the following content is provided under a Creative Commons license your support will help MIT open courseware continue to offer highquality educational resources for free to make a donation or to view additional materials from hundreds of MIT courses visit MIT open courseware at ocw.mit.edu today's lecture is not mathematically hard but I I but it's really important to establish vocabulary today we're going to talk about vibration for the rest of the term and vibration is essentially applied Dynamics so up until now we've been finding equations of motion but not solving them did you notice that we have
I've almost never asked you to F you know solve the equation of motion that you've just discovered using lrange or whatever the rest of the term we're actually going to be talking mostly about the result in motion the equations of motion are pretty easy to find you have all the techniques that you need to know for finding them and now we're going to talk about how things vibrate so why do we choose vibration vibration is one is a is an incredibly common phenomenon you we wouldn't have speech without vibration you wouldn't have musical instruments without
vibration you it's a positive thing when it's making good music it's a negative thing when it's keeping you awake at night because the air conditioner in the Next Room is causing something to rattle in the room and it's driving you nuts so it's good it can be you can want it it can be desirable and you can not want it and you need to know ways of getting rid of it and so we're going to talk about uh vibration about making vibration about suppressing vibration about isolating sensitive instruments from the vibration of the floor things
like that so that's the topic of the rest of the term and today we're going to talk about single degree of Freedom systems and you might you're you might think that we're spending an awful lot of time on single deegree Freedom systems but there's a actually there's a reason for that lots of things in real life like this is just an aluminum rod this this will vibrate and continuous systems which this is have an theoretically infinite number of degrees of freedom yet when it comes to talking about its vibration it is conceptually easy to think
about the vibration of an object like this one natural frequency one natural mode at a time and in fact you can model that natural mode with its equ single degree of Freedom equivalent and that's the way I approach vibration so if you can isolate one particular mode you can literally model it as a mass spring dashbot so you need to understand the mass spring dashbot Behavior inside and out okay because we're going to it's the vocabulary we use to to do much more complicated things so single degree of Freedom system you know like the simple
pendulum has a natural frequency in this case has a mode shape here's an here's another one kind of fun single degree of Freedom this obviously involves rotation and you can figure that out using G or whatever single degree Freedom systems but now I'm going to excite one mode of vibration of this here the real high pitch get it down here by the mic so the people at home can hear it about a kilohertz way up there and that's one natural mode of this thing in longitudinal vibration when I thump it sideways you hear a lower
tone hear that rather than you know that's bending vibration of this thing but each mode of vibration I can think of by its in terms of its equivalent single degree Freedom oscillator so we'll get to talking about these things a little bit continuous systems in the last couple lectures of the term but for today then we're really going to develop this vocabulary around the vibration of single degree of Freedom systems so let's start all right so to keep it from being to boring I'm going to start with a little Mass spring dashbot that has two
Springs and they're of such a length that unstretched they're they just meet in the middle and then I'm going to take a mass and I'm going to squeeze it in between these two Springs can't draw a spring very well today and this is k1 and this is K2 and here's M and we'll put it on roller so it's obviously constrained to Motion in One Direction and I'll pick this point here as the place I'm going to put my inertial coordinate so my inertial coordinates just measured from or happens to be where the end points of
these two Springs were now to squeeze this spring in here I have there's clearly pre is pre-compression in these Springs so there is we are no longer at a zero Force state right so and I want to get the equations of motion this and moreover I want to predict want to find out what's the natural frequency of this spring so let's check here let's check check your intuition so write down on your piece of paper whether or not the natural frequency will be different because there's pre-compression or whether or not that precompression in the springs
