Question

4. Consider the mechanical system shown below with a spring with stiffness, k (N/m), in parallel with a viscous damper with coefficient, h (Nós/m) and an externally applied force, Fexi(t) (N). u(t) a. Find the equation that relates the applied force, Fext(t) and the displacement, u(t). b. If the spring component has a stiffness of k = 75 N/m, the damper component has coefficient h = 50 N s/m and the externally applied force is a constant 4.5 N applied instantaneously at some time t = 0; for t 0 and 0 otherwise Determine the exact equation for the displacement, u(t), as a function of time. Assume that the initial displacement u(0) = 0. c. Use your Matlab skills to plot the displacement function, u(t), as a function of time. Note: use your judgement to appropriately choose the start point, end point and sample spacing of the discretized vector t in Matlab. d. Use Matlab/Simulink/Simscape to model this mechanical system with the given forcing function (hint: use a step function source) and visualize the displacement function, u(t), as a function of time. Note: you will also need to appropriately choose the end point and sample spacing of the discretized vector t in the Simulink model. Also Note: For the Simulink-PS Converter Block Parameters, you may need to change the input handling to Filter input, derivatives calculated. You may use the default values for the filter order and time constant. e. Compare and contrast your plots from parts c and d. Note: You may find it easier to output the displacement from the Simulink model to a Matlab workspace variable (see tutorial) and plot it in Matlab.

Answer 1

a) As the spring and damper are acting against the external force hence,

The equation is,

$F_{ext}-h\frac{du}{dt}-ku=0$

$h\frac{du}{dt}+ku=F_{ext}$

b) here h=50 N.s/m, k= 75 N/m, and Fext= 4.5 N hence the equation becomes,

$50\frac{du}{dt}+75u=4.5$

$\frac{du}{dt}+1.5u=0.09$

multipling both sides with exp(1.5t) we get,

$e^{1.5t}\frac{du}{dt}+1.5e^{1.5t}u=0.09e^{1.5t}$

$\frac{d}{dt}(e^{1.5t}u)=0.09e^{1.5t}$

$d(e^{1.5t}u)=0.09e^{1.5t}dt$

integrating we get,

$e^{1.5t}u=\int 0.09e^{1.5t}dt$

$e^{1.5t}u= \frac{0.09}{1.5}e^{1.5t}+c=0.06e^{1.5t}+c$ {where c is a constant}

at t=0 , displacement u=0 hence from the eq we get,

$0.06+c=0$

c=-0.06

$e^{1.5t}u=0.06e^{1.5t}-0.06$

$u=0.06(1-e^-1.5t)$

c) to plot the system accurately we must use the transfer function method here we use laplace transformation on the equation and find out the output/ input ratio called the transfer function from this function we can see the step response of the system( when the input [here external force] is constant )

from the equation

$\frac{du}{dt}+1.5u=\frac{1}{50}F_{ext}$ here F_ext = 4.5 U(t) {U(t)=1 for t>=1 ,0 otherwise}

using laplace transformation

$su(s)+1.5u(s)=0.09U(s)$

transfer function = u(s)/U(s)

$=\frac{0.09}{s+1.5}$

Matlab program,

>> num= 4.5/50; >> den= [1 1.5]; >> g=tf (num, den) Transfer function: 0.09 --- s + 1.5 >> ltiview

here to plot the response we will use "ltiview" here we have to import the transfer function by clicking on "file"> import and you will see the plot. and click on the binocular to autoscale the plot .

Step Response 0.06 0.05 0.04 Amplitude 0.02 0.01 0.5 1.5 2 2.5 3.5 Time (sec)

d) The simulink model is given below,

4.5 1/50 5+1.5 Step Transfer Fan Scope Gain

and the plot will open when you double click on the scope and click on the binocular icon to autoscale it

0.06 0.05 0.04 0.03 2.02 01 .............. offset: 0