Today, I am delivering the lecture 2 of module 1 in the course Vibration of Continuous System. In the last class, I have introduced the course, told you about the different technical terms used in vibration study and also told you what is the difference between the discrete system and continuous system. Then I started a modeling system of dynamic problem.
that was of single degree freedom model and I discussed about the undamped vibration and damp vibration. In this connection, I have shown how the decay of the time response takes place when the system is having a damping. Different types of damping I have introduced.
Now, today I want to cover the time domain analysis of linear system subjected to harmonic input. So, outlines of today's lecture is first I will discuss the general approach for the time domain analysis of linear system, then I will show you the decaying nature of the response of under damp model. Then I will discuss about the damped oscillator under harmonic motion, especially the trimester analysis. Then I will introduce the magnification factor and resonance phenomena. Then a special technique known as half power bandwidth that is used for determining the damping ratio will be discussed.
Now, let us see how a time domain analysis of linear system can be performed. A linear system as you see is characterized by the m c k which are the system parameters and these are time invariant and x y z are the spatial domain in which we measure the response as well as the exciting force and t is the time instant. So, these are the parameters operators involved in linear systems and this L represents a differential operator, because the linear system is represented by a second order. ordinary differential equation for the discrete system, whereas it will be a partial differential equation in continuous system. But the order of differential equation in continuous system may be 4 or may be 2 also depending on whether the vibration of beams and plates are considered or the axial vibration of bar or torsional vibration of the shaft is considered.
out In the linear analysis, you can see that input is given to the system. Input is in terms of exciting force or it may be in terms of initial conditions. So, input again it is defined in terms of the space coordinates as well as in the time t. So, once the input is given to the system, it will respond to this input. and it will produce a output which is again measured at a some point or some location and therefore, output is again a function of space that is x y z as well as the function of time.
In the linear system, it is shown that output R can be obtained after convoluting the input with the a function I, which is known as impulse response function. Actually, impulsation function contain the system parameters or physical parameters of the system you can call it just like your mass or mass moment of inertia then damping and stiffness. So, when this convolution is carried out then we get the output, but. Special feature of the linear system is that the input characteristics will be reflected in the output of the system.
So, let us see how a general approach can be outlined for the time domain analysis in linear system. Here let us see the vibration analysis of continuous system. Actually it involves the boundary value as well as initial value problem, The continuous system again for the solution purpose is reduced to a discrete system of infinite number of generalized coordinates. So, therefore, the knowledge of discrete system is necessary which will aid the solution of the continuous system.
Therefore, we shall first see how the discrete system. can be analyzed to initial value problem, initial conditions as well as the excitation of the external agency. Now, since continuous system is a special case of discrete system, the initial value problem has common method of solution. In both the cases, you will find that similar type of solution appears in the initial value problem.
The initial value problem in time domain method of dynamic analysis involves solution of second order differential equation subjected to initial condition. However, if the second order system is non-homogeneous means in absence of exciting force or driving force the system equation is homogeneous. That means you can see here a differential equation is written which is having constant coefficient a b c for our purpose, but it may have also, the coefficient variable in space.
But now, at this moment, we are considering the inertia properties, stiffness and the damping are uniform in the domain of the structure or the system. So, here you can see a second order linear equation, because it is linear, you can see that the variable x and its derivative are with a single power. That is no second order power of the x or x dot or x double dot. So, therefore, we call it a linear equation and F t is the exciting force which may be of any form. It is not necessary that F t will be always a harmonic function that is it represents a oscillation or wave it may be of any form.
So, given this linear system we have to find the solution of for x using the theory of differential equation. The theory of differential equation says that for non-homogeneous system in case of linear differential equation, the general solution is obtained after finding the solution of the homogeneous equation and superimposing this to the solution of the non-homogeneous equation. The solution of the non-homogeneous equation is called the particular integral. So, it involves two steps. In the first step, we have to obtain the solution of the homogeneous equation that means right hand side of the equation is taken to be 0 and then with the given exciting force F t whatever may be its nature, we have to find the particular integral and then we will superimpose.
