 Phyllis Fleming Physics Physics 104
 Outline - Wave Motion
1. Descriptive Terms

1. Period T equals the time for one complete oscillation of the source that produces the wave.

2. Frequency f equals the number of oscillations per second.

3. Wavelength λ equals the distance moved by the wave in one period or in one oscillation.

 λ x f = v

(distance/oscillation) x ( number of oscillations/s) = (distance/s)

4. Velocity v equals distance moved by the wave per second.

2. Equation of motion for a traveling wave

1. The displacement y of a point on a traveling wave depends on the distance x from the source and the time t.  y is a function of x and t.

2. Each point on the wave vibrates with simple harmonic motion of the same amplitude and frequency, but there is a phase difference between the motion of a point at x and a point at x’ ≠ x  if  x’ ≠ (x ± n λ) where n = 0, 1, 2, . . . .  Points separated by λ vibrate in phase; points separated by λ/2 vibrate 1800 out of phase.

3. y(x,t) = A sin (2πx/ λ ± 2 πt/T), where the - sign is used for a wave traveling to the right and the + sign for a wave traveling to the left. If the wave is traveling to the right, you want the function at x > 0 to be equal to the function at x = 0 and t = 0. To insure this you subtract vt and then f(x - vt) = f(0). If the wave is traveling to the left then the x is getting smaller and you have to add vt.

4. Letting k = 2 π/ λand ω= 2 π/T = 2 πf, we may write y(x,t) = A sin (kx ± ωt).

5. Fig. 1a below shows a travelling wave with λ= 4 m at t = 0. A crest C appears at x = 1 m. The following figures show the shape of the wave as it travels to the right. Since a wave travels one-wavelength λin one period T, it travels λ/4 in time T/4. In Fig. 1b below, at time t = T/4 the crest C has moved from x = 1.0 m to x = 2.0 m a distance of 1 m = λ/4.  At t = T/2, t = 3T/4, and T, the crest has moved λ/2, 3 λ/4, and λ, respectively.   6. In Fig. 2 below, I have plotted y as a function of time for three values of x. If you look back at Fig. 1 above and concentrate on the point at x = 0 and find the value of y for t = 0, T/4, T/2, 3T/4, and T, you will see that a plot of y as a function of t looks like Fig. 2a below. Now concentrate on the point at x = 1.0 m and you will find that a plot of y as a function of t looks like Fig. 2b below. A plot of y as a function of t for x = 2.0 m is shown in Fig. 2c below. Notice that all points on the wave vibrate up and down with the same amplitude and frequency, but they are not in phase. If you look at Fig. 2a and Fig. 2c below, you find the particles are 180o out of phase. 7. Sample problem 1. For the wave shown in Fig. for #1a below, the displacement of the wave as a function of x is shown for t = 0. In Fig. for #1b below, y is shown as a function of x for t = 1/4 s. This is the first time the wave looks this way after t = 0. Find (a) the amplitude of the wave, (b) the wavelength of the wave, (c) the period of the wave, (d) the frequency of the wave, (e) the velocity of the wave, and (f) the direction in which the wave is moving. Explain your answer to (f). 8. Sample Problem 2. Given y(x,t) = 4 cm sin ( π m-1 x - 4 πs-1 t) find (a) the amplitude , (b) the wavelength, (c) the period, (d) the frequency (e) the velocity of the wave and (f) the direction in which it is moving.

