Math 2650A. J. MeirCopyright (C) A. J. Meir. All rights reserved.This worksheet is for educational use only. No part of this publication may be reproduced or transmitted for profit in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system without prior written permission from the author. Not for profit distribution of the software is allowed without prior written permission, providing that the worksheet is not modified in any way and full credit to the author is acknowledged.Periodic functionsrestart:with(plots):f1(t):=sin(2*t);f2(t):=cos(3*t);f3(t):=cos(5*t+2);f4(t):=sin(2^(1/2)*t);f5(t):=cos(Pi*t);f1(t);f2(t);f3(t);f4(t);f5(t);plot(f1(t)+f2(t),t=0..8);Phase amplitude formIn order to better see the behavior of solutions of second order equations we can use the phase amplitude form of of periodic functions. That is given a periodic function of the formNiMvLSUiZkc2IyUidEcsJiomJSJhRyIiIi0lJGNvc0c2IyomJSZvbWVnYUdGK0YnRitGK0YrKiYlImJHRistJSRzaW5HRi5GK0Yrwe can write it asNiMvLSUiZkc2IyUidEcqJiUiQUciIiItJSRjb3NHNiMsJiomJSZvbWVnYUdGKkYnRipGKiUmZGVsdGFHISIiRio=whereNiMvJSJBRy0lJXNxcnRHNiMsJiokJSJhRyIiIyIiIiokJSJiR0YrRiw=andNiMvJSZkZWx0YUctJSdhcmN0YW5HNiMqJiUiYkciIiIlImFHISIi(note this last equation has two solutions which differ by NiMlI1BpRw==, you must chose the correct solution).Also recall the two trigonometric identitiesNiMvLCYtJSRjb3NHNiMlJmFscGhhRyIiIi1GJjYjJSViZXRhRyEiIiwkKigiIiNGKS0lJHNpbkc2IyomLCZGKEYpRixGKUYpRjBGLUYpLUYyNiMqJiwmRihGKUYsRi1GKUYwRi1GKUYtandNiMvLCYtJSRzaW5HNiMlJmFscGhhRyIiIi1GJjYjJSViZXRhRyEiIiooIiIjRiktJSRjb3NHNiMqJiwmRihGKUYsRilGKUYvRi1GKS1GJjYjKiYsJkYoRilGLEYtRilGL0YtRik=using these identities we can often rewrite periodic functions to better see the phenomenon of beats.For example, consider the equation NiMvLCYtLS0lI0BARzYkJSJERyIiIzYjJSJ4RzYjJSJ0RyIiIiomIiIqRjAtRi1GLkYwRjAtJSRzaW5HNiMqJkYrRjBGL0Yw with initial conditions NiMlIyU/Rw==restart:with(DEtools):sol:=rhs(dsolve({D((D)(x))(t)+9*x(t) = sin(2*t),x(0)=1,D(x)(0)=0},x(t)));Using the above identities we have that the solution can be written asNiMsJiooIiIiRiUtJSRzaW5HNiMqJiIiI0YlJSJ0R0YlRiUiIiYhIiJGJSooLSUlc3FydEc2IyIkSCNGJS0lJGNvc0c2IywmKiYiIiRGJUYrRiVGJSQiJmJLIiEiJkYlRiUiIzpGLUYlsincesqrt((2/15)^2+1);evalf(arctan(2/15));Free oscillationundampped oscillationsConsider the differential equation NiMvLCYqJiomKSUiZEciIiMiIiIlInhHRipGKiokKSUjZHRHRilGKiEiIkYqKiYiIztGKkYrRipGKiIiIQ==. With initial conditions NiMvLSUieEc2IyIiISIiIg== and NiMqJiUjZHhHIiIiJSNkdEchIiI=(0)=1.The characteristic equation is NiMvLCYqJCUickciIiMiIiIiIztGKCIiIQ==. This equation has the solutions:restart:solve(r^2+16=0);As you know, the homogeneous problem has the general solution NiMvJiUieEc2IyUiaEcsJiomJiUiY0c2IyIiIkYtLSUkY29zRzYjKiYiIiVGLSUidEdGLUYtRi0qJiZGKzYjIiIjRi0tJSRzaW5HRjBGLUYt.We