So, this is the general method. Now, because of this damping present in the system. Every material has some amount of damping. Even the vehicle tire which is inflated with air has some damping. So this is due to air damping.
But the nature of damping may be different. Sometimes the damping is linearly varying with the velocity. Sometimes it is non-linearly varying with the velocity.
But for our purpose, we will take the linear damping which is known as viscous damping because this If I call the damping force F d, which will be proportional to the velocity that is F d equal to c into x dot. So, x dot is the velocity. If x is the displacement, its time derivative, first derivative is the velocity, second derivative is the acceleration. Now, we have seen in my earlier lecture, you have also noted it that the system may have three types of damping.
One is critically damped system and another is over damped system. And then third one is underdamped system. So, overdamped system you will find some specialty that c by 2m this parameter squared is greater than k by m whereas in critical damping case it is equal to k by m and underdamped case it is equal to k by m.
So, in underdamped case if I define the damping ratio xi is equal to less than 1. Most of the engineering system are under damped. So, therefore, it is showing the oscillatory nature of the motion. However, the increase of damping may cause earlier decay of the motion if the force is withdrawn. Suppose, for example, in a system, the force is applied and it will go on oscillating until the force exists, but After the force is withdrawn, then the motion will be under free vibration and decay under free vibration will be fast if the damping parameter is higher. So, that you should know if the damping is increased decay will be faster compared to lower damping ratio.
Now, for free vibration case we take F t equal to 0 and after dividing both sides. by m and taking f t equal to 0, we now get this equation x double dot plus 2 xi omega n x dot plus omega n square x equal to 0. The solution of this equation is known to us and it is equal to e to the power minus xi omega n t, where omega n is the natural frequency of the system and it is equal to root over k by m. Then, multiplied by A cos omega dt plus B sin omega dt.
Omega d is the damped natural frequency. So, there is a difference between the damped natural frequency and undamped natural frequency that you have noted in my earlier lecture, in which case the damped natural frequency is related to the undamped natural frequency by a factor 1 minus xi square under the root. So, for lightly damped system if xi is very small then 1 minus xi square is also equal to approximately 1. So, in that case the un damped natural frequency damped natural frequency will be equal, but there is a difference because truly speaking the natural frequency in case of damped vibration omega d is equal to omega n root over 1 minus omega n square.
Now, you see The earlier equation, this equation can be written including the phase angle. So, that is written here, capital X is the amplitude and phi is the phase angle. So, it is written as X capital X e to the power minus xi omega n t into sin omega d t plus phi. So, at any instant of time the amplitude xi, x that is the amplitude of the motion, but it is decaying.
This amplitude will not be constant as in case of free vibration. So, here you will find the decay of the amplitude and this forms an envelope. The envelope is x e to the power minus xi omega n t, where you can see that this amplitude depends on the initial condition of the motion and the natural frequency. Natural frequency damping and initial condition of the motion.
Similarly, the phase angle also depend on the initial condition, the damping ratio as well as natural frequency. Because once you know the damping ratio and natural frequency, you can get the omega d. The capital X e to the power minus j omega and t is the envelope and you can see that how the motion is decaying. Now, our intention is to use this decaying curve for certain cases.
Say for example, a free vibration record is available that is motion is gradually decaying and it is coming to the rest. So, from that record we are interested to find what is the damping of the system or what is the damping ratio. So, that can be done by using this the expression.
for Xt at any two instant. So, for example, if I take the amplitude at t is equal to t1, which is X e to the power minus xi omega n t1 into sine omega d t1 plus phi, then one period apart, another peak, but because of damping, you can appreciate that this peak will be lower than the earlier peak. So, therefore, we get this at a time instant t2, t2 is nothing but t1 plus td, because it is a periodic motion. So, therefore, we can write x t2 equal to x e to the power minus omega n into t1 plus td. into sin omega d and in place of t2, I have written t1 plus capital Td plus phi, phi is the phase angle.