has nothing to do with the natural frequency to write down on your paper natural frequency is different or natural frequency is the same let's how come a give have a prediction here and then we'll set about figuring this out and in the course of doing it we'll develop a little vocabulary we've uh all through the course so far when we've done equations of motion we've usually picked you know the zero spring Force position and we sort of LED you down this Rosy path that suggests you know that's the way we do it but there are
other pre other ways that you're going to find that are preferable to that sometimes so that's one of the reasons I'm doing this example so let's do a free body diagram and if I held this mass for example right at the center when I put the springs in it's obvious that this spring gets compressed by half of the length of the mass and this spring gets compressed by half of the length of the mass right so this is going to be L long so if I held it right in the middle it would compress L
over2 and L over2 but then when I relase it if these Springs are of different spring constant it's going to move a little bit so the force on this side pushing back is some K1 * l/ 2 minus the distance that I move in that direction which would Rel it and the force on this side also pushes back it's K2 time L2 over 2 + x because when I go in that direction I'm compressing any even further and those are the total forces in the X Direction on this body there's an n and an mg
which we know we don't have to deal with because we're only interested in motion left and right all right so we can say some of the forces in the X Direction mass times the acceleration and those forces are K1 1 L2 - x - K2 l/2 plus X and that's the complete equation of motion for this problem okay I'm going to rearrange it so that I get the the functions of X together here so MX dot plus K1 + K2 * X equals L over2 * K1 minus K2 and that's your equation of motion it's
nonhomogeneous this is all constants on the right hand side and on the left hand sides are the functions of X stuff that's functions of X right so what's the natural frequency of the system I hear a square root of the quantity K1 plus K2 the stiffness divided by m you know K Over M usual Mass spring dashbot system did the pre-compression have anything to do with the natural frequency I won't to ask you to embarrass yourselves but a few of you probably got that wrong right so uh there's a lesson in this and that I
want you to go away with and we'll go uh and I'll say it once and that is when an external Force has nothing to do with the motion coordinates in the problem it doesn't affect the natural frequency does this these are extern these come from external forces these are these pre-c compressions right and I can separate them out and they are not functions of X the stuff on the the on the right hand side of the equation that's not function of the motion variable do not cannot affect the natural frequency okay so I give you
another one this is our common thing hanging in from a stick so I've taken my system that I built the other day for a different purpose but now it's just a mass hanging from a spring and it's right now at its equilibrium position are the forces in there's a net there's nonzero force in the spring clearly has a natural frequency and is that natural frequency a function of gravity and so if you go to write the equations of motion of this system you know you would find uh MX double dot plus KX equals mg but
the mg has not a function of X and natural frequencies again square root of K Over M okay now we want to talk about the solving this differential equation and because it's got this constant term in the right hand side it's non-homogeneous which is kind of a nuisance term in terms of dealing with the differential equation be a lot nicer at the right hand side were zero so I want to make this the right hand side of this one zero and there's a draw a can useful conclusion from that first thing I need to know
is I'd like to know what is the static equilibrium position of this and when you go to compute static equilibrium you look at the equation of motion and you say make all motion variables things that are functions of Time Zero so no acceleration you're left with this so you just solve this for whatever the value of x is and I'll call it x sub best for xstatic and you'll find that oh well it's that term divided by K1 plus K2 K1 minus K2 all over K1 + K2 and that's the static position so now let's
let's say ah well we started off with this motion variable that was arbitrarily defined at the at the middle and let's say that well it's made up of a static component which is a constant it's just a Value Plus a dynamic component I'll call X subd which moves this is the function of time this is a constant it's not a function of time and that means if I take its derivative I might need a value for x dot that goes away it's just XD Dot and X double dot is XD double dot and let's substitute
this into my equation of motion so it becomes M XD double dot plus K1 + K2 times and now this term has got two pieces now times XD Plus K1 + K2 time XS equals L /2 K1 minus K2 all right now if I say let's examine the static case then this goes away for the for the static equilibrium case this term is zero this term is zero because the the uh Dynamic motion is zero in the static case this is motion about D at XD is motion about the static equilibrium position so for static
