Question

1.- Starting from the differential equation for a 1-degree of freedom system with mass M, damping c and spring stiffness k: a.- Show that the particular solution for the equation with an applied force fo cos(ot), i.e., Mä+ci+kx=f, cos(or) can be expressed as x )= A cos(ot) + A, sin(or) and find the values of A, and A, that solve the differential equation in terms of M, c, k and fo. 5 points. b. Use the result from part a to find the value of frequency w of the applied force that maximizes the amplitude of the oscillations of the forced response and show that the value differs from the natural frequency of the undamped system by a term that is equal to one when C=0. 5 points.

Answer 1

Schematically of single-Degree of freedom, viscously - Damped system under forced vibration no shown? TM focosut Egn of motion & Apply Newton's Tord law Mr = focos wt-ka-ch mat citka = do caruto This a non-homogeneous linear differential egh. having indepondent parakroter as t 4 dependent parameter as a M, L,K, to are anotants. It has soon of the form aith= (1 + xp (t? The term X(t) accounts for the transcient response accounts Conrooponds to the free-vibration equation permit ci tkx =o). However, it is very snall & so generally at is damped out 4 at steady-state, displacement agh. ie xerth = X(t) Determination of x&ti & For finding xat method of undetermined coofficients in used which applies to eons of the formt Tyltay't by = rx 1 (here indopondent variable = to dependent variable=y/ how if rai= K coswa then Yp(x) = Acos w x t Azsah wn Comparing with (23 with (2) here retn= to cor wt . Sup(t) = A, co wt+ Azsinct), Al, Az are constants

Now as regth is a solution of egn(in ie Mätca+kx=to coswt put rath in this ean to determine A, & A₂ MITA, cor wt+Azsinwt) + cd (Arcos wft Azsincet +K(Arcosw++ Arsine t = to coswt MA, W / 806 wt) + A₂ w²C-sincut)] + c(-A, wsincut +A₂ w cos with & K (A Coswt & Az sincet) = to cos ut Erco Equating coetficients of cowt, sincet on both sides & for coefficient of cosut : - MAI ²+ (A₂W +KA, = fo IA (K-M W 2) + Az co = to of for coefficient of sincet it - MA, C02 – CA, W + KA2 = 0 . (K-MW2) Az = CWA, =cw All komwa Substituting A in egn & Ak -MW2) & (co22 A1 = to (k-mw²) A / (K-Mw 2² (CwiZ = to [ (K-M02) = folk-Mw3 (komu?)² + (ow)2 from agh , Az = CwAi = to (CW) K.Mw² (k-mw 2)2 + (W) ² A

= VARA (C) u (t) = A cocott Adin wt A coswt+ A2 sincoth 2 VA+A2 VA, 24 AZ here A al & A A VA 2TA VARAZ therefore substituting A = cos & & VARGAZ A = sind VARTAR (ti= VARA (cos wt cord & sincet sind) - VARTA cos (ut-1) = x cos(wta) This amplitude of forced vibration = VARAZ Now A to (k - Mwa), Az- fo che con (K-M02) + (cco)2' (K-Mw2²+ ((W22 x= Valentines.com. to Vk Mw2)+(con? olm

x = to V(kamw2²+6w) x = to ky (1 MW 2)2 + cage = tolk Vente mewn a who undarsped natural frequency (= critical damping coefficient - 2 km - 5 2 5 km w = 32 w kim = 25 w = 250 un x = folk Vp-p 2) + (25072 Now Xo/Folk) = magnificates factar as to ist Xsr = zero frequency deflection of a spring mass sestem under a steedy force to ML one V 11-722 24(2352 Therefore from the plot of musr it is clean that at P= I are o = 1 ie w=wn, amplitude af vibration become excessive for small damping f decreases with increase in damping .3 increases