Now, in order to utilize this expression to find out the damping, we first define a parameter which is known as logarithmic decrement. So, logarithmic decrement is denoted by delta and it is given by log. with the natural base of Xt1 by Xt2.
So, we write now after substituting Xt1 here and Xt2 from here in the denominator. First, I will write this Xt1 and then I will write the Xt2. Now, you can see here that After elapsing a period, time period of the motion, the repetition of motion is seen, because it is a harmonic vibration, free harmonic vibration.
All free vibrations are harmonic vibration of harmonic nature. So, therefore, utilizing the property of periodic function, we can now see that delta becomes xi is the damping ratio omega n Td, because you can see here omega d T 1 plus T d will be again omega d T 1 because you see that sin phi equal to sin 2 pi plus phi if 2 pi is a period. Now, in that case T d is a period. So, T 1 plus T d sin of omega d T 1 plus T d plus phi will be nothing but sin omega d T 1 plus phi.
So, therefore, after canceling some common term we will get psi omega n T d. Now, T d is the time period of the motion and time period is defined by 2 pi by omega d. So, 2 pi into omega d, omega d is omega n root over 1 minus psi square.
Now, since omega n will appear in the denominator, so this omega n will be cancelled. So, we are left with with the expression delta equal to 2 pi xi divided by root over 1 minus xi square. For light damping when xi is very very less than 1, we can neglect xi square and therefore delta this logarithmic decrement is equal to 2 pi delta. So, we obtain the logarithmic decrement in terms of damping ratio and therefore damping ratio is obtained as xi equal to delta by 2 pi.
So, let us illustrate this with an example. Now, in example of this logarithmic decrement, I have selected a problem in which case the amplitude after certain period is recorded. In experimental observation, you will get sometimes very noisy data at some time instant, but you will find some clear data.
So, in that case, some data may be ignored. So, it is not necessary that you have to take the conjugative peaks. Peak seen clearly after say jth cycle or nth cycle can also be taken. So, here is an example. It is observed that the vibration amplitude of a damped single degree freedom system had been reduced by 50 percent after five complete cycles.
So, conjugative cycle information is not given. After completing fifth cycle, fifth cycle, the amplitude reduction is known to us. Now, assuming that it is a viscously damped system, we are required to find the damping ratio.
Now, let us see how we can solve this problem. So, let x1 be the amplitude after one cycle and x2 be the amplitude after jth complete j complete cycle after elapsing the jth cycle we can now measure the amplitude x 2 or clearly this record is visible. So, therefore, the ratio x 1 by x 2 here x 2 will not be there because this will be your amplitude at after jth complete cycle. So, this amplitude Instead of x2, please see in the original formula, it was a ratio of x1 by x2.
But here after complete j cycle, we are now getting the amplitude. So, therefore, the ratio is written as x1 divided by x j plus 1 and it becomes e to the power j omega n xi td. Why it is?
Because after multiple of integral multiple of any period again we are getting the repetition of the cycles. So, therefore, this formula that I have written is similar to that we have got in the earlier cases only here the j term is coming that is it is seen after jth cycle. Now, this ratio x1 by x j plus 1 can be written like that x1 by x2. x2 by x3 into x3 by x4 like that up to xj divided by xj plus 1. Now, take log of both sides.
So, if I take log of both sides, then it is log x1 by xj plus 1 and then log x1 by x2 log x2 by x3 log x3 by x4 like that up to log xj by x j 1 and you can see that each of this term is nothing by delta by definition that is the logarithmic decrement which is defined as the ratio of the successive amplitude. So, in that case each of the term here in the right hand side is delta, but there are j number of terms, so it is j into delta. Now, we get delta that is the logarithmic decrement equal to 1 by j log x1 by xj plus 1, because we are knowing only these two peaks x1 and xj plus 1, other peaks, intermediate peaks are not recorded. So, for the present problem we have j is equal to 5 and x1 by xj1 j plus 1 that is 50 percent reduction, so we are getting it is 2. So, therefore, delta becomes 1 by 5 log base E of 2. So, therefore, after putting the numerical value of that the xi the damping ratio becomes 0.022. So, this is the concept of logarithmic decrement and which can be applied in practical field to measure the damping from experimental data or recorded data.