case these two terms go away and we know that this equals that but if that's true we can get rid of these they cancel one another these terms cancel and I'm left with mxd double dot plus K equivalent I'll call it XD equals z so the K equivalent just the total stiffnesses in the system whatever it works out right in this case it's K1 plus K2 and the natural frequency Omega n is the square root of K equivalent divided M so most often when you're doing if you're interested in vibration you're interested in natural frequencies
you're interested in solving the differential equation you will find it advantageous to write your equations of motion around the around the static equilibrium position so I could have started this problem by saying whatever the static equilibrium position is of this thing that's what I'm measuring X from and then I would have I would have come to this equation eventually you'd have to figure out what is a static equilibrium position and know what you're doing but once you know it then you have the answer now the same thing is true of that problem that's non-homogeneous differential
equation for the for the uh hanging mass and we've derived the equations of motion things for this many different ways this term right but we usually said zero spring Force but now if you started from here and said this is the static equilibrium position what's the motion about this position then you'd get the equation with zero in the right hand side lots of advantages there to using that okay once for all all single degree of Freedom oscillators will'll boil down to this equation this is one involving translation but for a simple pendulum uh this object
for example is a pendulum but it's you know it's a rotational so it's a pendulum but it's one degree of Freedom all all pendulum problems if you do them about equilibrium positions boil down to some I with respect to the point that they're rocking about Theta double dot plus some K T torsional spring constant theta equals z they take the same form so all translational single degree Freedom systems all rotational single degree Freedom system it's the same differential equation just this involves mass and linear acceleration this involves Mass moment of inertia and rotational acceleration so
everything that I say what that I say about the solution to single degree Freedom systems applies to both types of problems so let's look let's look into the uh solution of this equation briefly mostly I'm doing this to establish some terminology so the solution a solution I know or I can show that XD of T the solution to this problem notice are there any external forces by the way excitations F of TS or anything no so this thing has no external excitation it's going to make it move so it's only source of of vibration or
motion is what comes from I hear initial conditions right you have to do something to perturb it and then it will then it'll vibrate so here it is it's about in about its equilibrium position I give it an initial deflection and let go or it's around its initial condition and I give it an initial velocity it also responds or some combination of the two so initial conditions are the only things that account for motion of something without external excitation and that motion I can I can write that solution as a uh a cosine Omega T
you'll find is a is a possible solution B sin Omega T is another possible solution some a cosine Omega T minus phase angles also solution and some a e to the I Omega T you find is also a solution any of those things you can throw them in and the precise values of these things A's the B's the F's and so forth depend on the initial conditions okay so let's do this one quickly all right and I'll choose and I'm going to stop writing the subd here this is now my position from the equilibrium point
so X of T I'm going to say let it be an A1 cosine Omega t plus a B1 s Omega T and plug it in when I plug it into the into the equation of motion X double dot requires you to take two derivatives of each of these terms two derivatives of cosine gives you minus Omega cosine two derivatives s minus excuse meus Omega s cosine minus Omega s s so the answer comes out minus M Omega 2ar plus K equivalent here times A1 cosine plus B1 s Omega T is obviously in them equals zero
so I've just plugged it into that equation of motion I get this back this is what I started with that's X in general it is not equal to zero can take on any sorts all sorts of values so that side that's not generally zero and that means this must be and from this then when we solve this we find that Omega what we call n squar is K Over M and that's of course where our natural frequency comes from this is called the undamped natural frequency there's no damping in this problem yet we get to
square root of K Over m is the natural frequency of the system okay let's find out but what are A1 and B1 well at let's let X notot be X at tal Z here and if we just plug that in here put T equals z here cosine goes to one this term goes away so this implies that uh A1 equals X knot so we find out right away that the A1 cosine Omega T takes care of the response to an initial deflection and we need a uh x dot here minus A1 Omega sin Omega t
