or Structural Damping:- Solid damping is also called structural damping and is due to internal friction within the material itself. Experiment indicates that the solid damping differs from viscous damping in that it is independent of frequency and proportional to maximum stress of vibration cycle. The independence of solid damping frequency is illustrated by the fact that all frequencies of vibrating bodies such as bell are damped almost equally. Page 15 of 40

MODULE-II --- SINGLE DOF FREE S Slip or Interfacial Damping Energy of vibration is dissipated by microscopic slip on the interfaces of machine parts in contact under fluctuating loads. Microscopic slip also occurs on the interfaces of the machine elements having various types of joints. The amount of damping depends amongst other things upon the surface roughness of the mating parts, the contact pressure and the amplitude of vibration. This type of damping is essentially of a non-linear type. QUADRANT-2 Animations https://www.google.co.in/#q=animations+of+single+degree+of+free+vibration www.thirdmill.org/mission/bts.asp acoustics.mie.uic.edu/simulation/sdof%20undamped.htm acoustics.mie.uic.edu/simulation/sdof%20damped.htm www.brown.edu/.../vibrations_free.../vibrations_free_undamped.htm se.asee.org/proceedings/asee2009/papers/pr2009011erv.pdf www.efunda.com/formulae/vibrations/sdof_free_damped.cfm https://dspace.uta.edu/bitstream/.../deshmukh_uta_2502m_11706.pdf?... www.vibrationdata.com/matlab.htm www.vibrationdata.com/animation.htm www.acs.psu.edu/drussell/demos.html facultad.bayamon.inter.edu/.../chapter%202%20free%20vibration%20o... web.itu.edu.tr/~gundes/2dof.pdf Page 16 of 40

MODULE-II --- SINGLE DOF FREE S www.youtube.com/watch?v=vbluoxbfzjk www.youtube.com/watch?v=pfxltw3bs7g www.youtube.com/watch?v=bda8ghm9arw ww.youtube.com/watch?v=_rn68hc4rlc www.youtube.com/watch?v=jzwf2sdkhs8 www.youtube.com/watch?v=psxlbphwnue www.youtube.com/watch?v=derlagaj1d0 www.youtube.com/watch?v=rkfz081epsm www.youtube.com/watch?v=muwi-yi9y2s www.youtube.com/watch?v=ddkiai5oqtu www.youtube.com/watch?v=v_lj4pun_wm Videos Illustrations 1. Derive an expression for an Equation of Motion and Natural Frequency of Vibration of a Simple Spring Mass System. Page 17 of 40

MODULE-II --- SINGLE DOF FREE S Consider a spring mass system as shown in fig constrained to move in a collinear manner along with the axis of spring. The spring having stiffness is fixed at one end and carries a mass m at its free end. The body is displaced from its equilibrium position vertically downwards. This equilibrium position is called static equilibrium. The free body dia of the system is shown in fig. In equilibrium position, the gravitational pull mg is balanced by a spring force such that mg = kδ. Where δ is the static deflection of the spring. Since the mass is displaced from its equilibrium position by a distance x and then released, so after time t as per Newton s II law. Net Force = mass x acceleration mg k (δ + x) = m x m x = mg - kδ - kx (:- mg = k δ) m x = -kx m x + kx = 0 x + k /m x = 0 ----- (1) Equation (1) is a differential equation. The solution of which is x = A sin K/m t + B cos K/m t. Where A and B are constant which can be found from initial conditions. The circular frequency ω n = k/m The natural frequency of vibration ƒn = ω n /2 ƒn = 1 /2 π k /m = 1/2 π k / kδ/g = 1 / 2 π g / δ Where δ = static deflection 2) Explain a) Energy method b) Rayliegh Ritz method of finding Natural Frquency. Ans) a) Energy Method: Consider a spring mass system as shown in fig: 2.2 (a), assume the system to be conservative. In a conservative system the total sum of the energy is constant in a vibrating system the energy is partly potential and partly kinetic. The Page 18 of 40

MODULE-II --- SINGLE DOF FREE S 5) Viscous damping 6) Coulomb damping 7) Structural damping or solid damping 8) Slip or Interfacial damping Viscous Damping: Viscous damping is encountered by bodies moving at moderate speed through a liquid. This type of damping leads to a resisting force proportional to the velocity. The damping force. F d α dx dt F d = cx When c is the constant of proportionality and is called viscous damping Co-efficient with the dimension of N-s/m. Coulomb Damping: - This type of damping arises from sliding of dry surfaces. The friction force is nearly constant and depends upon the nature of sliding surface and normal pressure between them as expressed by the equation of kinetic friction. F = µ N When µ = co- efficient of friction N = normal force 4) Explian the principle of Virtual Work.And derive an expression for equation of Motion for a spring Mass System and Simple Pendulam. Page 21 of 40