Now, let us start. the time domain analysis of damped oscillator under harmonic motion. So, here I will consider the non homogeneous solution of the equation, because here this forcing function is existing, it is not the free vibration case.
Although some initial condition may be there in this system, but now we are mostly interested to find the forcing part that means, force vibration part that is called the particular integral. So, the model is described by this differential equation M x double dot plus C x dot plus k x equal to F t, substituting F t is equal to P cos omega t, because we have assumed that the system is under harmonic excitation. So, harmonic force can be a sine function or can be a cosine function. So, substituting this F t here and then divide both sides by M.
We, get x double dot plus 2 xi. n into x dot plus omega n square x equal to p by m cos omega n t because c by m equal to 2 xi omega n. So, therefore, we get this term and omega n square is nothing but k by m.
So, general solution of this equation has to be obtained in two steps. First, we have to obtain the homogeneous solution. and then we have to obtain the particular integral. Now, homogeneous solution we know in terms of integral constants of integration. So, we will focus our discussion to the particular integral now and then we will write the general solution superimposing the homogeneous solution also with the particular integral.
Now, let us see how this differential equation, non-homogeneous differential equation can be solved. There is a method In the theory of differential equation, which is known as method of undetermined coefficients. So, this method based on the concept that in the linear case, you have to choose a particular integral which reflects the characteristics of the input. So, in this case we are having two types of function may be there in the non-homogeneous part that is a forcing part, but here we are mostly concerned with the harmonic vibration.
So, that is a sine function and cos function. So, I have given a general table here, general table here in which by following which the functions can be chosen. So, you can see here if the type of the function is k into x to the power n or k into e to the power p t exponential function, then we can select the particular solution as another constant k 1 e to the power p t, because particular solution will reflect the characteristics of the input. So, that is the basic theory of linear differential equation. Now, in the instant case where the system is vibrated by a harmonic force, We are now assuming the function either as cosine function or sin function.
So, in that case the particular function or particular integral can be chosen as c 1 cos p t plus c 2 sin p t. These two trigonometrical function that cos and sin will lead you to a expression which will give amplitude as well as phase angle. Then if a So, particular integral is a constant force, choice of x p will be also a constant. As it is a linear differential equation, the particular integral will be chosen based on the characteristics or features of the forcing function, non-obvious part. So, in that case we have chosen this.
Xp equal to C1 sin omega t plus C2 cos omega t. After time differentiation, because we have the terms x dot and x dot p. So, x dot is determined as C1 cos omega t minus C2 sin omega t into omega. Then, x double dot p is determined as minus C, C1 sin omega t plus C2 cos omega t omega square. So, these are determined after substituting this value x p here in this differential equation and you can see after differentiation these terms are obtained.
So, when x p, x dot p and x double dot p are substituted in the differential equation, then we get the left hand side as c 1 sin omega t plus c 2 cos omega t of course, negative sign is there. omega square that is x double dot, then we are getting 2 xi omega n omega into c 1 cos omega t minus c 2 sin omega t, because it is with the velocity term. So, therefore, first derivative was taken and omega was coming here.