now find NiMmJSJjRzYjIiIi and NiMmJSJjRzYjIiIjc1c2:= {c1,c2};sol:=c1*cos(4*t)+c2*sin(4*t);eq1:=subs(t=0,sol);eq2:=subs(t=0,diff(sol,t));solve({eq1=1,eq2=1},c1c2);sol:=cos(4*t)+(1/4)*sin(4*t);plot(sol,t=0..4);underdampped oscillationsConsider the differential equation NiMsJiomKiYpJSJkRyIiIyIiIiUieEdGKUYpKiQpJSNkdEdGKEYpISIiRilGKEYpNiMvLCYqJiUjZHhHIiIiJSNkdEchIiJGJyomIiM7RiclInhHRidGJyIiIQ==. With initial conditions NiMvLSUieEc2IyIiISIiIg== and NiMqJiUjZHhHIiIiJSNkdEchIiI=(0)=1.The characteristic equation is NiMvLCgqJCklInJHIiIjIiIiRikqJkYoRilGJ0YpRikiIztGKSIiIQ==. This equation has the solutions:restart:solve(r^2+2*r+16=0);As you know, the homogeneous problem has the general solution NiMvJiUieEc2IyUiaEcsJiooJiUiY0c2IyIiIkYtLSUkZXhwRzYjLCQlInRHISIiRi0tJSRjb3NHNiMqJi0lJXNxcnRHNiMiIzpGLUYyRi1GLUYtKigmRis2IyIiI0YtRi5GLS0lJHNpbkdGNkYtRi0=.We now find NiMmJSJjRzYjIiIi and NiMmJSJjRzYjIiIjc1c2:= {c1,c2};sol:=c1*exp(-t)*cos(sqrt(15)*t)+c2*exp(-t)*sin(sqrt(15)*t);eq1:=subs(t=0,sol);eq2:=subs(t=0,diff(sol,t));solve({eq1=1,eq2=1},c1c2);sol:=exp(-t)*(cos(4*t)+(2/sqrt(15))*sin(4*t));plot(sol,t=0..4);critically dampped oscillationsConsider the differential equation NiMsJiooJSJkRyIiIyUieEciIiIqJCUjZHRHRiYhIiJGKCIiKUYoNiMvLCYqJiUjZHhHIiIiJSNkdEchIiJGJyomIiM7RiclInhHRidGJyIiIQ==. With initial conditions NiMvLSUieEc2IyIiISIiIg== and NiMqJiUjZHhHIiIiJSNkdEchIiI=(0)=1.The characteristic equation is NiMvLCgqJCUickciIiMiIiIqJiIiKUYoRiZGKEYoIiM7RigiIiE=. This equation has the solutions:restart:solve(r^2+8*r+16=0);As you know, the homogeneous problem has the general solution NiMvJiUieEc2IyUiaEcsJiomJiUiY0c2IyIiIkYtLSUkZXhwRzYjLCQqJiIiJUYtJSJ0R0YtISIiRi1GLSooJkYrNiMiIiNGLUY0Ri1GLkYtRi0=.We now find NiMmJSJjRzYjIiIi and NiMmJSJjRzYjIiIjc1c2:= {c1,c2};sol:=c1*exp(-4*t)+c2*t*exp(-4*t);eq1:=subs(t=0,sol);eq2:=subs(t=0,diff(sol,t));solve({eq1=1,eq2=1},c1c2);sol:=exp(-4*t)+5*t*exp(-4*t);plot(sol,t=0..2);overdampped oscillationsConsider the differential equation NiMsJiooJSJkRyIiIyUieEciIiIqJCUjZHRHRiYhIiJGKCIjNUYoNiMvLCYqJiUjZHhHIiIiJSNkdEchIiJGJyomIiM7RiclInhHRidGJyIiIQ==. With initial conditions NiMvLSUieEc2IyIiISIiIg== and NiMqJiUjZHhHIiIiJSNkdEchIiI=(0)=1.The characteristic equation is NiMvLCgqJCUickciIiMiIiIqJiIjNUYoRiZGKEYoIiM7RigiIiE=. This equation has the solutions:restart:solve(r^2+10*r+16=0);As you know, the homogeneous problem has the general solution NiMvJkkieEc2IjYjSSJoR0YmLCYqJiZJImNHRiY2IyIiIkYuLUkkZXhwR0YmNiMsJComIiIjRi5JInRHRiZGLiEiIkYuRi4qJiZGLDYjRjRGLi1GMDYjLCQqJiIiKUYuRjVGLkY2Ri5GLg==.We now find NiMmJSJjRzYjIiIi and NiMmJSJjRzYjIiIjc1c2:= {c1,c2};sol:=c1*exp(-2*t)+c2*exp(-8*t);eq1:=subs(t=0,sol);eq2:=subs(t=0,diff(sol,t));solve({eq1=1,eq2=1},c1c2);sol:=(3/2)*exp(-2*t)-(1/2)*exp(-8*t);plot(sol,t=0..2);