MODULE-II --- SINGLE DOF FREE S Ans) Draw the system in the displaced position x and place the forces acting on it, including inertia and gravity forces. Give the system a small virtual displacement δ x and determine the work done by each force. Using the fact that virtual work done by external forces equals virtual work done by inertia forces, we then obtain the equation of motion for the system. δw = m. x. δx The virtual work done by inertia forces is δw = kx. δx Equating the two quantities above and canceling δ x, we have the equation of motion m. x + kx = 0 Example: Simple Pendulum Use the virtual work method; determine the equation of motion for the system below. δw = δw inertia + δw c + δw nc δw nc = 0: Neoconservative force (external force or damping force) δw inertia = mlθ (lδθ) δw c = (mg sinθ) (lδθ) Page 22 of 40

MODULE-II --- SINGLE DOF FREE S δw = 0; If θ is small, sinθ = θ δw = δw inertia + δw c + δw nc δw = mlθ (lδθ) + (mg sinθ) (lδθ) + 0 δθ(t): arbitrary δw = mlθ + mg sinθ (lδθ) mlθ + mg sinθ = 0 Nonlinear equation θ + g l mlθ + mg θ = 0 lθ + g θ = 0 θ = 0 Linear equation 5) Determine the natural frequency of a compound pendulum. Solution: Figure below shows a compound pendulum in the displaced position. Let m = Mass of the rigid body = w g l = Distance of point of suspension from G O = Point of suspension G = Centre of gravity I = Moment of inertia of the body about O = mk 2 + ml 2 = m(k 2 + l 2 ) k = Radius of gyration of the body If OG is displaced by an angle, Restoring torque = -mglθ since θ is small sin θ θ According to Newton s second law Accelerating torque = Restoring torque i.e., Iθ = -mglθ i.e, θ + mgl θ = 0 I mgl i.e, θ + m(k 2 + l 2 θ = 0 Page 23 of 40

MODULE-II --- SINGLE DOF FREE S θ + gl (k 2 + l 2 θ = 0 ω n = gl k 2 +l 2 rad/sec Hence natural frequency f n = 1 ω 2π n = 1 gl Hz. 2π k 2 +l 2 6) Determine the natural frequency of a spring mass system where the mass of the spring is also to be taken into account. Solution: Figure shows a spring mass system If the mass of the spring is taken into account then, let x = Displacement of mass x = Velocity of the free end of the spring at the instant under consideration. m' = Mass of spring wire per unit length l = Total length of the spring wire. Consider an elemental length dy at a distance y measured from the fixed end. Velocity of the spring wire at the distance y from the fixed end = x y l Kinetic energy of the spring element dy = 1 2 (m dy) x 7) A block of mass 0.05 kg is suspended from a spring having a stiffness of 25 N/m. The block is displaced downwards from its equilibrium position through a distance of 2 cm and released with an upward velocity of 3 cm/sec. Determine (i) Natural Frequency (ii) Period of Oscillation (iii) Maximum Velocity (iv) Maximum acceleration (v) Phase angle. Solution: y l 2 Data: m = 0.05 kg; k = 25 N/m; x (0) = x 0 = 2 cm x R0 = ν 0 = 3cm/sec. The differential equation of the motion is given by x + k m x = 0 The general solution for the above differential equation is, When t = 0, x(t) = A cos ω n t + B sin ω n t = X cos (ω n t - φ) x (0) = x 0 = A = 2cm Page 24 of 40