Similarly, in case of this third term omega n square and then x whatever we assumed. we have written here right hand side is delta s t how delta s t is coming you can see p by m. Now, P by K is delta S t and m is K by m is omega n square. So, utilizing these two values P is equal to K into delta S t and K by m is equal to omega n square, we can find here the right hand side as delta S t omega n square cos omega n t. So, we get this step after that we rearrange.
in this fashion, so that we can separate coefficient of like terms in both the sides. We are interested to find the coefficient of this sin and cosine, because our left hand side that is forcing function is cosine function. So, we will equate the coefficient of like terms from the both sides and this will lead to the formation of two.
simultaneous algebraic equation, which needs to be solved to find out the unknown c 1 and c 2. Now, after manipulating this, we are writing that omega n square minus omega square c 1 minus twice xi omega n omega c 2 sin omega t. So, sin omega t coefficient is now omega n square minus omega square into c 1 minus 2 xi omega n omega c 2. Similarly, the coefficient of cos omega t is now, cos omega t is you can write is this is the coefficient omega n square minus omega square into C 2 plus 2 xi omega n omega C 1 cos omega t equal to delta s t omega n square cos omega n t. You can note here that in the left hand side We have sin omega t as well as cos omega t with some constant term as coefficients.
In the right hand side, we have only cos omega t. So, equating the coefficient of like terms, coefficient of sin omega t, whatever you get here will be 0 because there will be no terms like sin omega t in the right hand side. Whereas, the coefficient of cos omega n t From both the sides, if we equate, we will get omega n square minus omega square into C 2 plus 2 xi 2 xi omega n omega into C 1 equal to delta s t delta s t omega n square delta s t omega n square.
So, therefore, we get two equations here after equating the coefficient of sin omega t and cos omega t and this equation can be solved by any standard rule. Now, here I will choose the Kramer's rule to solve this equation. If I choose the Kramer's rule, then first let us find the determinant which will be formed by the coefficient of c 1 and c 2. So, this determinant is this d is equal to omega n square omega square. that is the first row first column and first row second column is minus 2 xi omega n omega like that second row first column is written and second row second column is written.
And after expanding we will get omega n square minus omega square whole square plus 2 xi omega n omega whole square. Then constant C 1 using the Kramer's rule will be C 1 equal to determinant 0 delta s t omega n square. Why this is 0? First column, the first column, first row element is 0, because the coefficient of sin omega t is 0, and coefficient of cos omega t we are getting. So, therefore, here 0 is written delta s t omega n square in the first row, second column minus 2 psi omega n omega.
and second row second column omega n square minus omega square divided by d. So, this is C 1 an expression for C 1 after simplification it is found as 2 xi omega by omega n into delta s t divided by 1 minus omega by omega n square whole square plus 2 xi omega by omega n whole square. Now, here You should note very carefully that omega i is the driving frequency, this is the frequency exciting force, whereas omega an is the natural frequency of the system which depends on the physical parameters of the system that is mass and stiffness. So, that is inherent in the system and if you change the mass or if you change the stiffness only this omega an will change, but Whether you increase the amplitude of the force or frequency of the force, this will not influence the omega n.
So, here I have written this C1 as with this ratio that is very important that is called frequency ratio omega by omega n. So, in terms of omega by omega n, I have written and delta S t is the static displacement, which is here in that case is the P is the amplitude of the force. divided by stiffness k.
Introducing a term r which is called frequency ratio, I write this equation c 1 again in this form twice xi r delta s t divided by 1 minus r square whole square plus 2 xi r whole square. Similarly, we can find other constant C 2. So, C 2 is found as determinant of this omega n square omega square 2 xi omega n omega and 0 delta s t omega n square divided by the determinant formed by the coefficient of this C 1 and C 2. And here you can note the second column is the non-homogeneous part that is 0 and sigma s t omega n square. After expanding this determinant and with the division of D, we ultimately arrive at this expression C 2 equal to 1 minus r square, r is nothing but omega by omega n into delta s t divided by 1 minus r square whole square plus 2 xi r square.
Now, we obtain c 1 and c 2. So, we can find the particular integral. So, x p is c 1 plus c 1 sin omega t plus c 2 cos omega t. So, substituting c 1 and c 2 here, we can now write x p that is the particular integral delta s t divided by root over 1 minus r square whole square plus 2 xi r whole square into 2 xi r sin omega t plus 1 minus r square cos omega t.
