MATH 101 - Section 203 - Lecture Notes - Week 9avners/courses/documents/MATH101W2018/Week… · you...

MATH 101 - Section 203 - Lecture Notes - Week 9

Instructor: Avner Segal (avners@math.ubc.ca)

Applications of Integration

Work

Example (Final, 2012)A tank in the shape of a hemispherical bowl of radius R = 3 m is fullof water. It is to be emptied through an outlet extending H = 2 mabove its top. Using the values ρ = 1000 kg/m3 for the density ofwater and g = 9.8 m/s2 for the acceleration due to gravity, find thework (in Joules) required to empty the tank completely. There is noneed to simplify your answer but you must evaluate all integrals.

SolutionWe "slice" the water into horizontal slabs perpendicular to they -axis. Each slab has a circular cross-section.

Solution Continued

The slab at height y above the bottom has distance R −y from thecenter of the sphere. Here, 0≤ y ≤R . The radius of the slab is thenr =

√R2− (R −y)2 =

√2Ry −y2. Since we consider slabs of width

dy , the volume of this slab is πr2dy =π(2Ry −y2)dy [m3]. Weconclude that the slab of water is weighting πρ(2Ry −y2)dy [kg ].

Solution Continued

A slab of water weights dm=πρ(2Ry −y2)dy [kg ]. The work doneby lifting this amount of water H −y meters up is

dW = (H +R −y)gdm=πρ(H +R −y)g(2Ry −y2)dy .

Recalling that R = 3 [m], H = 2 [m], ρ = 1000 [kg/m3] andg = 9.8 [m/s2] and integrating yields

W =∫ 0

−Rπρ(H +R −y)g(2Ry −y2)dy = ... = 9,800 ·π · 225

4[J]

Example - Hook’s LawA spring has a natural length of 20cm If a 25N Force is required tokeep it stretched at a length of 30cm, how much work is requiredto stretch it to a length of 35cm?

SolutionHook’s law states that an ideal spring has a constant k such thatthe force required to keep the spring stretched x meters from itsnatural length is

F = k ·x ← NOT A CONSTANT FORCE!

We first calculate the springs constant.

x = .3− .2= .1m, F = 25N ⇒ k = F

x= 250

N

m.

It follows that work put into stretching the spring 5cm more wouldbe:

W =∫ .35

.3F (x)dx =

∫ .35

.3250 ·x dx = 125x2

∣∣∣.35

.3= 4.0625J .

Example - GravityAccording to Newtorn’s universal law of gravitation, the forcebetween a planet of mass M and a probe of mass m is F = GMm

r2,

where r is the distances between them and G ≈ 6.67 ·10−11 m3

kg ·sec2 isthe gravitational constant. Find the work required to lanuch a probefrom the surface of the planet (radius R) all the way to infinity.

SolutionThe work is

W =∫ ∞

RF (r)dr =

∫ ∞

R

GMm

r2 dr =GMm

[−1r

]∞r=R

= GMm

R.

Escape VelocityIn order to get to infinity, the probe should start with a velocityv > v∞, where

12mv2

∞ = GMm

R⇒ v∞ =

√2GMR

Example (Final, 2010)A colony of ants builds an anthill that is in the shape of a conewhose base, at ground level, is a circle of diameter 1 m and whoseheight is also 1 m. How much total work, in Joules, is done by theants in building the anthill? For the density of sand, use the value1,920 kg/m3.

SolutionOnce again, we use the circular cross-section at height x above theground.

Solution Continued

Using similar triangles, the radius r of the cross section satisfy1−xr = 1

1/2 , i.e. r = 1−x2 . The volume of this cross-section is

πr2dx =π(1−x)2

4 dx . It follows that this slice of sand weightsdm= 480π(1−x)2dx . The work done to lift it isdW = gxdm= 480 ·9.8 ·πx(1−x)2dx . Integrating yields∫ 1

0480 ·9.8 ·πx(1−x)2dx = ... = 480 ·9.8

12π= 392π.

Averages

Average of n numbersThe average (mean value) of a sequence of numbers y1, ...,yn isgiven by

yaverage = y = ⟨y ⟩ = y1+ ...+ynn

.

Now consider a Riemann sum for f (x)

b−a

n

n∑i=1

f (x∗i )= (b−a)⟨f (x∗i )⟩.