plus B1 Omega cosine Omega t that's the derivative of x we know the solution is that so it's first derivative the velocity must look like this and let's let V KN equal x dot at tal Z and we plug that in this term goes away and we get B1 Omega and cosine is one so therefore B1 is v0 over Omega but in fact the Omega is Omega n because that's because it's that we already found out that the only only frequency that satisfies the equation of motion when you have only initial conditions in the system
the only frequency that is allowed in the answer is the natural frequency so we now know B1 is uh V over over Omega n and A1 is X knot so if I give you any combination of initial displacement and initial velocity you can write out for me the exact time history of the motion xot cosine Omega t plus v over Omega n sin Omega T is the complete solution for a response to initial conditions so Annie translational oscillator one degree of Freedom where you have a translational coordinate measured from its equilibrium position has the equation
of motion actually we could write you could you've done this enough but if we added a force here and we added some damping and I wanted the equation of motion of this you know that it's m x dot plus BX dot plus KX equal F of T and so you'll often you know you're going to be confronted with problems find the equation of motion of the system it comes up looking like that and they say what's the natural frequency and I've been a little sloppy I really mean what's the undamped natural frequency and so to
find the undamped when one says that what's the undamped natural frequency you just temporarily let B and F be zero just temporarily and solve then for Omega n equals < TK of K Over M it's what you do and then so we know this is a parameter that helps that that tells us about the behavior of the system which you always want to know for the single degree Freedom systems what is the natural frequency of the system okay and we know for b equals 0 and F of0 then the response can be only due to
initial conditions so we have X of T we know it's be some Co x notot cosine Omega n t plus v0 over Omega n sin Omega NT and every simple vibration system in the world behaves basically like this from initial conditions be some part responding to the initial displacement some part to the initial velocity and damping is going to going to make it a little bit more complex but not actually by much the same basic terms appear even when you have damping in it this can be expressed as Su a cosine Omega in this case
NT minus a phase angle and it's useful to know this trigonometric uh identity to be able to put things together into an expression like that and you'll find find out that a let's a is just a square root of the two pieces it's a sign and cosine term so you have an xot squar plus a v0 over Omega n squar square root remember this is any A and B it's just a square root of a square plus b^ s that's what we're doing here and the uh phase angle Le the tangent inverse If This Were
been calling this like an A and this is the B quantity it's a tangent inverse of my sign's right B over a which in this case then is tangent inverse of x v0 over X Omega n that's all there's to it and finally another trig thing that you need to know we're going to use it quite a bit is that if you have an expression a CO sin Omega T minus V that's equal to the real part of a e to the I Omega T okay and if a is real and uh I don't want
to write it that way when when a is real then this is just the real it's just the real part it's a Time e to the I Omega T minus V because Oiler formula says e to the I theta equals cine theta plus I sin of theta so if you have an i Omega T minus V here you get back a cosine Omega T minus V and another term I sin Omega T minus V so you can always express that as the real part of that so we're going to need that little trig identity as
we go through the term now I found in many years of teaching vibration that one is something that for some that many students find a little confusing is this notion of phase angle what does phase angle really mean so there's I'm I'll try to explain it to you in a couple different ways so let's look at what this vibration that we're talking about here x cosine Omega t plus v KN over Omega n s what's it look like so that's we've just got our you know we see what it looks like but if you plot
the motion of this thing just versus time what's it look like and where does phase angle come into it well this is now X of T and this is tal 0 and this undamped system essentially going to look like that and this is the value X knot the amplitude the initial condition on X that you begin with and right here the slope V is the slope the initial slope of this curve right because the time derivative is of x dot if we were plotting x dot the initial velocity is Omega X knot and so it's