You should note that this denominator is under the root. So, frequency ratio is a very important parameter which will influence the particular integral. Now, let us introduce the phase angle. So, phase angle is defined as phi and in that case, we assume that The parameter the coefficient of sin omega t is sin phi. So, we have written here this divided by this is sin phi of course, delta h 2 we have taken separately.
Similarly, coefficient of cos omega t 1 minus r square divided by root over 1 minus r square whole square plus 2 xi r square is again another term that is assumed as cos phi, because when you square it and add you will get 1. So, this may represent sin phi and cos phi. So, writing this as sin phi and cos phi, then X p is written as sigma s t divided by root over 1 minus r square whole square plus 2 psi r square cos omega t minus phi, because you know that cos phi cos omega t plus sin phi sin omega t can be written as cos omega t minus phi. So, now the response is written in terms of phase angle and amplitude. So, this is the amplitude of the motion.
So, one thing you can note the frequency ratio and damping is very important which will influence the amplitude of the motion. However, if the damping is small, the effect of damping in amplitude of the harmonic response will be not significant. Now, what we get actually?
We got earlier that is the particular integral, but this is not the full solution. Full solution is obtained after superimposing. homogeneous part.
So, homogeneous part is due to initial condition and therefore, a and b contains the initial condition, should contain the initial condition and other parameters also. So, to apply the initial condition, one should not use this expression in isolation, that will be wrong. When you want to find the constraints of integration a and b, you you have to consider the general solution which includes the homogeneous part as well as the particular integral. Now, irrespective of the nature of the forcing part, generally we get a transient part first. So, transient part you will not get any characteristics of the driving frequency.
So, driving frequency will not influence the transient part. This is the transient part where the single frequency that is the natural frequency of the system or if it is a system with several degrees, then combination of two three frequencies may also influence the initial portion that is the transient portion, but transient portion will gradually decrease. depending on the magnitude of the damping.
If the damping is large, the transient part will be very small, so for small duration and then in case of harmonic vibration, a steady state oscillation will take place. So, you can see the steady state oscillation is again a harmonic function with the driving frequency as omega, which is nothing but the frequency of the exciting force. So, response frequency is again omega that is the characteristics of the linear system and therefore, you are getting the steady state part and you can see here that frequency time period of steady state part is twice pi by omega which is equal to the time period of the your exciting force or driving force.
The second part is mainly influenced by initial conditions, natural frequency and damping ratio. Whereas, steady state part depends mainly on the amplitude of force and driving frequency. Although some influence of damping is there because tau is there, but for small damping, this damping influence will be very small in the steady state part.
So, in a damped system, steady state motion is not much altered by the change of damping if it is within a limit. And then you will get the full motion, but now we are interested to know the steady state part what happens if the driving frequency approaches the natural frequency of the system. You can see here in the steady state part the delta s t is the static response and it is multiplied by a factor 1 upon root over 1 minus r square whole square plus 2 xi r square. 2 xi r whole square. So, this factor 1 upon under root 1 minus r square whole square plus 2 xi r square is called the magnification factor, because it is magnifying the altering the static response delta s t.
So, we identify the magnification factor as M f equal to 1 by root over 1 minus r square whole square plus 2 xi r square and phase angle phi. is 2 xi r divided by 1 minus r square. Very interesting characteristics of the magnification factor is noted when we change the damping ratio from 0 to other finite value including the natural frequency in this range.
So, here is a plot which shows the variation of magnification factor with the change of frequency ratio. Now, you can see for an undamped system, when the xi is 0 and r is 1, there is no damping, so omega n is equal to omega d. So, frequency ratio is omega by omega n or omega n by omega is 1. So, in that case, x becomes unbounded.