MODULE-II --- SINGLE DOF FREE S x R(0) = ν 0 = B ω n ; B = θ 0 ω n ω n = k m 25 = = 22.36 rad/sec 0.05 Maximum amplitude of vibration X = A 2 + B 2 = x 0 2 + θ 0 2 = 2 2 + 32 = 2.0045cm 22.36 2 i) Natural Frequency f n = 1 ω 2π n = 1 22.36 = 3.56 Hz 2π ii) Period of oscillation T = 1 f n = 1 3.56 = 0.28 sec. iii) Maximum Velocity x Rmax = X ω n = 2.0045 22.36 = 44.82 cm/sec. iv) Maximum Acceleration x Rmax = X ω n 2 = x Rmax.ω n = 44.82 22.36 = 1002.2 cm/sec 2 v) Phase angle φ = tan -1 θ 0 = tan 1 ω n x 0 3 22.36 2 ω n 2 = 3.838o 8) An oscillating system with a natural frequency of 3.98 Hz starts with an initial displacement of x 0 = 10 mm and an initial velocity of x R0 = 125 mm/sec. Calculate all the vibratory parameters involved and the time taken to reach the first peak. Data: f = 3.98 Hz; x 0 = 10 mm; x R0 = ν 0 = 125 mm/sec. Solution: The differential equation of the motion is given by x + k m x = 0 The general solution for the above differential equation is, When t = 0, x(t) = A cos ω n t + B sin ω n t = X cos (ω n t - φ) x (0) = x 0 = A = 10 mm x R(0) = ν 0 = B ω n ; B = θ 0 ω n Frequency f = 1 ω 2π n i.e, 3.98 = 1 2π ω n ω n = 25 rad/sec i) Maximum amplitude of vibration X = A 2 + B 2 = x 2 0 + θ 0 2 ω2 = 10 2 + 1252 = 11.18 mm. n 252 ii) Period of oscillation T = 1 f n = 1 3.98 = 0.251 sec. Page 25 of 40

MODULE-II --- SINGLE DOF FREE S iii) Maximum Velocity x Rmax = X ω n = 11.18 25 = 279.5 mm/sec iv) Maximum Acceleration x Rmax = X ω n 2 = x Rmax ω n = 279.5 25 = 6987.5 mm/sec v) Phase angle φ = tan -1 θ 0 = tan 1 125 = 26.565o ω n x 0 25 10 vi) Time taken to reach the first peak = π = 26.565 π 180 = 0.018546 sec. ω n 25 vii) Lead angle Ψ = tan -1 ( ω nx 0 θ 0 = tan 1 25 10 125 = 1.107 radian = 63.435o QUADRANT-3 Page 26 of 40

MODULE-II --- SINGLE DOF FREE S Wikis en.wikipedia.org/wiki/vibration wikis.controltheorypro.com/single_degree_of_freedom,_free_undamp... Wikis.controltheorypro.com/Single_Degree_of_Freedom,_Free_Undamp... apmr.matelys.com/basicsmechanics/sdof/index.html www.structuralwiki.org Home Topics petrowiki.org/basic_vibration_analysis vibrationdata.com/python-wiki/index.php?title=runge-kutta_ode... en.wikipedia.org/wiki/energy_functional Open Contents: Mechanical Vibrations, S. S. Rao, Pearson Education Inc, 4 th edition, 2003. Mechanical Vibrations, V. P. Singh, Dhanpat Rai & Company, 3 rd edition, 2006. Mechanical Vibrations, G. K.Grover, Nem Chand and Bros, 6 th edition, 1996 Theory of vibration with applications,w.t.thomson,m.d.dahleh and C Padmanabhan,Pearson Education inc,5 th Edition,2008 Theory and practice of Mechanical Vibration : J.S.Rao&K,Gupta,New Age International Publications,New Delhi,2001 Problems QUADRANT-4 Page 27 of 40

MODULE-II --- SINGLE DOF FREE S 1. An instrument panel of natural period 0.1 second, is excited by a step function 0.5 cm magnitude for a period of 0.075 second. Determine the response of the system. Solution: Natural frequency = 1 = 10 Hz 0.1 Also f n = 1 ω 2π n i.e, 10 = 1 2π ω n ω n = 20π radian/ sec = 62.832 rad. sec With the reference to translated equilibrium position for the first part, x (t) = X cos ω n t; x R(t) = ν 0 = 0; x (0) = 0.5 = X x (t) = 0.5 cos 62.832 t With reference to the original mean equilibrium position of the mass i.e., x = x (t) - x (0) = 0.5 cos 62.832t 0.5 = 0.5[cos 62.832t-1] x 62.832 x = (0.5 62.832)(-sin 62.832t) = 0.5 sin 62.832t i.e., x = 0.5 sin 62.832t ω n At the end of first part, t = 0.075 sec x (0.075) = 0.5 [cos (20π 0.075) 1] = -0.5 cm x 0.075 For the second part with t as time, ω n = -0.5 sin (20π 0.075) = 0.5 cm = θ 0.075 x = A cos + B sin ω n t = X cos (ω n t - φ) X = A 2 + B 2 2 = x (0.075) + θ 0.075 = ( 0.5) 2 + 0.5 2 = 0.7071 ω n θ 0.075 φ = tan -1 ωn = tan 1 0.5 = 0.7854 rad = 2.3562 radian x 0.075 0.5 x = 0.7071 cos (62.832t 2.3562) cm. ω n Page 28 of 40

