2. Phase
• In radians
• dependent on variables
(x) and (t)
• represented with Greek
symbol: Φ (upper case
phi)
• in D(x,t) = Asin(kx±ωt+ϕ)
Φ=(kx±ωt+ϕ)
Phase
Difference• in radians
• difference between phase of two
points(positions) at the same time
• points that are in phase are an integer
multiple wavelengths apart with a phase
difference of an even multiple of π rad.
• they will have an even displacement
at all times
• points that are out of phase are an odd half-
integer multiple of wavelengths apart with a
phase difference of an odd multiple of π rad
• they will have an equal but opposite
displacement at equilibrium position
at all times
3. Deriving the Equation for
Phase Difference
• D(x,t) = Asin(kx±ωt+ϕ) is the equation for the displacement of a harmonic
wave.
• The phase difference ∆Φ is equal to the difference between the phases
at two points x1 and x2 at time t
• Therefore: ∆Φ=(kx2±ωt+ϕ)-(kx1±ωt+ϕ)
• Since time and the phase constant says the same, they will cancel out,
giving: ∆Φ=(kx2)-(kx1)
• k can be factored out, giving: ∆Φ=k(x2-x1) or ∆Φ=k(∆x)
• sometimes k is substituted with k=2π, giving:__
λ
4. Given the equation of a harmonic wave
at position x and time t:
D(x,t) = 2sin(8x-3t+ )
A. Find the phase constant
B. Find the phase at x= and time t=0s
C. Find the phase difference between two points,
x1= and x2= at time t=0s
D. Using the phase difference, find the largest
possible wavelength
E. Using the phase difference, find the
displacement when x1= at time t=0s
5. D(x,t) = 2sin(8x-3t+ )
A. Find the phase constant
The phase constant is ϕ in the function of a
harmonic wave D(x,t) = Asin(kx±ωt+ϕ).
Therefore, in the function D(x,t) = 2sin(8x-3t+ ), the
phase constant is ( ).
__π
2
6. B. Find the phase at x= and time t=0s
Since the phase in the function of a harmonic wave
D(x,t) = Asin(kx±ωt+ϕ) is equal to (kx±ωt+ϕ).
Plugging in the known variables, we have
• Φ=(kx±ωt+ϕ)
• Φ=(8(3π/8)-3(0)+π/2)
• Φ= 7π__
2
7. C. Find the phase difference between two
points, x1= and x2= at time t=0s
We can find the phase difference by using the equation for phase
difference ∆Φ=k(∆x)
∆Φ=k(∆x)
We know k=8 since in the function of a harmonic wave, k is a
constant multiplied to position x.
• ∆Φ=8(x2-x1)
• ∆Φ=8( - )
• ∆Φ=8( )
• ∆Φ=π
π__
8
8. D. Using the phase difference, find the largest possible
wavelength
The phase difference equation can be used once again to find the wavelength, now that we know that
the phase difference is π.
∆Φ=k(∆x)
Since we are trying to find wavelength, we substitute k with since k=
∆Φ=
Rearrange this to solve for λ
• λ=
• now input the known variables derived from the previous questions
• λ=
• λ=
9. E. Using the phase difference, find the
displacement when x1= at time t=0s
We have solved that ∆Φ=π and we know that
∆Φ=(kx2±ωt+ϕ)-(kx1±ωt+ϕ).
We can rearrange this into
(kx2±ωt+ϕ)=(kx1±ωt+ϕ)+∆Φ
Since D(x,t) = Asin(kx±ωt+ϕ), then we have
Asin(kx2±ωt+ϕ)=Asin(kx1±ωt+ϕ+∆Φ)
To solve for displacement D(x1,t), we can use D(x1,t)=Asin(kx1±ωt+ϕ+∆Φ), and after substitution
D(x1,t)=2sin(8( )-3(0)+ +π)
Using trig identities, we know sin(θ+π)=-sin(θ)
so D(x1,t)=-2sin(8( )-3(0)+ )
and D(x1,t)=-2sin( ) = 2