just the slope is V KN here so this is your initial velocity this is the and I didn't uh yeah that's right this is the initial displacement the total the total written out mathematically it looks like that and I'm plotting this function a cosine Omega T minus V yeah I see hand up no you know I just you you I was just looking at myself and this can't be right this has got to be the initial condition on X and this has to be the initial condition on B and whatever this turns out to be
is whatever turns out to be you have you have some initial velocity you have some initial displacement the system can actually Peak out sometime later at a maximum value right and that maximum value is that so this is uh this over here is the square root of xot 2ar plus b0 over Omega n squar square root that's what the peak value is and this system's undamped so it just goes on forever so the question is though what is this Gap here between when it starts and when it makes its maximum well you can when we
use an expression like uh we said we can express this as some a cosine Omega T minus V it's just the point at which the cosine then reaches is its maximum so if this axis here is Omega T if we plot this actually versus Omega T then one full cycle here is 2 pi or 360° so if you plot it versus Omega T then this Gap in here is just Fe that's the delay in angle if you will that the system goes through between getting from the initial conditions to getting to the peak of the
cosine so and Fe must also then be equal to some omega n times a Delta tow I'll call it some time delay so if I plot if this is plotted if this axis is time not omega T but time then this same plot the same plot this delay here this is a Delta this is a time delay in this when you plot it against time it's a delay in time to get to the peak and Omega n Delta to this delay must be equal to the phase angle so the Delta to this time delay is
f over Omega n so you can think about this as a delay in time or as a shift in Phase angle depending on whether or not you want to plot this thing as a function of Omega t or a function of time but we're going you're going to use need this concept of phase angle the rest of the term want to ask any questions about phase well the other the this because one of the things doing uh because we're doing uh vibration for the remainder of the term the this is an introduction to a topic
called linear systems and so this is basically the fundamental stuff that in which you then when you go on to two4 which is controls and that sort of thing this is the basic intro to it okay and we'll talk more about linear system Behavior as we as we go along okay now we're going to do something that you've much of this stuff I know you've seen before some of the new parts is just vocabulary and ways of thinking about vibration that Engineers do that mathematicians tend not to so you have seen most of the stuff
before where 1803 right you've done all this and uh a year two year a year ago last May um I about in May I I taught the 1803 lecture with Professor Haynes Miller now if you had 1803 last spring I think you had somebody different but he invited me to come the same classroom and we taught this we taught the second order ordinary differential equation together it was really a lot of fun he did it so he said well here's here's what we do and then I said oh well Engineers look at it in the
following way so I'll give you what I'm going to show you is what he and I did in class that day you can go back and watch that on video it's kind of fun but I I'll give you my take on it today so this is the engineer's view of what you've already seen in 1803 so we have that system and we have that equation of motion and the engineers and mathematicians more or less agree to that MX double dot plus BX but I went and looked at the web page last night last spring the
person used C instead of B haes Miller the year before used B so the every you can't you can't depend on any absolute consistency but so let's um let's start off with our homogeneous equation here and I'm looking now for the response to initial conditions with damping you've done this in a 1803 you know that you can solve this by assuming a solution of the form AE to the St plugging it in gives you a quadratic equation that looks like S2 Plus SB plus k equal Z this has roots uh I am left out my
M here so it starts off looking like that you divide through by the M S2 + B over M s+ K Over m equals zero and that's where Haynes would leave it and he'd give you the entire answer in terms of B over M and K Over M and that kind of thing Engineers we like to call that the natural frequency squared and this terms we modify to put it in a terminology that that is more convenient to engineering so I'll show you where that how that works out when you solve this quadratic just using
the quadratic equation you get the following you get that the roots you get that the roots there's two of them I'll call them S1 and two the roots to this equation look like minus B over 2 m plus or minus Square Ro T of b^2 over 4 m^ 2us K Over M and that's you'd get that's what you'd get doe in 1803 and an engineer would say well let's change that a little bit so my roots that I would use for S1 and two I just I factor out this is that's that's Omega n squar