So the following makes sense:

Average of a functionLet f (x) be an integrable function on [a,b]. The average value of fon the interval is

faverage = f = ⟨f ⟩ = 1b−a

∫ b

af (x)dx .

ExampleThe average of f (x)= x2 over [1,5] is

⟨f ⟩ = 15−1

∫ 5

1x2dx = 1

4· 5

3−13

3= 31

3.

ExampleThe average of f (x)= sinx over [0,π/2] is

⟨f ⟩ = 1π/2−0

∫ π/2

0sinxdx = 2

π(−cos(π/2)+cos(0))= 2

π.

Example (Final 2012)Let k be a positive constant. Find the average value off (x)= sin(kx) on [0,π/k].

Solution

faverage = 1π/k

∫ π/k

0sin(kx)dx = k

π

[−cos(kx)k

]π/k0

= 2π

Example - Average VelocityGiven a moving object whose position is given by x(t) and velocityis given by v(t)= x ′(t) the average velocity is

vaverage = 1b−a

∫ b

av(t)dt = 1

b−a

∫ b

ax ′(t)dt = x(b)−x(a)

b−a

due to FTC.

Example (Final 2010)Find a number b > 0 such that the function f (x)= x −1 hasaverage value 0 on the interval [0,b].

Solution

faverage = 1b

∫ b

0(x −1)dx = 1

b

(b2

2−b

)= b

2−1

We solve b2 −1= 0 for b to get b = 2.

Weighted AvergaeThe average (mean value) of a sequence of numbers y1, ...,yn withweights w1, ...,wn > 0 is given by

⟨y |w ⟩ = w1y1+ ...+wnynw1+ ...+wn

.

It is customary to take w1+ ...+wn = 1 so that

⟨y |w ⟩ =w1y1+ ...+wnyn.

The usual average is given by w1 = ... =wn = 1, or otherwisew1 = ... =wn = 1

n .

ExampleGiven two school classes, one with 20 students, and one with 30students, the grades in each class on a test were:

Morning class (20 students) = 62, 67, 71, 74, 76, 77, 78, 79, 79,80, 80, 81, 81, 82, 83, 84, 86, 89, 93, 98

Afternoon class (30 students) = 81, 82, 83, 84, 85, 86, 87, 87, 88,88, 89, 89, 89, 90, 90, 90, 90, 91, 91, 91, 92, 92, 93, 93, 94, 95,96, 97, 98, 99

The average of the morning class is 80 and the average of theafternoon class is 90. One can calculate the average of all studentsin two ways:

Ï Directly.Ï Weighted average

20 ·80+30 ·9020+30

= 86.

Center of Mass

Center of MassIf you support a body at its center of mass (in a uniformgravitational field) it balances perfectly. That’s the definition of thecenter of mass of the body.

If the body consists of a finite number of masses m1, ...,mn

attached to an infinitely strong, weightless (idealized) rod withmass number i attached at position xi , then the center of mass isat the weighted average value of x:

xc .m. =∑n

i=1mixi∑ni=1mi

ExampleThe mass of the Earth is about 6 ·1024 kg . The mass of the Moonis about 7.2 ·1022 kg . The distance between the centers of theEarth and the Moon is 3.8 ·105 km. Where is the center of mass ofthe Earth–Moon system? (aside: the radius of the Earth is about6400 km).

ExampleThe mass of the Earth is about 6 ·1024 kg . The mass of the Moonis about 7.2 ·1022 kg . The distance between the centers of theEarth and the Moon is 3.8 ·105 km. Where is the center of mass ofthe Earth–Moon system? (aside: the radius of the Earth is about6400 km).

For simplicity, let’s put the Earth center at xEarth = 0 andxMoon = 3.8 ·105 km. We also put MEarth = 6 ·1024 kg andMMoon = 7.2 ·1022 kg . The center of mass of the system is at

xc .m. = xEarth ·MEarth+xMoon ·Mmoon

MEarth+Mmoon

= Mmoon

MEarth+Mmoon·xMoon ≈ 4.5 ·103 km.

So the center of mass is inside the Earth’s crust.

ExampleA tenderizing hammer consists of a 1 kg head attached to a30 cm-long shaft massing 400 g .

(a) Find the center of mass of the hammer.

