Question

QUESTION 9 15 p Find a square root of 7 using bisection method using a prescribed absolute relative approximation error, Es - 10%. Using lower bound xy - 2 and upper bound xu - 3, a few iteratively calculated parameters are given in Table below. Iteration Number Axpm) Ea, % What is the Im when the absolute relative error is below 10%? Note: to receive the credit, the final numeric answer can be within the range of plus/minus 0.1. QUESTION 10 20 points Displacement of undamped spring, x, is modelled as 2nd order Ordinary Differential Equation (ODE) given below. Solve it using Finite Difference Method (FDM), A1-1/4s, and given initial conditions below. mi + kx0, x(-0)-0.1. and X - 0) = 0 1. Index Number 11/4 n /2 31/4 What is the displacement at 314, L. ? Notes to receive the credit, the final numeric answer can be within the range of plus/minus 0.1. Click Save and submit to save and submit. Click Save All A t o e alla Save All Answers

Answer 1

Using bisection method and doing iteration we can solve question no. 9. I have shown below work and table-
2 - Sol: we have to find square root of 7 using 'Bisection method . > let, n=55 si f(2)=4- 7 = -3 E-ve ov, a-720 f(3) = 32732 = +ve So, f(n) = n²_7 So, root lies b/w 2 &3. using Bisection method ,am= net . where he = lower bound. х = ИРРеут km = middle " Now, Absolute relative érror is given by €= 1 km, it -km, il alm, ito I - Where, ami= midpoint in the previous root - search iteration km.iti = midpoint in a new root - search S operation. Iteration Number o :- X100 2 2. km = dettu = 2 +3 = 2. Now, f(25) = 2.5?7=-0.95 = - ve sa, root lies between 2.5 f3 1::$(2-5) =-ver [ +(3) = tres . Iteration number li- Xm = 2.5+3 = 2.75 So, fr = 1207575 12.3 xloo - 9.09% 2.75 -2.51

$(2775) = 2.75?. 7 = 0.5625 - +ve so, root lies blu 2.5 and 2.75. o Iteration Number 2 :- Wm= 2.5 +2.25 - 2.625 $(2.625) = 2.625 - 7 = -0.11 = -ve so, root lies b/w 2.625 of 2.75. Er = 1 2625–275 x 100 = 4.76%. 2.625 e Iteration number 3 :- um - 2.625 +2.75 - = 2-6875 2 $(2.6875) = 0.22 = tre so, root lies b/w 2:625 and 2.6875 12.6875-2.625 vioo 1. E=12.6875-2.62 2-327, 2.6875 @ Iteration number 4 :- Non = 2.625+2.6875 = 2.6562 $(2-4562) = 0.055 = tve so, root lies b/w 2.625 & 2.6562 Erol 2.6562-2.6875 l. 1 x 100 = 1.18% 2.6562 • Iteration number Si Xm = 2.625+ 2.6562 2_ - = 2-6406 f(2-6400) = -0.025 = -ve

so root lies b/w 2-6406 and 2.6562 Eyal 26406 - 2.6562 2.64061 *100 = 0.59 Iteration number 6 :- um= 2.6406+ 2.6562 - = 2-6484 2 $12.6484) = 0.014 = tve so, root lies between 2-6406 4 2.6484 2-6484 - 2.6406 Ey = 1 2.64841x100 = 0.29%, Iteration number 7 :- um = 2.6406 42-6484 = 2.6445 +(26445) = -ve so root lies b/w 2.6445 4 2.6484 Er | 276445-2.6484 I- 216445- x100 = 0.14% 0 Iteration number 8 :- alm = 2.6445+2.6484 2 = 2.646 f 2.646 +(2-646) = tve so, root lies b/w 2.6445 Era 2.696-2.6445 po 2.646 Iteration number 9 :- IX - 0*05% am= 2.6445+ 20646 2

2.645- Ey = 12001322505 |x100 = 0.03%. $(2.645) = -ve : So, root lies b/w 2.645 42.646 Iteration number 10 :- Am = 21645+2=646 = 2.645 fra 12-645572675 1x100 = 0 so, root of 7 = 20645) a La Ans. Now, if we fill up the table then it will be - 2.645 Square dom Iteration Number If (ale)xf(lm) 3 1 2.5 tre 9.09 4.76 -ve 13 2175 -ve | ais 2,75 2.625 tve 2-625 2.75 2.6875 -ve 2.625 1. 2.6875 2.6562 2-625 2.6562 2.6406 2.6406 2.6562 2.6484 -ve 12.6406 1 2.6484/2.6445 | tve 2-6445 2.6484 2.646 12.6445 2.646 2.645 tve 2.32 1:18 0.59 0. 2g tve -ve 0.14 oios 0:03 10 2.645 1 20646 2.6451 tve 0.00