i can factor that out it becomes Omega n on the outside I'm not putting Omega n in the numerator and denominator here as well so I get roots that look like so I just manipulated that a little bit I have a name for this term I use the Greek letter Zeta is b/ 2 Omega n m is the way I remember it my brain it's called the damping ratio okay and if I say that then the roots S1 and 2 for this look like minus Zeta Omega n plus or minus Omega n * Square <
TK of Zeta 2us 1 those are the roots that a vibration engineer would use to describe this second order linear differential equation solution homogeneous solution okay those are the roots of the equation and when you have no damping then this term goes away and you're left with and I've left an i out of here I think no I'm fine the eye comes out of here okay so for one thing to absolutely take away from today is to remember this that's our definition of damping call the damping ratio when that's one it's a number we call
critical damping I'll show what that means in a second and what CR when it's greater than one you will the system won't vibrate it it just has exponential decay if it's less than one you get vibration and that's why we like to use it this way it has it's meaningful it's value instantly know if it's greater or less than one it's going to change the behavior of the system from vibrating to not vibrating okay so now there's uh four possible solutions to this I'm not going to elaborate on all of them but Z equals z
we've already done know the answer to that response to initial conditions simple we know that one we have another solution when Zeta is greater than one zeta is greater than one this quantity here is inside's greater than one so it's a real positive number and this is the all the both roots of this thing are completely real okay and you know that the remember the response we hypothesized in the beginning that response looks like some AE to the St so now we just plug back in this is our St values we can plug them back
in and we will get the motion of the system back so for Zeta greater than one St comes out looking like minus Zeta Omega n t plus or minus Square < TK of Zeta 2 -1 time T and you just plug this in and X is just e to the St but these are just pure real values and you'll find out that the system from initial conditions on velocity and displacement just says and dies out Zeta equals to one then St is just minus you get a double root minus Omega n t twice and the
solution for this I can I'll write out the whole thing X of T here is just some uh A1 + T A2 e minus Zeta Omega n t and again it looks it's just some kind of damp not very interesting response no oscillations and the finally Zeta less than one and this is the only one this one produces oscillation and the solution for St is plus or minus minus is Thea Omega and t a real part plus or minus I Omega n t time the < TK of 1us Zeta 2 now I've turned around this
Zeta 2 minus one this is now a negative number square root a negative number gives me I and now I turn this around so this is just a real positive number so when you get I into this answer what does it tell you that the uh solution looks like Signs and cosiness right so now this gives you signs and cosiness with a DEC this is an exponential e to the minus Zeta Omega n t multiplied by a s or and a cosine and so this is the interesting part so the most of the uh work
of the rest of this term we only interested in this Final in that solution and what it looks like for this one so for Zeta less than one X of T is some a e the minus Zeta Omega n ttimes a cosine Omega DT Oma D * T minus the phase angle come out looking like that and if you draw it depends on initial conditions so again a positive velocity in a positive displacement it does this but then it dies out so it's very similar to the undamped case except that it has this damp damping
causes to die out with time but this right here this is still the initial slope is V KN and the initial displacement here is xot okay and I'll give you I'll I'm going to give you the exact expressions for this and we'll talk about it another way of writing this then in terms of the initial conditions is this looks like X knot cosine Omega DT plus v0 over Omega d for okay so expanding this out this result clearly has to depend on the initial displacement and on the initial velocity and now what's this I keep
writing this Omega D so notice in here in this solution it's Omega n * the < TK of 1 - th^ 2 so the frequency that's in here isn't exactly it's Omega n it's Omega n altered by a bit Omega subd is called the damped natural frequency and it's equal to Omega n * the < TK of 1 - Theta 2 okay system actually oscillates at a slightly different frequency and for most systems uh that vibrate at all this damping term is quite small and when you square it it gets even smaller so this is
usually this is a number that's like 09 9 often times or even bigger than that this is very close to one for all small amounts of damping but being really careful about this in including it everywhere that's what this result looks like and this little thing sigh this little phase angle here is tangent inverse of of theta over the square < TK of 1us theta 2qu and this number turns when when damping is small this is a very small number and most of the time the problems that we deal with the damping will be small
