Oscillatory Behaviors of Dynamic Systems - Lab Report Example

1.0. Introduction Oscillatory behaviors of dynamic systems encompass a lot. It is important to note that bodies that have elasticity and mass are having the capability of vibrating. With mechanical vibration in this case therefore, bodies, pulley for instance, are treated as one that are elastic instead of rigid. Second aspect to note is that bodies have mass as well. The mass they posses, pulley, by virtue of their velocity, can have kinetic energy. As this is one aspect to be noted, analysis of a single degree of freedom system for free vibrations becomes another aspect. This is why it is advisable that before the analysis of the system is done, simplification of the system by modeling should be done. (Ryder and Bennett, 23) also advises that when handling pulley system having one degree of freedom a lumped mass is easier to deal with as compared to distributed mass. This is because determination of dynamic behavior of a lumped mass by one independent principal coordinate is easier than the former. With all these aspects, this project will critically analyze, through elements of mechanical vibration, a single degree of freedom pulley for free vibrations as shown in the diagram below. 2.0. Mechanical Expressions The mechanical expression of this project is going to be analyzed based on the figure below. Before embarking on practical calculations that entail the aspect of Pulley system having one degree of freedom, definition of terms that will be commonly referred to is important. To begin with, (Ryder and Bennett, 23) defines degree of freedom of a pulley as the number of independent coordinates that the pulley 2.1. Preamble The figure above consists of a pulley that has been assembled to pull a load. The spring labeled K is used to impose the motion that is going to pull the load M2. Another point is that as per the figure above, K represents stiffness which also represents a flexible connection to the pulling force. Then there is the damping coefficient at M2 represents drag between the mass pulled and the shaft walls. 2.2. Definition of terms related to the figure above 2.2.1. Natural frequency Within the context of a pulley system above, this project will need to recognize the fact that there is only one degree of freedom since movement of mass is dependent of the motion of mass. The second aspect is the natural frequency, according to Beards (207), natural frequency is the point when the system above can be able to execute free vibration which are not damped. Natural frequency is the point at which the pulley is able to vibrate naturally once it is set to move. 2.2.2. Static Equilibrium Position The project will set up the figure above in such a manner that the Static Equilibrium Position (SEP) will mean that there is no motion of the mass (M2) or external agent holding the spring. Therefore at the SEP spring K as represented above will be deflected since it will have to support at least 50% of the mass (M2) 2.3. Free body diagrams and kinetic constraints of the above figure To help in the process of calculation regarding free body and kinetic constraints of the figure above, the project will consider the following terms regarding figure above: T = Tension from cable connecting block to spring. Cable is not extensible Fs = force in spring connecting external drive agent to inextensible cable FD = viscous drag force δs = static deflection for spring From this position the project will draw the (Free Body Diagram) FBD of the figure above. Starting with the mathematical statement means block will be moving up, and will be transmitting a force towards the direction represented by X3. (See diagram above) Where Fdrive = Fspring = T = K δs + K (Z-Y) (in this case, Z and Y will been fixed to represent the distance between the beginning and the end of the spring shown on the diagram). Note that (K δs = W/2) will be the static force in the spring. This will be the static force necessary to hold the system statically, i.e. without motion. Hence δs = 0.5 W/K. From the FBD diagram the project will assume that the body/the mass (M2) will not move. That is, M2 >0 and applying Newton's 2nd law to the figure above, From the expression above, T will be the cable; δs will be the spring static deflection. From the pulley above, under kinematic constraint the length of the pulley above is going to be regarded as constant. Let us take an assumption that in the above pulley, the block and one of the disk will be that whose mass value is m and that disk will have a radius of r. It therefore means that this project will determine the natural frequency of the above spring mass pulley system. That will be: First, the Energy (T) of this system will be equal to kinetic mass m + rotational kinetic energy of pulley + potential energy of spring K. That is, T=½mx2 + ½Iθ2 +½K x2=c From free body diagram, it can be concluded that x=r θ Therefore to express moment of inertia of this pulley then this will be the expression: (I) = ½Mr2 (T) on the other hand will be: ½mx2+½x½Mr2 θ2+½ Kx2=c….let this be equation one of the pulley above (T) can as well be: ½Mr2 θ2+1/4Mr2 θ2+K r2 θ2 =c…this can be equation two of the same pulley Worked example 1: calculation of natural frequency and viscous damping ration of the figure Let us assume the following details will be introduced: K = 105 lb/in, Mg=5000 lb, and C=1500 lb.s/in The stiffness K represents a flexible connection of the spring to the wall or object pulling the mass. The damping coefficient (C) represents the viscous drag between the block and shaft walls. The damping ratio in this case will be rather large—meaning that the motion of the load will be oscillatory but that will be quickly damped. Therefore the damped natural frequencies and period of motion will be: and respectively. 2.4. Set up of the apparatus above and its design The project as shown above will entirely be designed with respective to laws of mechanical vibration and Newton’s second law to be specific. The project will be designed in such a manner that it will be easy to make mathematical calculations out of it. The project will also try to apply a “moving frame of reference as described” by (Beards, p. 34). In the project, the distances will be measured from a moving origin located at a fixed distance Lmax. Just like any other project or manufactured model well elaborated shock absorbers will be included so that mass that will be involved in the project shall be guarded against any effects of damper. That is, depending on the maximum weight that will be used, the damper will be chosen dependent on that fact. The figure below shows the project. 3.0. Experimental setup The experiment for this project will consist of a pulley system having different masses and spring as the diagram illustrates. The objective of this experiment will be to determine the effect of varied masses, spring and damper on the one degree of freedom pulley system. Typical example of the project is shown in table above. According to the diagram above, there will be different spring exhibiting different constants, different mass values and dampers so as to ascertain the relationship. Different masses and springs will be introduced at different point and the result tabulated as shown below. Table 1: The table matrix for the experiment The project will assume the following masses to be used in the pulley: m1 =0.3kg, m2=0.4kg and m3=0.5kg. These masses as will be used, will assume that there is negligible masses of the spring. On the other hand, dampers that will be used in the assignment will be having different constants of proportionalities as shown in the table. The table shows these dampers in terms of under-damped, correctly damped and over-damped respectively. And the spring constants will be arranged in terms of k1 =10kgs-2, k2 =12kgs-2 and k3=22kgs-2. 4.0. Result From the set up above, it can be noted that the varied masses applied on the system at different points, there subsequent change in the spring and force used. That is, the effect of mass and the force of gravity take effect. Mathematical expressions are as follows: In the set up above, if the masses are varied vis-à-vis the constants of the spring then the equation become: -kx=m that is kx=0 therefore the natural frequency becomes: To measure x from its static equilibrium: mg-k (x+x0)=m Please note that in this expression, k x0=mg…that is m+kx =0. Works Cited Beards, C.F., Engineering Vibration Analysis with Application to Control Systems, Arnold, 034063183X (1999). Print Ryder, G.H, and Bennett, M.D. Mechanics of Machines, 2, Industrial Press, Inc., 083113030X. (2000). Print Read More