So, at r is equal to 1, for a lightly damped system, or you can see that this frequency ratio is 1, then resonance peak is So, at r is equal to 1, resonance peak is infinite, but if damping is present, the peak is bounded. So, you will get a finite value of the peak and one thing you can see that if you increase the damping, the sharpness of the peak is decreased. So, peak this curve starts flattening.
So, this is the characteristic if you increase the damping the curve will start flattening. So, for light damping that is when the resonance peak is sharp you will get the magnitude of the resonance peak is equal to delta s t divided by 2 xi. Similarly, if we see the variation of the phase angle you can see that phase angle here for r is equal to 1 for undamped system it represents a discontinuous curve, because at r is equal to 1, you will get unbounded, that is unbounded response corresponding to phi is equal to pi by 2. Now, for undamped system, phase angle for r less than 1 is 0, and you will see that for again for undamped system frequency ratio greater than 1. the phase angle is pi.
So, the motion is in phase with the forcing function if the frequency ratio is less than 1 and it is out of phase if the frequency ratio is greater than 1. But when the damping is present, you will get this the phase angle variation and one thing you can note that all the curves passes through this point which is a straight line it is shown. as phi is equal to pi by 2. Peak amplitude at frequency ratio. Now question arises whether for damped system the peak is exactly at the r is equal to 1. Let us investigate it. Seeing the nature of the amplification factor, we can obviously tell that for light damping the peak is at the 1, approximately at r is equal to 1. But let us find theoretically.
Differentiating the magnification factor with respect to r and equating to 0, we get this ratio r is equal to root over 1 minus 2 xi square. That means the peak occurs at a frequency ratio root over 1 minus 2 xi square, but for lightly damped system it is approximately r is equal to 1. Some important points on the behavior of the dynamic system subjected to harmonic force. With increase of damping, frequency ratio at which resonance occurs shifts towards left.
For damped system, resonance frequency is omega n equal to omega n root over 1 minus xi square. At r is equal to 0, the body undergoes rigid body motion, magnification factor is 1. For undamped system, the magnification factor is unbounded. For phase angle plot, all curves passes through phi is equal to pi by 2 at omega by omega n equal to 1. For undamped system r is equal to 1, the curve has discontinuity jumping from phi is equal to 0 to phi is equal to pi.
For undamped system r less than 1, the motion is in phase with the exciting force since phi is equal to 0, but for undamped system r greater than 1, the motion is out of phase with exciting force with phi is equal to pi. Now, let us find the magnification factor by another method that is a graphical method, because sometimes the solution of the differential equation or the step by step solution a rigorous algebra you may commit mistake there is a possibility. So, a graphical solution if it is available it will give the quick results and it will be accurate also, because there will be less chance of error. So, mx double dot plus Cx dot plus kx equal to P cos omega t. Now, here it is inertia force mx double dot Cx dot is the damping force and kx is the spring force.
P is the amplitude of the exciting force and we know that steady state motion or displacement is represented by capital X cos omega t minus phi. So, we have written these derivatives of X. Derivative of x that is the velocity is minus c omega capital X sin omega t minus phi and acceleration is minus omega square X cos omega t minus phi. So, if this is the spring force. and this is your displacement.
So, velocity is at 90 degree phase C omega x, where P is also 90 degree with the displacement for any ratio not equal to not exactly at the resonant frequency. So, k x is the spring force and the acceleration the inertia shear force is just out of phase with the spring force. So, if I draw a vector diagram, then this is k x, then C omega x is the damping force and omega m omega square x is the inertia force. So, complete the polygon like that and let us find now resultant.
So, the base of the triangle is k minus m omega square. square x whereas, the altitude is C omega x. So, hypotenuse is the p. So, we can use the Pythagoras theorem and can write p square is equal to k minus m omega square whole square x square plus C omega x square. So, from this expression we can arrive x equal to p by root over k minus m omega square whole square plus C omega whole square.
