Learning object 1
- 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