MODULE-II --- SINGLE DOF FREE S 2. The solution to the differential for single degree freedom motion is given by x = X cos (100t + φ) with initial condition x R(0) = 1250mm/sec and x (0) = 0.25 mm find the values of X and φ, and express the given equation in the form sin ω n t + B cos ω n t. Data: x R0 = ν 0 = 1250 mm/sec; Solution: x (0) = x 0 = 0.25 mm Given equation x = X cos (100t + φ) = X cos (ω n t + φ) Also ω n = 100rad/sec x (t) = A sin ω n t + B cos ω n t x (t) = x 0 = 0 + B B = x 0 = 0.25 mm x R(t) = A ω n cos ω n t - B ω n sin ω n t x R(0) = ν 0 = A ω n i.e., 1250 = 100 A A = 12.5 mm Maximum amplitude X = A 2 + B 2 = θ 0 2 2 ω n 2 + x 0 = 12.5 2 + 0.25 2 = 12.5025 mm. Now X cos (ω n t + φ) = A sin ω n t + B cos ω n t (given) i.e., X cos ω n t.cosφ - X sin ω n t sinφ = A sin ω n t + B cos ω n t A = -X sin φ; B = X cos φ tan φ = - A B Phase angle φ = tan -1 A = B tan 1 θ 0 ωn = tan 1 12.5 = 1.55 radian x 0 0.25 x = A Hence the given equation is, = 1.5908 radian = 91.146 o x = 12.5 sin 100t + 0.25 cos 100t = 12.5025 cos (100t + 1.5908) = 12.5025 cos (100t + 91.146 o ). Page 29 of 40

MODULE-II --- SINGLE DOF FREE S 3. Determine the natural frequency of an Spring Mass System where the Mass of the Spring is also to be taken in to account consider a spring mass system as shown in fig. let L be the length of the spring under equilibrium condition. Consider an element dy of the spring at a distance y from the support as shown. If ρ is the mass per unit length of the spring in equilibrium condition, then the mass of the spring m s = ρl and the mass of the element dy is equal to ρdy. At any instant, let the mass be displaced from the equilibrium position through a distance x, then the P.E of the system is P.E = ½ kx 2 The K.E of vibration of the system at this instant consists of K.E of the main mass plus the K.E of the spring. The K.E of the mass is equal to ½ m x 2 The K.E of the element dy of the spring is equal to 1 /2 (ρdy) (y/ L x x) 2 k m y dy Therefore the total K.E of the system is given by L K.E = 1/ 2 m x 2 + 0 ½ (ρ dy) (y/l x ) 2 = 1/ 2 m x 2 + ½ ρ x 2 / L 2 0 y 2 dy x l = 1/ 2 m x 2 + ½ ρ x 2 / L 2 [y 3 / 3] L 0 = 1/ 2 m x 2 + ½ ρ x 2 / L 2 [L 3 /3] = 1/ 2 m x 2 + ½ ρ L /3 x 2 Page 30 of 40

MODULE-II --- SINGLE DOF FREE S = 1/ 2 m x 2 + ½ m s /3 x 2 K.E = 1/ 2 x 2 [m + m s /3] We have by energy method P.E + K.E = Constant 1/ 2 kx 2 +1/ 2 x 2 [m + m s / 3 ] = Constant Differentiating the above equation ½ k (2x) ( x ) + ½ (2 x ) ( x ) [m + m s /3] = 0 Or kx + (m + m s /3) x = 0 (m + m s /3) x + kx = 0 f n = 1/2π k / (m + m s /3) ω n = k / (m + m s /3) Hence the above equation shows that for finding the natural frequency of the system, the mass of the spring can be taken into account by adding one third its mass to the main mass. 4) Explain a) Energy method b) Rayliegh Ritz method of finding Natural Frquency. Ans) c) Energy Method: Consider a spring mass system as shown in fig: 2.2 (a), assume the system to be conservative. In a conservative system the total sum of the energy is constant in a vibrating system the energy is partly potential and partly kinetic. The K.E, T is because of velocity of the mass and Potential energy V is stored in the spring because of its elastic deformation As per conservation law of energy. T + V = constant Differentiating the above equation w.r.t. t Fig:2.2 (c) Page 31 of 40

