My Physics 101 Learning Object.

1. Learning Object3:
Waves in a Wave Pool
UBC hasdecidedtoupgrade theircurrentpool witha hightechnew wave machine andDr. Batesis
taskedwithsettingupthe newmachine’ssoftware.All he hastodois inputa wave functionandthe
fancymachine will create the waves. The swimmingpool is50mlongand the wave machine will be
installedonthe leftedge of the pool soyoucan assume thatX₀=0 andincreasesfromlefttoright.You
can alsoassume that the energyof the wavesis only dissipatedatthe farendof the pool so the waves
do notlose any amplitude astheygetclosertothe endof the pool.
Dr. Batessetsup the machine to produce simple harmonic
wavesthatlooklike this snapshotgraph:
a) If Dr. Bateswants the wavestohave a wave speedof
2.0 m/s,whatwave functionwouldgive Dr.Bateswaves
that looklike thisforanytime,t,assumingthatthe waves
picturedare at t = 0?
b) What wouldthe maximumpositivevertical segment
velocityof the wavesbe inthe pool?
c) Wave pool enthusiastFirazbelievesthatthe wavesinthe pool are not highenough,sohe adjuststhe
settingstogive a newmaximumcrestheightof 2 m.What is the new maximumvertical velocity of the
waves?
d) Unfortunately,the newwavesinthe pool have avertical segmentvelocitythatiswaytoohighfor
some swimmers.Luckily,Dr.Rottlerisrelaxinginthe whirlpoolwhenhe noticesthisanddecidesto
adjustthe settingstofix it.If he wants to keepthe amplitude the same (2m),increase the wave length
to 4 m,and lowerthe maximumvertical velocitytoπ m/s, whatwave equation musthe inputintothe
wave machine tofix it?
Thisis the formof the wave equationthatthe wave machine accepts:D(x,t) =Asin(kx - ωt+ φ)

2. Solutions
a) First,we needtofindthe constantsthat describe the wave fromthe position graph. Youcan see that
at the topof a crest the displacementis1m, so that isthe amplitude,A.The wavelength,λ,isdistance of
one full wave,whichyoucansee is2m.
You can alsosee the startingpointof the graph,whichwill give usthe phase constant(φ).If the
displacementat(x,t) =(0,0) is 0, thensin(φ) mustbe equal tozero.Thishappensatsin(0), givingusour
phase constant, φ = 0.
We needtoalsofindthe wave constant,k,and the angular frequency,ω.Tofindk,we will use the
formula:
K = 2π/λ rad/m  k = 2π/2m  k = π rad/m
To findω, we will use the formula:
v = ω/k 2.0 m/s = ω/π rad/m  2.0 * π = ω  ω = 2π rad/s
Nowwe can plugin all of our valuesintothe wave function!Leavinguswithan answerof:
D(x,t) = 1 * sin(πx– 2πt)
b) The formulaforthe vertical velocityof awave segmentis:V(x,t) =(-ωA)cos(kx - ωt+ φ)
We knowwe wantthe maximumvelocity,socos(kx - ωt + φ) mustequal 1. We alsoknow that thiswave
has harmonicmotionandthat since we wantthe positive velocitythatthe signmustbe positive,leaving
us with: Vmax(x,t) = ωA
From part a, we can see that the amplitude is1m and the angularfrequencyis 2π rad/s.
Puttingthese valuesbackinthe Vmax formulaleavesuswithamaximumvelocityof 2π m/s,whichis
roughly 6.3 m/s
c) The onlythingthathas changedisthe amplitude,sotofindthe new Vmax all we needto dois use the
same formulafrompart b withthe newamplitude.
Vmax = 2 * 2π  Vmax = 4π m/s  12.6 m/s
d) For thispart,we needtofindthe newA,k, andω valuesforthe wave equation.We know A is2m, so
to findk andω we will use similarformulasasparta.
Firstwe will findk:
k = 2π/λ rad/m  k = 2π/4  k = π/2 rad/m
To findω, we will needtodosomethingdifferent.We know thatVmax mustbe π m/s,so knowingwhat
the amplitude iswe will use the formula:
Vmax = ωA = π m/s  ω = π/2 rad/s
Puttingthese newvaluesintothe wave functionwill provide use withouranswer:
D(x,t) = 2sin(π/2*x - π/2*t)