And in the similar fashion after certain manipulation. population, we can get the magnitude of the force steady state response as omega s t root over 1 minus r square whole square plus 2 xi r square. So, this is the graphical method of finding this magnitude of the steady state response and the phase angle you can note here tan phi is given by C omega x divided by this K m omega square into x and after introducing the frequency ratio. another relation k by m is equal to omega n square and your this c by m is equal to 2 xi omega n.
We can arrive at this phase angle tan phi equal to 2 xi r divided by 1 minus r square. Frequency response function, this function is defined as the response of the force subjected to unit amplitude of harmonic excitation. That means subjected to harmonic excitation excitation of unit temperature.
The harmonic excitation with frequency omega, but amplitude is 1. Now, I have written it in the complex form, because you know that harmonic excitation may be represented by cosine or may be represented by sine. So, e to the power i omega t represents both sine and cosine. Of course, a phase angle is involved here.
We assume the steady state response as x is equal to h omega e to the power i omega t and substitute substituting this here in the system equation, we can now find h omega is equal to 1 divided by K m omega square plus I c omega. So, modulus of h omega that is the amplitude or mod of frequency response function is 1 upon K minus m omega square whole square plus I c omega square. Now, we have seen the frequency response function.
function or this the curve how the magnification factor varies with the frequency ratio. So, from that curve it is also possible to determine the damping. Now, select a curve with some arbitrary damping and this is the resonance peak.
The resonance peak is given by this sigma s t divided by root 2 xi because this is the expression that we have found. and we select two other points which are known as half power points and here the magnitude of the response is x max divided by root 2 that is 0.707 times of that resonance peaks. So, we get this two half power points on this curve why it is called half power because if you square it then it will be x square by root 2. So, since the power is related to energy energy or x square represents energy like term.
So, therefore, it is called half power points. Now, here omega 1 corresponding to first half power point on the left hand side of the peak is known as lower cutoff frequency and on the right hand side of the peak the frequency corresponding to the half power point is known as the upper cutoff frequency. Now, it is shown that at half power point sigma s t divided by 1 root over 1 minus r square whole square plus 2 xi r square equal to 1 by root 2 sigma s t. T delta S T divided by 2 xi.
So, from this equation if we attempt to solve R square we get R square equal to 1 minus 2 xi square plus 2 xi root over 1 minus xi square neglecting this. square of the xi square, we ultimately got the two frequency ratio that is lower frequency ratio and upper frequency ratio as R 1 square equal to 1 minus 2 xi square minus 2 xi from where we get R 1 equal to root over 1 minus 2 xi plus xi square. And another this frequency ratio that is upper frequency ratio is R 2 square equal to 1 minus 2 xi square plus plus 2 xi which is equal to R 2 equal to root over 1 plus 2 xi minus xi square. Now, this expression inside the square root term can be written as 1 plus or minus something raised to the power half and this can be expanded by binomial theorem and neglecting the higher order terms and retaining only the first term, we can now write that R 1 equal to to 1 minus xi minus xi square and R 2 is equal to 1 plus xi minus xi square.
So, these two frequencies ratios are obtained. Then the difference of these two frequencies ratio is obviously 2 xi. So, therefore, we can write now R 2 as omega 2 by omega n minus R 1, R 1 we can write omega 1 by omega n equal to 2 xi. So, this is omega 2 minus omega 1 that term in the numerator is nothing but bandwidth.
So, delta omega is the bandwidth of the curve and omega n is the natural frequency. So, from this expression we can get the damping ratio. So, from frequency response curve also we can find the damping ratio as well as from logarithmic decrement of the free decay we obtain the damping ratio.
So, in summary let us Let us see what we have discussed today. In this lecture, a general procedure of superposition of homogeneous solution with particular integral for second order time invariant system representing damped oscillator was discussed. Following the procedure, a case of harmonic excitation in a damped oscillator was illustrated.
Magnification factor, phase angle relationship with frequency ratio was described. Logarithmic decrement measurement of free decay, time history and half power bandwidth method in frequency response were illustrated for determination of damping factor. Thank you very much.