MATH351 Introduction to Ordinary Differential Equations, Fall 11-1 Hints to Week 7 Worksheet: Mechanical Vibrations 1. (Demonstration) ( 3.8, page 3, Q. 5) A mass weighing lb streches a spring by 6 in. If the mass is pulled down an additional 3 in and then released, and if there is no damping, determine the position u of the mass at any time t. (Plot u against t) and find the frequence, period, and amplitude of the motion. Hints: It is easy to construct the governing equation of this vabration process without damping and external force mü + ku =. The characteristic equation is mr + k = and its solutions is r 1, = ±i k/m. So the general solution u = c 1 cos( k/m t) + c sin( k/m t). units system. So we have the gravitational acceleration near earth s surface is about g = 3 ft/sec. Then the spring constant k is = 3 6/1 lb/sec = 18 lb/sec, where is the stationary stretched length when the spring is in equilibrium position. So k/m = 8 rad/sec. Then we consider the initial condition: u() = 3 1 ft and u() = ft/sec, from which we deduce the value of c 1 and c. So: u = 1 cos8t ft 4 From the expression of u, we see that the frequency is ω = k/m = 8 rad/sec and the period is T = π/ω = π/4 sec. The amplitude of motion is R = 1 4 ft. We remark that the unit of the spring constant k is in weight.5..15.1.5.5.1.15..5.5 1 1.5 per lenght, that is, the same in wight per second square.. (Demonstration) ( 3.8, page 3, Q. 1) A mass weighing 16 lb stretches a spring 3 in. The mass is attached to a viscous damper with a damping constant of lb-sec/ft. If the mass is set in motion from its equilibrium position with a downward velocity of 3 in./sec, find its position u at any time t. (Plot u against t). Determine when the mass first retruns to its equilibrium position. Also find the time τ such that u(t) <.1 in. for all t > τ. Hints: Construct the governing equation of this vabration process with damping but without external force: mü + γ u + ku =. The characteristic equation is mr +γr+k = and its solutions is r 1, = γ ± 1 γ 4km. Then let us consider the quantitative solution of this problem, all the quantities here are in lb-ft-sec units 1
system. We easily deduce that the spring constant k and damping constant γ are, respectively, 16 3 = lb/sec = 48 lb/sec, 3/1 γ = g lb/sec = 64 lb/sec, where is the stationary stretched length when the spring is in equilibrium position. So γ ± 1 γ 4km = ± 31i and we have the general solution: u = c 1 e t cos 31 t + c e t sin 31 t. Then consider the initial condition: u() = ft and u() =.5 ft/sec, the values of c 1 and c can be solved. So: u = 1 8 31 e t sin 31t ft..15.1.5 Position u (in.).1 X: 1.59 Y:.15.1 1.4 1.5 1.6 1.7 1.8.5.1.5 1 1.5.5 3 Suppose the mass first return to its equilibrium position at u(t 1 ) =, so t 1 = π sec. From the 31 figure, we know τ = 1.59 sec. We note that the length.1 is in inch, that is 1/1 of a foot. 3. (Demonstration) ( 3.9, page 14, Q. 5/7) A mass weighing 4 lb stretches a spring 1.5 in. The mass is displaced in. in the positive direction from its equilibrium position and released with no initial velocity. Assuming that there is no damping and that the mass is acted on by an external force of cos3t lb. (a) Find the position u(t) of the mass from its equilibrium position. Plot a graph. (b) If the given external force is replaced by 4 sinωt, then determine the frequency ω so that resonance occurs. Hints: (a) The governing equation of this vibration process with external force: mü + ku = F. = 4 3 1.5/1 lb/sec = 14 lb/sec. where is the stationary stretched length when the spring is in equilibrium position. The external force can be expression as g cos3t lb ft/sec. So the original equation can be transformed into: 4ü + 14u = 64 cos3t. The characteristic equation of homogeneous ODE is 4r +14 =. So the general homogeneous solution u h = c 1 cos16t + c sin 16t.