Let the axis run along the shaft, starting at the handle. Thecenter of mass of the shaft is 15 cm in, so the center of massof the hammer is at

xc .m. = 0.4 ·15+1 ·300.4+1

= 1807

≈ 25.7 cm.

ExampleA tenderizing hammer consists of a 1 kg head attached to a30 cm-long shaft massing 400 g .

(a) xc .m. = 1807 ≈ 25.7 cm

(b) What fraction of the mass of the hammer is on each side of thecenter of mass? To the left of the center of mass there is180/7

30 = 67 of the shaft’s mass, i.e. 6

7 ·0.4≈ 342 gr .

In particular only67 ·0.41.4

1249 ≈ 0.25 of the mass (1.4 kg) of the

hammer is to the left of the center of math.Conclusion (and a WARNING): It is not true that there is halfof the mass on either side of the center of mass!.

ExampleThree masses are placed at the points (−1,0) , (1,0) , (0,5). Findthe center of mass of the configuration.(a) When the masses are equal.

Denote the mass of each point by m.

xc .m. =m · (−1)+m ·1+m ·0

m+m+m= 0

yc .m. = m ·0+m ·0+m ·5m+m+m

= 53

So the center of mass is at(0, 5

3).

(b) When the mass at (−1,0) is twice as much as the others.Denote by m the mass at the points (1,0) and (0,5). The massat (−1,0) is 2m. The center of mass is given by

xc .m. =2m · (−1)+m ·1+m ·0

2m+m+m=−1

4, yc .m. = 2m ·0+m ·m+m ·5

4m= 54

.

So the center of mass is at(−1

4 , 54).

ExampleCalculate the center of mass of the semicircle bounded by thecurves y2+x2 = 1 and y = 0.

It is assumed (implicitly in the question) the mass of the semicircleis spread evenly; this is called a constant mass density.Let m (in kg) be the mass density, i.e. the mass of a unit area (i.e.the mass of a square with area 1 m2). Note that the mass of thesemicircle is given by M =m ·A, where A is its area.Also note that since the semicircle is symmetric about the y -axis, itis obvious that xc .m. = 0.

Solution

We approximate the center of mass using Riemann sums. Let f (y)be the width of the semicircle at y , i.e. f (y)= 2

√1−y2. The

center of mass of the rectangles is given by:

n∑k=1

y∗k ,n

(m1nf (y∗

k ,n)

)n∑

k=1

(m1nf (y∗

k ,n)

) =

1n

n∑k=1

y∗k ,nmf (y∗

k ,n)

1n

n∑k=1

mf (y∗k ,n)

.

Taking the limit as n→∞ yields

yc .m. = limn→∞

1n

n∑k=1

y∗k ,nmf (y∗

k ,n)

1n

n∑k=1

mf (y∗k ,n)

=

∫ 1

0myf (y)dy

M=

∫ 1

0myf (y)dy

M.

Now we integrate∫ 1

0ymf (y)dy = 2m

∫ 1

0y√1−y2dy = 2m

[−(1−y2)

3/2

3

]1

0

= 2m3

.

Now, recalling that M =m ·A and A= π·12

2 = π2 we find that

yc .m. = 2m/3mπ/2 = 4

3π .

In general, the center of mass of a semicirlce of radius r will be at4r3π of its central radius.

A list of centers of mass of simple shapes:https://en.wikipedia.org/wiki/List_of_centroids, youshould try to do as many of them as you can at home!

Center of MassReminderThe center of mass of pointed masses m1,...,mn located at x1,...,xn(respectively) is given by

xc .m. =∑n

i=1mixi∑ni=1mi

In what follows, we use this to compute the center of mass ofregions in the plane with a constant mass density. In this case, thecenter of mass, is also called centroid or geometric center.

ExampleCalculate the center of mass of the semicircle bounded by thecurves y2+x2 = 1 and y = 0.

It is assumed (implicitly in the question) the mass of the semicircleis spread evenly; this is called a constant mass density.Let m (in kg) be the mass density, i.e. the mass of a unit area (i.e.the mass of a square with area 1 m2). Note that the mass of thesemicircle is given by M =m ·A, where A is its area.Also note that since the semicircle is symmetric about the y -axis, itis obvious that xc .m. = 0.

Solution

We approximate the center of mass using Riemann sums. We chopthe shape along the y -axis into rectangles. The width of thesemicircle at height y is 2

√1−y2 so the area of each rectangle is

1n2

√1−y∗ 2

k ,n and the mass of each rectangle is m 1n2

√1−y∗ 2

k ,n .

