1.15 Linear superposition applied to two harmonic waves

1. 1.15 Linear superposition applied to two harmonic waves • Can use the idea of linear superposition to understand • Standing waves • Beats in sound • Interference effects in light • Will consider two harmonic waves, y1(x,t) and y2(x,t) of equal amplitude A. • y1(x,t) = Asin(k1x-w1t + 1) • y2(x,t) = Asin(k2x-w2t + 2)

2. 1.15 Linear superposition applied to two harmonic waves • From principle of linear superposition we have • yt(x,t) = y1(x,t) + y2(x,t) • yt(x,t) = Asin(k1x-w1t + 1) + Asin(k2x-w2t + 2) • To simplify let F1 = k1x - w1t + 1 and F2 = k2x - w2t + 2 • So yt(x,t) = Asin(F1) + Asin(F2) • This can be written as

3. 1.15 Linear superposition applied to two harmonic waves • For two harmonic waves interacting at the same point and time the resulting displacement is • From this equation it is evident that the phase relationship between the two waves affects the resulting displacement. • Understanding the phase relationship explains • Standing waves • Beats in sound • Interference in light

4. 1.16 Standing waves on a string • Consider a perfect string. Two harmonic waves are excited on the string, each with the same amplitude A, propagating with same frequency w. One wave travels left to right, the other right to left. • Here F1 = kx - wt + 1 and F2 = kx + wt + 2 • So F1 + F2 = 2kx + (1 + 2) F1 - F2 = -2wt + (1 - 2)

5. 1.16 Standing waves on a string • If we let 1 = 2 = 0 we get • From this equation it can be seen that the time and space coordinates have been separated. • The time and space coordinates now influence the displacement of the wave independently. • This is the equation for a standing wave

6. ---Resultant Initially the two waves are in phase. So from principle of linear superposition the two waves add together to give maximum displacement. Time advances and the two waves move in opposite directions. A phase difference occurs between the waves so the displacement due to each individual wave is different. Waves continue to move away from each other. The phase difference increases so net displacement is reduced.

7. Waves continue to move away from each other. The phase difference increases reaches π so peak of one wave overlaps trough of the other wave. The net displacement is zero. ---Resultant The two waves keep moving opposite and the phase difference increases so the peak and trough no longer overlap. Hence the resulting displacement increases. Waves continue to move away from each other and reach a point where the phase difference is 2π again leading to a maximum in the displacement

8. Key Features of Standing Waves Although the component waves move in opposite directions the resultant standing wave pattern does not move. The peak highlighted with the line is displaced up and down owing to the particle motion but the peak does not move sideways.

9. Key Features of Standing Waves • There are points where the displacement of the standing wave is always zero irrespective of the time evolution. • These points are known as NODES (no displacement). • The nodes occur when • sin(kx) = 0 • This is satisfied when • kx = nπ (n an integer 0,1,2….)

10. Why are nodes formed? • Let us consider the displacement due to the individual traveling waves waves at a node in the standing wave pattern. • At a node, irrespective of the phase difference between the two component waves, the displacement caused by one wave is equal and opposite to the displacement of the other wave. • y1(x,t) = - y2(x,t) • From linear superposition we find • Yt(x,t) = 0

11. What is the distance between successive nodes? ∆x x1, n x2, n+1 • Let a node occur at a point x1 and have integer value n • The neighbouring node occurs at x2 and has integer value n+1. • kx1 = nπ • kx2 = (n+1)π • k(x2 - x1) = k∆x = π • But k = 2π/l • 2π∆x/l = π • ∆x = l/2

12. Key Features of Standing Waves • There are points on the string where the displacement of the standing wave is a maximum. • These points are known as ANTI-NODES and occur spatially when • sin(kx) = ±1 • This is satisfied when • kx = (n+1/2)π (n an integer 0,1,2..) • To see these point the time must satisfy wt = mπ with m an integer.

13. What is the distance between successive anti-nodes? x2, n+1 ∆x x1, n • Let an anti-node occur at a point x1 and have integer value n • The next anti-node occurs at x2 and has integer value n+1. • kx1 = (n+1/2)π • kx2 = (n+1+1/2)π • k(x2 - x1) = k∆x = π • But k = 2π/l • 2π∆x/l = π • ∆x = l/2

14. Example: Separation between adjacent nodes A string of mass per unit length 2 g/m is placed under a tension of 20 N. Two counter propagating harmonic waves that vibrate at the same frequency of 1000 Hz are excited on the string and a standing wave is formed. What is the spacing between successive nodes? We know that ∆x = l/2 with l the wavelength. And l = v/f with v the wave speed and f the frequency. But ∆x with m the mass per unit length and F the tension

15. So the wavelength is given by Hence the spacing between adjacent nodes is As µ = 2 x 10-3 kgm-1, F = 20 N and f = 1000 Hz ∆x = 5 x 10 -3 m.