For the particular solution u p = Acos3t + B sin3t. Substituting the u p into the original equation to solve for the coefficients A = 16 47 and B =. Then consider the initial condition: u() = 1 ft and u() = ft/sec, the value of c 1 and c can be solved. So: u = 151 16 cos16t + cos3t ft. 148 47..15.1.5.5.1.15..5 1 1.5 (b) When resonance occurs, ω = 16 rad/sec. 4. (Class work) (Ex. 3.8, Q. 1 (B & D)): Write u = 3 cost + 4 sint in the form of u = R cos(ω t δ). Hints: Do the transformation u = 3 cost + 4 sin t = R cos(ω t δ) = R cosω t cosδ + R sin ω t sin δ. There are two equations to solve: ω =, R cosδ = 3, R sin δ = 4, so, we have R = 5 and δ = arctan 4 3 + nπ. Then u = 5 cos(t.97 nπ). Usually we choose n = here. 5. (Class work) ( 3.8, page 3, Q. 7) A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward, contracting the spring a distance of 1 in., and then set in motion with a downward velocity of ft/sec, and if there is no damping, find the position u of the mass at any time t. Determine the frequency, period, amplitude, and phase of the motion. Hints: It is easy to construct the governing equation of this vabration process without damping and external force. mü + ku =. The characteristic equation is mr + k = and its solutions is r 1, = ±i k/m. So the general solution u = c 1 cos( k/m t) + c sin( k/m t). = 3 3 3/1 lb/sec = 384 lb/sec. where is the stationary stretched length when the spring is in equilibrium position. So k/m = 8 rad/sec. Then consider the initial condition: u() = 1 1 ft and u() = ft/sec, the values of c 1 and c can be solved. So: u = 1 1 cos8 t + 8 sin 8 t ft. From the expression of u, the frequency: ω = k/m = 8 rad/sec. The period: T = π/ω = π/8 sec. The amplitude of motion: R.195 ft. 3
6. (Class work) ( 3.9, page 14, Q. 9) If an undamped spring-mass system with a mass that weighs 6 lb and a spring constant 1 lb/in. is suddenly set in motion at t = by an external force of 4 cos7t lb, determine the position of the mass at any tiem and draw a graph of the displacement against t. Hints: (a) The governing equation of this vabration process with external force is: mü + ku = F. k = 1g lb/sec = 384 lb/sec, where is the stationary stretched length when the spring is in equilibrium position. The external force can be expression as 4g cos7t lb ft/sec. So the original equation can be transformed into: 6ü + 384u = 18 cos7t. The characteristic equation of homogeneous ODE is 6r +384 =. So the general homogeneous solution u h = c 1 cos8t + c sin 8t. For the particular solution u p = Acos7t + B sin7t. Substituting the u p into the original equation to solve the coefficients A = 64 45 and B =. Then consider the initial condition: u() = ft and u() = ft/sec, the value of c 1 and c can be solved. So: u = 64 ( cos8t + cos7t) ft. 45 3 1 1 3 1 3 4 5 6 7. (Further work) ( 3.8, page 3, Q. 11.) A spring is stretched.1 m by a force of 3 newtons. A mass of kg is hung from the spring and is also attached to a viscous damper that exerts a force of 3 newtons when the velocity of the mass is 5 m/sec. (a) If the mass is pulled down.5 m below its equilibrium position and given an initial downward velocity of.1 m/sec, determine its position u at any time t. (b) Find the quasi frequency µ and the ratio of µ to the natural frequency of the corresponding undamped motion. Hints: (a) Construct the governing equation of this vabration process with damping but without external force. mü + γ u + ku =. 4
The characteristic equation is mr +γr+k = and its solutions is r 1, = γ ± 1 γ 4km. Then let us consider the quantitative solution of this problem, all the quantities here are in lb-ft-sec units system. Then the spring constant k and damping constant γ are = 3.1 N/m = 3 N/m, γ = f v =.6 N sec/m =.6 kg/sec, where is the stationary stretched length when the spring is in equilibrium position. So γ ± 1 γ 4km =.15 ± 3.871i and we have the general solution: u = c 1 e.15t cos3.871t + c e.15t sin 3.871t. Then consider the initial condition: u() =.5 m and u() =.1 m/sec, the values of c 1 and c can be solved. So: u =.5e t cos3.871t +.8e.15t sin 3.871t m. (b) The quasi frequency: ν = 3.871 rad/sec. The governing equation of this vibration process without damping effect: mü + ku =. The natural frequency: ω = k/m = 15 3.873 rad/sec. And the ratio: ν/ω =.9993. 8. (Further work) ( 3.9, page 14, Q. 6.) A mass of 5kg stretches a hanging spring.1 metres. The mass is acted on by an external force of 1 sin(t/) newtons, and is damped by a force which is proportional and opposite to the motion, and which is newtons when the speed is.4 metres/second. (a) Find the spring constant k; so, force = k stretch. (b) Find the damping constant γ; so, force = γ speed. (c) Formulate the initial value problem if the mass is set in motion from its equilibrium position with a velocity of.3 metres/second. Hints: (a) All the quantities here are in kg-m-sec units system(si unit system), where the gravitational acceleration near earth s surface is about g = 9.8 m/sec. Then the spring constant k is = 5 9.8 kg/sec = 49 kg/sec = 49 N/m,.1 where is the stationary stretched length when the spring is in equilibrium position. (b) The damping constant can be expressed as: γ = f v = N sec/m = 5 kg/sec..4 (c) The governing equation of this vabration process with damping effect: mü + γ u + ku = F. Then substituting all the quantitative values, the original equation can be transformed into: 5ü + 5 u + 49u = 1 sin(t/), or ü + 1 u + 98u = sin(t/) with the initial condition: u() = m and u() =.3 m/sec. 5