Let f (y) be the width of the semicircle at y , i.e. f (y)= 2√1−y2.

The center of mass of the rectangles is given by:

n∑k=1

y∗k ,n

(m1nf (y∗

k ,n)

)n∑

k=1

(m1nf (y∗

k ,n)

) =

1n

n∑k=1

y∗k ,nmf (y∗

k ,n)

1n

n∑k=1

mf (y∗k ,n)

.

Taking the limit as n→∞ yields

yc .m. = limn→∞

1n

n∑k=1

y∗k ,nmf (y∗

k ,n)

1n

n∑k=1

mf (y∗k ,n)

=m

∫ 1

0yf (y)dy

M=m

∫ 1

0yf (y)dy

M.

Now we integrate

m

∫ 1

0yf (y)dy = 2m

∫ 1

0y√1−y2dy = 2m

[−(1−y2)

3/2

3

]1

0

= 2m3

.

Now, recalling that M =m ·A and A= π·12

2 = π2 we find that

yc .m. = 2m/3mπ/2 = 4

3π .

Alternatively, taking the limit as n→∞ yields

yc .m. = limn→∞

1n

n∑k=1

y∗k ,nf (y

∗k ,n)

1n

n∑k=1

f (y∗k ,n)

=

∫ 1

0yf (y)dy

A.

Solving the integral as before yields yc .m. = 2m/3mπ/2 = 4

3π .

In general, the center of mass of a semicirlce of radius r will be at4r3π of its central radius.

A list of centers of mass of simple shapes:https://en.wikipedia.org/wiki/List_of_centroids, youshould try to do as many of them as you can at home!

Question (Final 2013)Find the centroid of the region below, which consists of a semicircleof radius 3 on top of a rectangle of width 6 and height 2.

SolutionThe region R consists of a semicircle of radius 3 on top of arectangle of width 6 and height 2. The shape is symmetric aboutthe y -axis and so xc .m. = 0. We now compute yc .m..

Solution IThe region R consists of a semicircle of radius 3 on top of arectangle of width 6 and height 2. The center of mass of thesemicircle is at y = 4·3

3π and the center of the rectangle is at y =−1.The mass of the semicircle is π32

2 m (where m is the mass density)and the mass ofthe rectangle is 6 ·2 ·m. We conclude that

yc .m. =4π · 9πm

2 + (−1) ·12m9πm

2 +12m= 48+3π

.

Solution II

We will now calculate yc .m. by integration again, but instead ofchopping the shape along the y -axis, we will chop it along the xaxis.

Solution II

Here, we will average of the centers of mass of the rectangles. Foreach of these rectangles

Ï Its area is given by 6n ·

(√9−x2

k ,n− (−2)).

Ï Its mass is given by m · 6n ·

(√9−x2

k ,n− (−2)).

Ï Its (y-coordinate of its) center of mass is located at√9−x2

k ,n+(−2)

2 .

The center of mass of all these rectangles together is

n∑k=1

√9−x2

k ,n−2

2·m · 6

n·(√

9−x2k ,n+2

)n∑

k=1m · 6

n·(√

9−x2k ,n+2

) =6n

n∑k=1

[(9−x2

k ,n)−4]

2 · 6n

n∑k=1

√9−x2

k ,n+2.

Taking the limit n→∞, the denominator goes to 2 ·A= 24+9π,while the numerator goes to∫ 3

−3(5−x2)dx = 12.

This yields

yc .m. = 48+3π

.

More GenerallyThe center of mass of the region (with area A){

(x ,y)∣∣a≤ x ≤ b, f (x)≤ y ≤ g(x)

}between the curves y = f (x) and y = g(x) (with a constant massdensity) is given by

xc .m. =

∫ b

ax (g(x)− f (x))dx

A

yc .m. =

∫ b

a

(g(x)2− f (x)2

)dx

2A.

Notice that the units here fits!

Example (Last year quiz v.M1)Express the x–coordinate of the centroid of the triangle withvertices (−1,−1), (−1,1), and (1,0) in terms of a definite integral.Do not evaluate the integral.

SolutionThe equations of the top and bottom of the triangle are

y = f (x)= 1−x

2, y = g(x)= x −1

2.

