Question

4. The mechanical system below represents a weight machine often used by Olympic downhill skiers to increase lower body strength. The skier attempts to displace a mass, M, by applying vertical force f(t). Friction occurs between the mass and vertical surfaces with frictional coefficients 7, and 12. 2, N = 2 = 1 Ns/m D, - R = 3 Ns/m K = K₂ * 13 N/m м. fit 40 M - 2kg Assuming that the skier is training in a zero gravity environment, a) Determine the frequency response (jw) of the system. b) Determine the impulse response (1) of the system.

Answer 1

Okay so here we first write the mechanical differential equation , then we take the Laplace tansform of the differential equation obtained ( which converts it into the frequency domain ) to solve it , then we take the inverse Laplace transform to convert it back into the time domain.

Now if you're not familiar with Laplace transform don't worry , you see ,until now you have solved real world problems which consists of equations in TIME domain, but complex differential equations are very difficult to solve in the time domain that's why we convert the equation into frequency domain ( take its Laplace transform ) solve it in that domain and then return the result in time domain again ( inverse Laplace transform ) .

Here we are also going to do the same . I've provided a link to the rules of Laplace and inverse Laplace transform in the end of the answer. Now lets start solving.

First we need to derive the mechanical differential equations in time domain , you can do it by drawing FBD diagrams but that's time consuming and involves a lot of thinking here I"ll show you a trick for the same , to write the differential equation of any block of mass in a mechanical system simply see to which components its directly associated with, in this case that's a damper , spring and friction. now write your equation in the given form .

M* + internal forces caused due to external forces = External forces

(d(h(t))/dt)

Below I've shown the step by step process of the above explanation , go through it you'll understand the Solution

HIDI EK Ina Y M11 L p het) € $11) + dia dt Md hết) + (Di + Be + Notha) dhct) + (ki+Ka ) thet) = f(t) dt as you can see Now, we can put the numerical values a drit) + 8 dhet) + 262ct) = $(t) -O take thek) Lablace transform g equation-0 25?H(s) + 85 H(S) + 26H(s) = F(s). Since No inital Now take H(s) common.ie a conditions stated also HS) xa ( så +45 + 13) - Fls) => HIS) - FG) - 213? + 45+13) Now we can Replace s with jw to get answer of (a) H (gw) = f(gw) 2 (4jw + (13-wr)).

Now from equation @ we have FIS 2 (524 45+ 13) Taking Laplace inverse to convert in time domain da His)) !! ! 5 +45 +13) . Form (4) Now to find a la desi conuent it to the farm lan Convert it to the form ) $2 +45 +13) sº +45 +13 Can be cevritten as = sạ+45+4 + 9 (+23° + 32. se stable to love Now Laplace inverse of form (a) is ebt sinut So d" (ottisti) j (849 sinut) . Now putting it back in eavation ③ we have. autredt. 51n3t) The+) = 547 x 62. sindt) -bt hct) = f(t) x le 2t sinut)

You can refer to the following formula sheet for the rules that I have used here

Table of Inverse Laplace Transform 3. C- th-1 U n =1, 2, 3,... lsh 1.67{}}=1 2.6-{=a} = cm 3.c-{}=1n = 1,2 4.6*{(–a)H} = (n = 1) 5.2+(227)} = sin bt 6.47{245} = cos bt 1 eat tr-1 70-1S 1 1 - =-sinh at 82 - as a 16+{2-e} = cosh at 1."{16-a32+ be} = {ch sin be 10. C-1 S. S-a = eat cos bt 1 (8 - a)2 +62 11.6"{@*+} = zze sin be

Table 2.1 Laplace Transforms nu(t) . U(S) s-1 S0 eat sa sa s > 0 52 faz and gaiak, 8 >0 2007 and 2 ani s > lal s-le-as, 8>0 e-as t", n a positive integer sin at and cos at sinh at and cosh at H(t - a) 8(t - a) H(t - a)f(t - a) f(t)e-at H(t – a) f(t) erft to exp (*) 1 - erf (2) exp ( ) F(8)e-as F(s + a) e-48L(f(t + a)) 8-+(1+8)-1/2, 8>0 /se-avs, (s > 0) s-de-avs, 8>0 Te-avs, 8>0

Hope This Helps !