 Namaste, welcome to the session Solving Problems Based on Mathematical Modelling of Mechanical Systems. At the end of this session students will be able to solve problems based on mathematical modeling of mechanical systems. So, let us see mathematical models, differential equation model and transfer function model. Differential equation model also referred as time domain model and transfer function model is also referred as frequency domain model. It is called as time domain model because representations of the system are in terms of function of time. Transfer function model is known as frequency domain model because the representations of the system are in terms of function of frequency. In differential equation model dynamics of the system represented in terms of differential equations and in case of transfer function model dynamics of the system represented in terms of Laplace transform expression. Let us see importance of Laplace transform and transfer function. The transformation technique relating the time functions to frequency dependent function of a complex variable is called as the Laplace transformation technique. It is useful in solving linear differential equations and analyzing control systems. The transfer function gives mathematical models of all system components and overall system. Induitional analysis of various system components is also easy and possible by using the transfer function approach. Transfer function, it is defined as the ratio of Laplace transform of output response of the system to the Laplace transform of input excitations with zero initial conditions. Now, let us consider a system with input in time domain as R of t and respective output response C of t. Then to find out Laplace transform input is denoted as capital R of s and output is denoted as capital C of s. So, the Laplace transform for this diagram is given as Laplace transform t of s is equal to Laplace transform of output divided by Laplace transform of input equal to C of s divided by R of s. Now, let us see mathematical modeling of translation motion. So, obtain differential equation and transfer function for given mechanical translation system. So, we have an example here which we have seen in the last session. So, take a pause here and recall and write differential equation. So, for the given diagram, f of t is equal to m d square x by dt square for the mass plus b into dx by dt for damper plus kx for the spring. So, taking Laplace transform on both sides we will get or taking x of s common we will get the equation as given below. Then to find out transfer function x of s divided by f of s where x of s is output and f of s is input equal to 1 upon m s square plus bs plus k. So, this is the equation of transfer function of the given system. So, let us see mathematical modeling of rotational motion. So, obtain differential equation and transfer function for given mechanical rotational system. So, this is the example which we have already seen in the last session. So, take a pause here and find out differential equation and free body diagram. So, writing the equation here we will get. So, taking Laplace transform on both sides we will get. So, taking theta of s common we will get the equation and to find out the transfer function output upon input that is theta s upon t of s equal to 1 upon j s square plus bs plus k. So, this is the equation of transfer function of a rotational system. Now, obtain differential equation and transfer function for given mechanical translation system. So, let us find out differential equations. So, applying the Newton's law of motion which we have already seen to the m 1. So, the equation becomes f of t m 1 d square x 2 by dt square plus here now we have spring which has a displacement x 2 as well as x 1. So, equation becomes k 2 into x 2 minus x 1. Now, taking Laplace transform on both sides we have f of s is equal to m 1 s square x 2 of s plus k 2 into the bracket x 2 of s minus x 1 of s taking x 2 of s common here we get. So, this is the equation 1 then applying Newton's law of motions to mass m 2 where you can find there is no force applied. So, 0 is equal to m 2 d square x 1 by dt square plus spring k 2 having displacement x 1 and x 2 as it is connected to m 2 here. So, we are considering k 2 x 1 minus x 2 then k 1 into x 1. But taking Laplace transform on both sides we get 0 is equal to m 2 s square x 1 of s plus k 2 into the bracket x 1 of s minus x 2 of s plus k 1 x 1 of s. So, by taking x 1 of s common here from the Bohr equation we get. Now, let us find out the value for x 2 of s. So, taking k 2 x 2 of s at other side we will give us x 2 of s is equal to x 1 of s into the bracket m 2 s square plus k 1 plus k 2 divided by k 2. Now, let us put equation 3 in equation 1 to find out the transfer function. So, let us consider equation 1 again that is f of s let us put value of x 2 into this equation. Now, let us find out transfer function by dividing x 1 of s by f of s. So, the equation gives us x 1 of s by f of s is equal to k 2 divided by m 2 s square plus k 1 plus k 2 into the bracket m 1 s square plus k 2 minus k 2 square bracket complete. So, this is the transfer function and let us have one more example based on rotational motion. So, obtain differential equation and transfer function for given mechanical rotational system. So, this is the given diagram. So, let us find out equation for the inertia j 1 so that will give us torque T of t is equal to j 1 d square theta 1 by dt square plus k theta 1 minus theta 2 let us take Laplace transform of the above equation so that will give us T of s so this is the equation 1. Now, let us find out the differential equation for inertia 2 that is j 2 now taking Laplace transform on both side that will give us now taking theta 2 of s common that will give us now let us take equation 2 again here and let us find out the value of theta 1 of s that will give us now to find out transfer function let us put the equation 3 in equation 1. So, let us bring equation 1 again here so equation becomes now to find out the transfer function let us divide theta 2 of s by T of s so that will give us k by j 2 s square plus b s plus k into j 1 s square plus k minus k square so in this way we can find out differential equations and transfer function and hence we can find out the mathematical modeling for translation and rotational system these are references thank you.