The area of the triangle is A= 12 ·2 ·2= 2 and hence

xc .m. = 1A

∫ 1

−1x (f (x)−g(x))dx = 1

2

∫ 1

−1x(1−x)dx=−1

3.

ExampleCalculate the centroid of the area{

(x ,y)|0≤ x , x2 ≤ y ≤ 6−x}

SolutionThe intersection point between the curves y = x2 and y = 6−x isx = 2 with y = 4. Using the formulas for centroids we get

xc .m. =∫ 20 x(6−x −x2)dx∫ 20 (6−x −x2)dx

= 16/322/3

= 811

yc .m. =12∫ 20 ((6−x)2− (x2)2)dx∫ 2

0 (6−x −x2)dx= 664/15

22/3= 166

55.

ExampleCalculate the centroid of the area{

(x ,y)|0≤ x , |y | ≤ e−x}

SolutionUsing the formulas for centroids we get

xc .m. =∫ ∞0 x(e−x − (−e−x))dx∫ ∞0 (e−x − (−e−x))dx =

∫ ∞0 2xe−x dx∫ ∞0 2e−x dx

= 22= 1

yc .m. =12∫ ∞0 ((e−x)2− (−e−x)2)dx∫ ∞0 (e−x − (−e−x))dx = 0∫ ∞

0 2e−x dx= 0.

ExampleCalculate the centroid of the area{

(x ,y)|0≤ y ≤ e−x2}

SolutionUsing the formulas for centroids we get

xc .m. =∫ ∞−∞ xe−x

2dx∫ ∞

−∞ e−x2dx

= 0 (xe−x2is odd)

yc .m. =12∫ ∞−∞ e−2x2

dx∫ ∞−∞ e−x2

dx=

{x = xp

2

}= 12p2

∫ ∞−∞ e−z

2dz∫ ∞

−∞ e−x2dx

= 12p2

.

Aside,∫ ∞−∞ e−x

2dx =p

π.

Separable Differential EquationsSome examples of "real-life" differential equations:

Newton’s Law of Motion describes motion of particlesMaxwell’s equations describes electromagnetic radiation

Navier–Stokes equations describes fluid motionHeat equation describes heat flowWave equation describes wave motion

Schrödinger equation describes atoms, molecules and crystalsStress-strain equations describes elastic materialsBlack–Scholes models used for pricing financial optionsPredator–prey equations describes ecosystem populationsEinstein’s equations connects gravity and geometry

Ludwig–Jones–Holling’s equation models spruce budworm/Balsam fir ecosystemZeeman’s model models heart beats and nerve impulses

Sherman–Rinzel–Keizer model for electrical activity in Pancreatic β–cellsHodgkin–Huxley equations models nerve action potentials

In this course, we will only look at only the simplest examples ofdifferential equations. Try reading the optional sections in the CLPnotes.

Example - Introduction IConsider the differential equation y ′ = 3y2. We are looking forsolutions of such equations.

Ï For which values of C and D is y(x)=CxD solution of theequation? Assume it is. Then

y ′(x)=C ·DxD−1 != 3y2(x)= 3(CxD

)2 = 3C 2x2D .

If C = 0 then this is a solution (admittedly, not an interestingone but it is important to remember the boring solution...) i.e.y(x)= 0. Otherwise, it follows that C ·D = 3C 2 andD−1= 2D. Hence D =−1 and C =−1

3 . So y(x)=− 13x .

y ′ = 13x2 = 3

(− 13x

)2 = 3y2 .

Example - Introduction IIConsider the differential equation y ′ = 3y2.

Ï Suppose y = f (x) is a solution. Show that y = f (x −a) is alsoa solution for any a. What is the solution with f (0)= 1?The assumption means that f ′(x)= 3f (x)2 for any x . Hence

[f (x −a)]′ = f ′(x −a)= 3 · (f (x −a))2

fo any x .Ï Applying part I, we see that for any a the following is a solution

f (x)=− 13(x −a)

.

We solve f (0)= 1 for a.

13a

= 1⇔ a= 13

.

We conclude that the solution we look for is y = 11−3x .

Seperable Differential EquationsA separable differential equation is an equation for a function y(x)of the form

dy

dx(x)= f (x) ·g(y(x)).

We look for solutions y(x) for this equation.Given such an equation, we write (and this just abstract nonsense)

dy

g(y)= f (x)dx .

