Arbitrary Excitations | Vibration Under Periodic Forces | Alison
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Vibration Under Periodic Forces

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Vibration under Periodic ForcesNow, let us look into Vibration of single degree of freedom systems under Periodic Forces. We have already learnt the vibration under harmonic forces; harmonic force is a special type of periodic force. So, now we will look into the vibrations under any periodic force.(Refer Slide Time: 00:38)So, in this case the single degree of freedom system is acted upon by a force which is a periodic function in time. So, what is a periodic function? We can say the periodic function has a specific portion of it, within a specific duration repeating itself indefinitely. So, we will understand this using an example. So, this is an example of a periodic function, this function is called sawtooth function and this has a period T naught. So, after the duration of T naught the function repeats itself. So, this much specific portion of this function gets repeated indefinitely.So, let us see one more example. So, this is also a periodic function and this period is T naught and this portion of this function gets repeated indefinitely. So, the function p t is said to be periodic with the time period of T naught if p t the value of the function at t is equal to p t plus j T naught that is p t is equal to the value of the function after multiples of T naught. So, p t is equal to p t plus j T naught where j ranges from minus infinity to plus infinity. So, this function continues indefinitely. So, now, let us see how a periodic function can be represented in terms of harmonic components.p(Refer Slide Time: 02:26)So, we can do that using Fourier representation. So, any periodic function p t can be represented like this using Fourier representation. So, here a naught aj and bj are constants and we have cosine and sin functions as well. So, this periodic function can be split into multiple harmonic functions that is functions of cos and sins. The series has infinite number of terms and as the value of j increases from 1 to infinity, that particular harmonic will have frequency equal to j times omega naught. Both these harmonics will have same frequency equal to j omega naught.pAnd the fundamental harmonic in this excitation has the frequency omega naught that is when j is equal to 1. So, omega naught has equal to 2 pi by T naught where T naught is the period of the periodic force. And, the coefficients of this Fourier series that is a naught aj and bj can be expressed in terms of the periodic function itself that is p t; a naught the first coefficient is expressed as this this 1 by T naught integral 0 to T naught p t dt so; that means, this is the average value of the periodic function.So, if you do this we will get the average value of this function p t. So, that is equivalent to the first coefficient and aj that is the coefficient of the cosine terms are calculated likethis 2 by T naught integral 0 to T naught p t cos j omega naught t dt. So, to find a j we have to integrate p t multiplied by the cos harmonic and it can be found out for all the values of j, j ranging from one to infinity.Similarly, the coefficient of the sin terms can be calculated as 2 by T naught integral 0 to T naught p t multiplied by sin j omega naught t dt. This also can be found out for all the values of j. So, if you can calculate a naught aj and bj we can split the function p t in terms of harmonic components.(Refer Slide Time: 05:28)We have seen that the coefficient a naught indicates the average value of p t and the coefficients aj and bj indicate the amplitudes of the j th harmonics and the j th harmonic will have frequency equal to j times omega naught and omega naught is the frequency of the fundamental harmonic that is equal to 2 pi by T naught. Theoretically this series has infinite number of terms; that means, we need infinite terms to represent this periodic function in terms of harmonics, but in reality only a few terms are sufficient for this Fourier series.pSo, with very few terms this series will converge to the periodic function. If this function has any discontinuity at the discontinuity this Fourier series will converge to an averagevalue that is the average of the values to the left and to the right of the discontinuity that is average of the neighboring values of the discontinuity.(Refer Slide Time: 06:56)Now, let us find the response of damped systems to periodic force. We have already learnt the response of damped systems under harmonic forces and we learned that transient responses due to initial displacement and velocity decays in time. So, most of the analysis are focused on steady state of the system. So, after some time the transient responses will die out and only the steady state response will be prevalent and the steady state response will be present as long as the force exists. So, we can focus on the steady state response of this system.Now, the response of a linear system to a periodic force can be determined just like that of a harmonic system and here we combine the responses to individual excitation terms in Fourier series. So, we just have seen that the periodic function can be separated into different harmonics using Fourier series. So, the response of the periodic force can also be treated as the sum of the responses of the individual harmonic forces.So, the response to a periodic force is equal to sum of the responses to many harmonic forces. So, when a constant force is applied to a single degree of freedom system, the steady state response is given by a naught by k. So, this is like a static response, the force is constant and we are looking at the steady state response. So, this is equal to the static response of the system that is a naught by k.(Refer Slide Time: 08:56)Now, let us see: what is the response of damped systems to sin force. This this is exactly what we have seen during harmonic vibrations, this is the equation of motion of a single degree of freedom systems acted upon by a sin force the forces p naught sin omega t.+t Damping ratio,

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