MODULE-II --- SINGLE DOF FREE S For a spring mass system shown K. E = T = ½ mx 2 P. E = V = ½ kx 2 d (T + V) = 0 dt d (½ mx 2 + ½ kx 2 ) = 0 dt d ( mx 2 + kx 2 ) = 0 dt Hence the natural frequency is m x x + kx x = 0 x + (k/m) x = 0 ωn = k/m ƒ n = 1 /2π k/m = 1/ 2π g / δ d) Rayleigh s Method: Consider the spring mass system as shown. In deriving the expression, it is assumed that the maximum K.E at mean position is equal to the maximum P.E at the extreme position. The motion is assumed to be SH Then x = A sin ω n t X= displacement of the body from mean position after time t A = Maximum. displacement from mean position to extreme position. Differentiating w. r. t x = Aω n cos ω n t Fig:2.2 (b) Maximum Velocity at mean position x = ω n A Maximum kinetic energy at mean position = 1 2 (mx 2 ) Page 32 of 40

MODULE-II --- SINGLE DOF FREE S = 1 2 (mω n 2 A 2 ) And maximum potential energy at Extreme position P. E = 1 2 (ka2 ) W.K.T, K.E = P.E = 1 2 (mω n 2 A 2 )P = 1 2 (ka2 ) Hence the natural frequency is ƒ n = 1 /2π k/m = 1/ 2π g / δ 1. Eriefly Explain different models of damping. ω n 2 = (k/m) ω n = (k/m) Ans) Different Models of Damping Damping is associated with energy dissipation. There are several types of damping. Four of which are important types which are discussed here. 9) Viscous damping 10) Coulomb damping 11) Structural damping or solid damping 12) Slip or Interfacial damping Viscous Damping: Viscous damping is encountered by bodies moving at moderate speed through a liquid. This type of damping leads to a resisting force proportional to the velocity. The damping force. F d α dx dt F d = cx When c is the constant of proportionality and is called viscous damping Co-efficient with the dimension of N-s/m. Page 33 of 40

MODULE-II --- SINGLE DOF FREE S Coulomb Damping: - This type of damping arises from sliding of dry surfaces. The friction force is nearly constant and depends upon the nature of sliding surface and normal pressure between them as expressed by the equation of kinetic friction. F = µ N When µ = co- efficient of friction N = normal force 6) Explian the principle of Virtual Work.And derive an expression for equation of Motion for a spring Mass System and Simple Pendulam. Ans) Draw the system in the displaced position x and place the forces acting on it, including inertia and gravity forces. Give the system a small virtual displacement δ x and determine the work done by each force. Using the fact that virtual work done by external forces equals virtual work done by inertia forces, we then obtain the equation of motion for the system. The virtual work done by inertia forces is δw = m. x. δx δw = kx. δx Equating the two quantities above and canceling δ x, we have the equation of motion m. x + kx = 0 Example: Simple Pendulum Use the virtual work method; determine the equation of motion for the system below. Page 34 of 40

MODULE-II --- SINGLE DOF FREE S δw = δw inertia + δw c + δw nc δw nc = 0: Neoconservative force (external force or damping force) δw inertia = mlθ (lδθ) δw c = (mg sinθ) (lδθ) δw = δw inertia + δw c + δw nc δw = mlθ (lδθ) + (mg sinθ) (lδθ) + 0 δw = mlθ + mg sinθ (lδθ) δw = 0; δθ(t): arbitrary mlθ + mg sinθ = 0 Nonlinear equation If θ is small, sinθ = θ mlθ + mg θ = 0 lθ + g θ = 0 θ + g l θ = 0 Linear equation 7) Determine the natural frequency of a compound pendulum. Solution: Figure below shows a compound pendulum in the displaced position. Let m = Mass of the rigid body = w g l = Distance of point of suspension from G O = Point of suspension G = Centre of gravity I = Moment of inertia of the body about O Page 35 of 40