so let's say for small damping and by that I mean beta say less than 10% what we call 10% 0.1 and if you want to if you have a little more you know to you don't care too much about the Precision it might even be 20% what's actually if it were 0. 2 squared is 0.04 right 1us 0496 Square t98 so even with 20 % damping the difference between the undamped natural frequency and the damped natural frequency is 2% so for most uh for most cases with any kind of small damping at all we can
write an approximation which is easier to remember and it's all I carry around in my head I can't remember this quite frankly don't try to and I would instead Express the answer to this as a just an X not cosine Omega d t plus v over Omega d s Omega damped time time time e the minus AA Omega n t so why do I bother to carry the Omega D's along if I just said that they're almost exactly the same for light damping that Omega n's approximately Omega D well you need to keep the this
one in here because even though it's only 2% difference at 20% damping if you say the solution is Omega n when it's really Omega D there'll be a this this thing will accumulate a phase error over time so that it gets bigger and bigger this error here because you haven't taken care of that little 2% that 2% can bite you if you go through enough Cycles okay so we keep I keep Omega D in the expression here but other than that it's almost exactly the same expression that we just came up to for the simple
response of an undamped system to initial conditions x cosine plus v over Omega n sin and now all all we've added to it is put the transient Decay the fact that it decays into the expression and change the frequ natur the frequency it oscillates at to Omega D instead of Omega n so I'm going to try to impress something on you if I took this pendulum and my stopwatch measured the natural frequency of this thing I could get a very accurate value if I do it carefully then I take the same object and I dunk
it in in water and it goes back and forth and it conspicuously goes back and forth but dies down now after a while because it's got that water damping it but I measure that frequency and it's you know it's 10% different 20% different I have and and I've seen people make this mistake dozens of times you say that's the experiment explain why what's the reason that the that measured frequency has changed got any ocean engineers in the audience all right so why does the if you put the pendulum in water and it's still oscillating now
so it's got It's it isn't so damp that it's okay so it's got some damping it's dying out and the natural frequency Changed by 15 or 20% what's the explanation and the answer you always get from people is damping why because everybody's been taught this thing right they all then assume that the change in the frequency is caused by damping but damping couldn't possibly be the reason because with 20% damping this thing will die out in about two swings and it's done that's a lot of damping actually but it only accounts for a 2% change
in natural frequency not 15 so what causes the change in the frequency no not buoyancy uh that could that that could actually have an effect that's actually it I should say yes you're partly right there's an even there's another reason when the thing is uh swinging back and forth there in the water it actually carries some water with it effectively the kinetic energy you now know how to do vibration problems finding the equations to motion accounting for the potential energy and the kinetic energy the kinetic energy changes because some water moves with the object and
it's called added Mass it literally there's water moving with the object that has kinetic energy associated with the motion and it it acts like it's more massive it is dynamically more massive there's water moving with it okay so trying to impress on you that damping doesn't cause much of a change in systems that actually vibrate really You observe the vibration if you can observe the vibration damping cannot possibly account for a very large shift in frequency okay what's the motion look like let's move on a little bit here so that's what this solution looks like
we know it depends on initial conditions the U the distance from here to here will make this a Time axis this is one period so this is to d That's the damped period of vibration and we know that X of [Music] T is some a e to the minus Theta Omega n t cine Omega DT minus a phase angle we could write that expression like this and this term this is just a cosine this term repeats every period right if it's at maximum value here exactly one period later it's again at its maximum so the
cosine term goes to one every 2 pi or every period of motion all right so I want to take going to take the I'm going to Define this as the value of at X at some time T I'll call it t knot and out here is X at T KN Plus n to d n periods later so this is a period to find this period remember Omega D is the same thing as 2 pi * the frequency in hertz and it is and frequency is one over period 2 pi over the period so remember there's
