1. R
ϕ
Simple HarmonicMotion – DeterminingPhase Constant
Phase constant (ϕ) isthe angle inradiansthat is dependentonbothpositionandvelocity
Position - accountsfora shiftor howmuchout of place the functionis
Velocity - accountsforwhetherϕispositive ornegative
Sample Question
Points1 and 2 are on a circularwheel withradius (R).Att=0,point1 isat pointA and point2 crosses
pointA 2 secondslater. The wheel isspinningcounterclockwiseat1/12 rev/s. Determinethe phase
constantfor the functionsof the positionof point1and 2 overtime.How far was point2 initiallyoutof
phase withpoint1?
A
Point1
Point2
2. Soution:
Uniformcircularmotionviewed fromthe side issimple harmonicmotion.
Positionx(t)=Acos(ωt+ϕ)
ω=angular frequency=2π/T
ϕ=phase constant
f=frequency =1/Twhere T isthe period
Centervertical of circle isthe equilibiumpoint
R = Amplitude (A)
f = 1/12 = 1/T; So,T=12
ω = 2π/T = 2π/12 = π/6
x(t)=Acos(π/6t+ ϕ) *missingphase constant
x(t)=0
POINT1:
point1 isat the maximum positionatt=0
x1(t)=Acos(ϕ) *plugginginfort=0
At thispointx(t)=A,so cos(ϕ)=1
we know cos(0)=1 so ϕ=0
Puttingittogether:x1(t)=Acos((π/6)t)
POINT2:
It takespoint2 2 secondsto reachthe maximumposition
x2(t)=Acos((π/6)2+ϕ) *pluggingint=2
At t=2, x(t)=A socos(π/6)2+ ϕ)=1
we know cos(0)=1 so (π/6)2+ ϕ=0
π/3+ ϕ=0
ϕ = - π/3
Puttingittogether:x2(t)=Acos((π/6)t-π/3)
To visualize,the graphsof the positionfunctionsare shownbelow (the amplitude isarbitrary;4was
chosen):
x1(t)=
x2(t)=
ϕ1 = 0 and ϕ2 = -π/3
Therefore,point2 is initially
π/3 out of place from point 1
Max x(t) +AMin x(t) -A
3. ϕ
ϕ
Note that in the positionpoint 2, phase constant isnegative
v<0 towardsEQ
v>0 towardsmax position
v=0 v max v=0
v > 0
v < 0
Since the directionof the velocityof the pointischangingasitmovesaroundthe circle,the signof the
phase constantis determinedbythe direction(right=positivedirection).
-Whenvelocity>0(movingtothe maximumposition), ϕ<0
-Whenvelocity=0, ϕ=0
-Whenvelocity<0(movingtothe minimumposition), ϕ>0