Tatu samarqand filiali

Mexanik tebranishlarga doir masalalar differensial tenglamalarini sonli yechish

1. 
   misol. Mexanik sistemaning ketma-ket yarim kosinusoidal kuchlar ta‟sirida vertikal tebranishlari quyidagi oddiy differensial tenglama bilan ifodalanadi:
   

d ² x

^m d t ²
kx F_m cos( ,

bu yerda x – sistemaning boshlang‟ich holatidan chetlashishi; t – vaqt; m – sistemaning massasi; - ishqalanish koeffisiyenti; k – amortizatorning bikrlik

koeffisiyenti;
F_m va
- majburiy kuchning parametrlari.

Misolni quyidagi ma‟lumotlar uchun yeching: sistema massasi m = 2 kg; ishqalanish koeffisiyenti = 1 kg/kuch; amortizatorning bikrlik koeffisiyenti k =4 N/m; F_m = 3000 N; = 0,1 rad/s. Boshlang‟ich shartlar: t = 0 da x=0 va dx/dt =0.

Sistemaning tebranishi ustivor bo‟lgan vaqt oraligi uchun yechimni aniqlang. F(t)
= F_m cos( t) va x(t) funksiyalarning bog‟lanishing grafigini quring.
Yechish. Berilgan tenglamani analitik usulda yechib bo‟lmaydi:

• 
  restart; m:=2; beta:=1; k:=4; Fm:=3000; omega:=0.1; de:=m*diff(x(t),t$2)+beta*diff(x(t),t)+k*x(t)=abs(Fm*cos(omega*t)); dsolve(de,x(t));
  

cond:=x(0)=0, D(x)(0)=0; dsolve({de,cond},x(t));


m := 2



k := 4
Fm := 3000
:= 0.1

x( t )
de := 2
x( t )
x( t )
4 x( t )
3000
e
cos( 0.1 t )

cos e dt sin


cos e dt cos
cond := x( 0 ) 0, D( x )( 0 ) 0

x( t )
cos
e d_z1 sin

t


sin


cos


e d_z1 cos

0
Ammo uni sonli usulda yechsa bo‟ladi:

• 
  restart; m:=2: beta:=1: k:=4: Fm:=3000: omega:=0.1: de:=m*diff(x(t),t$2)+beta*diff(x(t),t)+k*x(t)=abs(Fm*cos(omega*t)):
  

with(DEtools): DEplot(de,{x(t)},t=0..50*Pi,[[x(0)=0, D(x)(0)=0]],stepsize=0.1);

2. 
   misol. Asbob blokini titrashdan himoyalash uchun unga mxsus elastik tayanchlar (amortizatorlar) o„rnatilgan. Uning amortizatorlardagi harakati yonlama va buralma tebranishlari e‟tiborga olinmaganda ushbu
   

d ^2 x dx m kx 0 d t ^2 d t differensial tenglama bilan ifodalanadi,

bunda x – blokning dastlabki holatidan chetlanishi; t – vaqt; m – blok massasi; d²x/dt² – tezlanish; – amortizatorlarning ishqalanish koeffisiyenti; dx/dt – blokning tebranishidagi harakat tezligi; kx – elastik elementlar (prujinalar)ning qarshiligini ifodalovchi had; k – amortizatorlarning bikrlik koeffisiyenti; Prujinalarning yig„indi bikrligi x – deformatsiyadan quyidagicha bog„liq: k = k₀ (1+ax²). Berilgan oddiy differensial tenglamani tenglamani = 0,5 kg/kuch, m = 12 kg, k₀ = 0,5 N/m, a = 1 1/m² boshlang„ich shartlar: t=0 da x(0) = 0 sm va dx/dt = 1 hamda quyidagi jadval ma‟lumotlari bo„yicha yeching. Tebranishning kamida beshta davrini o„zida ifodalovchi yechim nuqtalarini toping va shu oraliq uchun x(t) bog„lanishning grafigini chizing.
Yechish. Avalo bu tenglamaning umumiy yechimini analitik usulda topaylik:

• 
  restart; m:=12; beta:=0.5; k:=0.5; de:=m*diff(x(t),t$2)+beta*diff(x(t),t)+k*x(t)=0; dsolve(de,x(t));
  

m := 12
:= 0.5
k := 0.5

de := 12
x( t )
0.5
x( t )
0.5 x( t ) 0

x( t )

_C1 e
_sin 95 t
48


_C2 e


cos

Tenglamaning xususiy yechimi quyidagicha: cond:=x(0)=0, D(x)(0)=1; dsolve({de,cond},x(t));

cond := x( 0 )


x( t ) e
0, D( x )( 0 ) 1


sin

Endu bu tenglamani sonli yechaylik:

• 
  restart; m:=12: beta:=0.5: k:=0.5: de:=m*diff(x(t),t$2)+beta*diff(x(t),t)+k*x(t)=0:
  

with(DEtools): DEplot(de,{x(t)},t=0..50*Pi,[[x(0)=0, D(x)(0)=1]],stepsize=0.1);

3. 
   misol. Konsol mahkamlangan bir jinsli balkaning sof og„irligi ostida egilishi ushbu
   

3/ 2 d ^2 y PL^2 1 x dy ^2 dx^2 EJ L L^2 ^1dx ^0

differensial tenglama bilan ifodalanadi, bunda L – balkaning uzunligi; P – balkaning solishtirma og„irligi (uzunlik birligiga mos); EJ – balkaning bikrligi; x – koordinata (0<x<1); L= 1 m; PL²/EJ = 0,001. Berilgan boshlang„ich shartlar: x=0 da y=0 va dy/dx=0 hamda jadvalda ko„rsatilgan parametrlar qiymatlari uchun balkaning butun uzunligi bo„ylab y(x) yechim nuqtalarini toping.
Yechish. Berilgan tenglamani analitik usulda yechib bo‟lmaydi:

• 
  restart; a:=0.001; L:=1;
  

de:=diff(y(x),x$2)+a*(1/L-x/L^2)*(1+(diff(y(x),x))^2)^1.5=0;
cond:=y(0)=0, D(y)(0)=0; dsolve({de,cond},y(x));
a := 0.001
L := 1

de :=


y( x )


0.001 ( 1 )


y( x )
₂ 1.5

cond := y( 0 ) 0, D( y )( 0 ) 0

x

y( x )

( _z1 _z1 ² ) d_z1

Ammo bu differensial tenglamaning sjonli yechimi quyidagi natijani beradi:

• 
  restart; a:=0.001: L:=1:
  

de:=diff(y(x),x$2)+a*(1/L-x/L^2)*(1+(diff(y(x),x))^2)^1.5=0:
with(DEtools): DEplot(de,{y(x)},x=0..L,[[y(0)=0, D(y)(0)=0]],stepsize=0.1);

3.