a relationship that you need to remember now it relates radian frequency to frequency in cycles per second at herts to to frequency expressed in Period right this would be to D here and this would be an FD all frequency that's for any frequency you can say that Omega D is 2 pi Omega 2 pi f is one 2 pi over to so you got to be good with that but now so here we are two peaks separated by n periods and I want to take the ratio of X of T to X of t plus
n to D here and that's just going to be then my when I take that ratio X of T has cosine Omega DT minus V in it and end periods later exactly the same thing appears right so the cosine term just cancels out this just is e and the A's cancel out that's the initial conditions it's e to the minus Zeta Omega n t and I guess I called it t KN over e the minus Zeta Omega n T KN plus n damped periods and if I bring this into the numerator the exponent becomes positive
the T terms minus Z Omega and t plus those cancel and this expression is just e to the plus Zeta Omega n * n TD and the last step that I want to do to this I'm want what I'm coming up with is a way of estimating purpose of doing this is this transient curve we know depends on this is controlled you know by damping by Zeta I want to have an experimental way to determine what is data and I do it by Computing something called the logarithmic decrement so if I take the natural log
of x of T over X of t + n periods it's the natural log of this expression so it's just of an ex I just get the exponent back this then is uh n Zeta Omega I guess better do it carefully Omega n n to D but to D is 2 pi over Omega I get some nice things to cancel out [Music] here so this natural log over the ratio this is n Theta Omega n and this is 2 pi over Omega D which is Omega n * the < TK of 1 - Thea 2
Omega n's go away and for Zeta small this term is approximately one in which case this then becomes n 2 pi Zeta and Zeta equal 1 over 2 piun n natural log of this ratio of X of t uh over X of t + NT so experimentally if you just go in and measure your you plot out the response you measure a peak value you measure the peak value end periods later compute the log of that ratio divide by 1 over 2 pi n the number of periods you have an estimate of the natural frequency
estimate of the damping ratio excuse me okay so I'm going to give you one quick little rule of thumb here so this is a a an experimental way that very quickly you can estimate the damping of pendulum or whatever by just doing a quick measurement so if it happens that after n periods this value is half of the initial value then this ratio is two right so X of T some end periods later this is only half as big this value is two the natural log of two is some number you can calculate so so
there's a little rule of if you just work that out you find that Zeta equals 1 over 2 pi n 50% not one uh times Well natural log of two and you end up here with zero do this carefully 1 over 2 pi n 50% natural log of 2 and that that is11 over n 50% that's a really handy little engineer tool to carry around in your head so if I have an oscillator this will end here I can do an experiment give it an initial deflection and and say you know it starts off at
6 in or 3 in amplitude and you let it oscillate until you see it die down to half of that value so let's say one two about four Cycles this thing decays by about 50 by 50% okay four Cycles plug four into that formula you get about [Music] 0.025 agree 2 and a half% damping really very conv little thing to carry around with you measure pendulum how much damping does it have and now this is why I was saying most things that have any substantial amount of vibration the damping is going to be way less
than 10% if it dies if it takes one cycle to for the amplitude to decrease one cycle for the amplitude to decrease by 50% how much damping does it have 11% so 11% damage a lot of damping the thing starts out here and the next cycle it's half gone and the next cycle after that it's half of that and so in about three Cycles it's gone so if you see anything that's vibrating any length of time at all this damping is way less than 10% and this notion of small damping is a perfectly good one
now I'll and I'll close by just saying one other one other thing if something vibrates a lot the damping small just D if you don't have if you don't if you have you need small damping for things to actually vibrate very much this thing this is vibrating that high-pitched one that's about a kilohertz how many cycles you think it's gone through to get down to 50% of that initial amplitude that you could hear few thousand how much damping do you think this r has really tiny really tiny okay all right so even though all we
talked about today was single degree Freedom oscillators I hope you learned a few things that we'll carry now through the rest of the term we use all this these Concepts that we did today to talk about more complicated vibration good luck on your two1 quiz see you on Tuesday