Integrating, we get ∫dy

g(y)=

∫f (x)dx .

Miraculously, this will give all solutions to the equationdydx (x)= f (x) ·g(y(x)).

ExampleSolve the separable differential equation y ′ = x3.

SolutionWe write dy

dx = x3. Using the idea in the previous slide, we write∫dy =

∫x3dx

Solving both sides,we have

y = x4

4+C .

ExampleSolve the separable differential equation y ′ = x3 such that y(1)= 1.The condition y(1)= 1 is called initial condition.We consider the general solution we found y(x)= x4

4 +C . Pluggingin the initial condition yields

1= y(1)= 14+C .

It follows that C = 34 and conclude that

y(x)= x4+34

.

QuestionsSolve the following separable differential equations:(a) y ′ = 5y .(b) (Final 2012) y ′ = xy , y(0)= e.

(c) (Last years quiz) dydx =−xy3, y(0)=−1

4 .

(d) (Final 2014) x dydx +y = y2, y(1)=−1.

Solution (a) - y ′ = 5y

dy

dx= 5y

∫dy

y=

∫5dx

log |y | = 5x +C

By continuity, either y(x)> 0 for all x or y(x)< 0 for all x .Ï If y > 0 then y = e5x+C =De5x (here D > 0).Ï If y < 0 then y =−e5x+C =De5x (here D < 0).

In both cases, we can write the solution as y(x)=De5x , D (and itssign) is decided by an initial condition.Note that y ′ = 5De5x = 5y .

Solution (b) - y ′ = xy , y(0)= e

dy

dx= xy

∫dy

y=

∫xdx

log |y | = x2

2+C

So y(x)=±e x22 +C =De

x22 , where D =±eC . Plugging in the initial

condition yields e = y(0)=De0 =D so

y(x)= ex22 +1 .

Indeed, y(0)= e and y ′ = xy .

Solution (c) - dydx =−xy3, y(0)=−1

4∫dt

y3 =−∫xdx

− 12y2 =−x2

2+C

y2 = 1x2−2C

y(x)=± 1px2−2C

−14= y(0)=± 1p−2C

, C =−8

y(x)=− 1px2+16

Indeed, y(0)=−14 , y

′ =−xy3

Solution (d) - x dydx +y = y2, y(1)=−1

∫dy

y2−y=

∫dx

x∫dy

y2−y=

∫dy

y −1−

∫dy

y= log

∣∣∣∣y −1y

∣∣∣∣+C

log∣∣∣∣y −1

y

∣∣∣∣= log |x |+C ⇒∣∣∣∣y −1

y

∣∣∣∣= eC |x |

y −1y

=Dx ⇒ y −1=Dxy ⇒ y (1−Dx)= 1

y(x)= 11−Dx

⇒ −1= y(1)= 11−D

⇒ D = 2

y(x)= 11−2x

Check that this is a solution!

Example - Harmonic OsciliatorA physical system satisfies the equation E = 1

2mv2+ 12kx

2, wherem, k and E are constants (mass, spring constant, energy,respectively) and v = dx

dt is the velocity. We solve for x(t).

1. Note that dxdt = v =

√1m · (2E −kx2)=

√2Em ·

√1− kx2

2E .

2. We solve ∫dx√

2Em ·

√1− kx2

2E

=∫dt

3. We make the change of variables z =√

k2E x , with

dz =√

k2E dx , to get

∫dx√

2Em ·

√1− kx2

2E

= 1√2Em ·

√k2E

∫dzp1−z2

= arcsin(z)√k/m

+C .

4. z = sin(√

km t+C

)⇒ x(t)=

√2Ek sin

(√km t+C

)

Let’s understand this solution

E = 12mv2+ 1

2kx2

x(t)=√

2Ek

sin

√k

mt+C

This is called an harmonic oscillator. Note that this is a periodicfunction with period T = 2π

√mk .

https://upload.wikimedia.org/wikipedia/commons/9/9d/Simple_harmonic_oscillator.gifhttps://universe-review.ca/I15-71-classical.gif

When√

km t0+C = π

2 we have x(t0)=√

2Ek , this is the maximal

distance between the object and the origin. This is when v = 0 andhence E = 1

2kx2.

The role of C in the solution is to fix a moment at which the objectpasses through the origin.