MODULE-II --- SINGLE DOF FREE S = mk 2 + ml 2 = m(k 2 + l 2 ) k = Radius of gyration of the body If OG is displaced by an angle, Restoring torque = -mglθ since θ is small sin θ θ According to Newton s second law Accelerating torque = Restoring torque i.e., Iθ = -mglθ i.e, θ + mgl I mgl θ = 0 i.e, θ + m(k 2 + l 2 θ = 0 gl θ + (k 2 + l 2 θ = 0 ω n = gl rad/sec k 2 +l 2 Hence natural frequency f n = 1 ω 2π n = 1 gl Hz. 2π k 2 +l 2 8) Determine the natural frequency of a spring mass system where the mass of the spring is also to be taken into account. Solution: Figure shows a spring mass system If the mass of the spring is taken into account then, let x = Displacement of mass x = Velocity of the free end of the spring at the instant under consideration. m' = Mass of spring wire per unit length l = Total length of the spring wire. Consider an elemental length dy at a distance y measured from the fixed end. Velocity of the spring wire at the distance y from the fixed end = x y l Kinetic energy of the spring element dy = 1 2 (m dy) x 9) A block of mass 0.05 kg is suspended from a spring having a stiffness of 25 N/m. The block is displaced downwards from its equilibrium position through a distance of 2 cm and released with an upward velocity of 3 cm/sec. Determine y l 2 Page 36 of 40

MODULE-II --- SINGLE DOF FREE S Solution: (i) Natural Frequency (ii) Period of Oscillation (iii) Maximum Velocity (iv) Maximum acceleration (v) Phase angle. Data: m = 0.05 kg; k = 25 N/m; x (0) = x 0 = 2 cm x R0 = ν 0 = 3cm/sec. The differential equation of the motion is given by x + k m x = 0 The general solution for the above differential equation is, When t = 0, x(t) = A cos ω n t + B sin ω n t = X cos (ω n t - φ) x (0) = x 0 = A = 2cm x R(0) = ν 0 = B ω n ; B = θ 0 ω n = k 25 = m 0.05 ω n = 22.36 rad/sec Maximum amplitude of vibration X = A 2 + B 2 = x 0 2 + θ 0 2 = 2 2 + 32 = 2.0045cm 22.36 2 vi) Natural Frequency f n = 1 ω 2π n = 1 22.36 = 3.56 Hz 2π vii) Period of oscillation T = 1 f n = viii) 1 3.56 = 0.28 sec. Maximum Velocity x Rmax = X ω n = 2.0045 22.36 = 44.82 cm/sec. ix) Maximum Acceleration x Rmax = X ω n 2 = x Rmax.ω n = 44.82 22.36 = 1002.2 cm/sec 2 x) Phase angle φ = tan -1 θ 0 = tan 1 ω n x 0 3 22.36 2 ω n 2 = 3.838o Assignment 1. An unknown mass m kg attached to the end of an unknown spring K has a natural frequency of 94 HZ, When a 0.453 kg mass is added to m, the natural frequency is lowered to 76.7 HZ. Determine the unknown mass m and the spring constant K N/m. 2. An unknown mass is attached to one end of a spring of shiftness K having natural frequency of 6 Hz. When 1kg mass is attached with m the natural frequency of the system is lowered by 20%. Determine the value of the unknown mass m and stiffness K. Page 37 of 40

MODULE-II --- SINGLE DOF FREE S 3. Find the natural frequency of the system shown in fig (1).given K 1 = K 2 = 1500 N/m K 3 = 2000 N/m and m= 5 kg. 4. A mass is suspended from a spring system as shown in fig (2). Determine the natural frequency of the system. Given k1= 5000N/m, K2=K3= 8000N/m and m= 25 kg. 5. Consider the system shown in fig (4). If K1= 20N/cm, K2= 30N/cm K4= K5= 5N/cm. Find the mass m if the systems natural frequency is 10 Hz. Page 38 of 40

MODULE-II --- SINGLE DOF FREE S 7. Find the natural frequency of the system shown in fig (5). K1= K2=K3=K4=K5=K6=K = 1000 N/m. 8. A mass m guided in x-x direction is connected by a spring configuration as shown in fig (1). Set up the equation of mass m. write down the expression for equivalent spring constant. Page 39 of 40

MODULE-II --- SINGLE DOF FREE S 9. Find the equivalent spring constant of the system shown in fig (2) in the direction of the load P. 10. Determine the natural frequency of spring mass system taking the mass of the spring in to account. 11. Drive the differential equation for an undamped spring mass system using Newton s method. 12. Derive the equation of motion of a simple pendulum having an angular displacement of θ. 13. Show that the frequency of undamped free vibration of a spring mass system is given by F n = 1/ 2π g δ. 14. Show that the natural frequency of undamped free vibration of a spring mass system is given by ω n = 1/ 2π k J. 15. Using the energy method derive the differential equation of motion of an undamped free vibration and show that frequency ω n = k 16. Using the Rayleigh method derive the differential equation of motion of an undamped free vibration and show that frequency ω n = k 17. Derive the natural frequency of torsional vibrations. m m Page 40 of 40