9. Sample Problems in 104 Problem Set for Wave Motion: 1-4, 7-10.

3. Standing Wave

1. A standing wave is formed by combining a wave to the right and to the left. For a standing wave,

y(x,t) = y1(x,t) + y2(x,t)
y(x,t) = A sin (2 πx/ λ- 2 πft) + A sin (2 πx/ λ + 2 πft)
= A[sin 2 πx/ λ cos 2 πft - cos 2 πx/ λ sin 2 πft}
+ A[sin 2 πx/ λ cos 2 πft + cos 2 πx/ λ sin 2 πft}
= 2A sin 2 πx/ λ cos 2 πft = (2A sin 2 πx/ λ) cos 2 πft

2. This reminds us of the equation for simple harmonic motion: y(x,t) = (Amplitude) cos 2 πft or y(x,t) = (Amplitude) cos ωt

3. For a standing wave, all points on the wave do not have the same amplitude.

1. In fact, when sin 2 πx/ λ = 0, the displacement of the rope at those points never have any displacement. These points are called nodal points and occur when 2 πx/ λ= 0, π, 2 π, 3 π, 4 π, etc. A neat way to write this is to say you get a nodal point when 2 πx/ λ = m π, where m = 0, 1, 2, 3, 4,. . .  or x = mπλ/2 π  or x = m λ/2. (A historical note here. This was the first time that a quantity took on integral values or discrete values. In other words, the values were not continuous. This is a foundation of quantum mechanics.) The distance between nodal points =Δx = (m + 1) λ/2 - m λ/2 = λ/2.

2. Positions of maxima with amplitude 2A, where A is the amplitude of a single wave, are called antinodes. They occur for sin 2 πx/ λ= 1 or 2 πx/ λ = (m + 1/2)λ or x = (m + 1/2) λ/2. The distance between antinodes is again λ/2.

3. All points, except the nodal points, on the standing wave vibrate with simple harmonic motion, but the amplitudes vary from zero at a nodal point to a maximum at an antinodal point. The entire string on which a standing wave exists has zero displacement twice each period.

4. Sample Problem 3. Below, I have drawn y1(x,t), y2(x,t), and y(x,t) = y1(x,t) + y2(x,t) for t = 0. You find y(x,t) by adding the displacements of y1(x,t) and y2(x,t) for a given time t. I have also drawn y1(x,t) for t = T/4, t = T/2, t = 3T/4, and T. Notice that the crest at x = 0.50 m for y1(x,t) at t = 0 moves to x = 1.0 m at t = T/4. In other words, the wave moves λ/4 in time T/4. You are to complete the figure by drawing y2(x,t) and y(x,t) for t = T/4, t = T/2, and T = T. How do the y(x,t) compare at time t = 0 and t = T? Is this what you would have expected? What are the distances between nodes in the y vs x curves? What are the distances between antinodes in the y vs x curves? 5. Sample Problems in 104 Problem Set for Wave Motion: 11,
and 21-24.

4. Solution to Sample Problems

1. From Fig. for #1 below, (a) the amplitude of the wave is 0.100 m. (b) the wavelength = 2.00 m. (c) In Fig. b, the crest has moved 0.50 m to the right in 1/4 s. Since 0.50 m = λ/4, the time elapsed = T/4. T = 1.0 s. (d) The frequency f = 1/T = 1.0 s-1. (e) λf = v = (2.00 m)(1 s-1) = 200 cm/s.
(f) Since the crest moves to the right, the wave is moving to the right. 2. Given
y(x,t) = 4 cm sin ( π m-1 x - 4 πs-1 t)

Compare with
y(x,t) = A sin (2 πx/  λ± 2 πt/T)

(a) A = 4 cm, (b) 2 π/  λ= πm-1 or λ= 2 m, (c) 2 π/T = 4 πs-1 or T = 0.5 s
(d) f = 1/T = 2 s-1 s, (e) v = λf = (2 m)(2 s-1) = 4 m/s (f) The wave travels to the left because f(x,t) = f(x –vt)

3. As illustrated in the figure below, equation of wave to the right is y1(x,t) = 0.10 m sin ( π m-1x - 2 πft). The equation of the wave to the left is y2(x,t) = 0.10 m sin ( π m-1x + 2 πft). The equation of the standing wave is y(x,t) = 0.20 m sin π m-1x cos 2 πft. Nodal points are 1.00 m apart = l/2 = distance between antinodal points. After one period T, the waves look the same as they did for t = 0. Homepage Sitemap
 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming September 25, 2002 April 4, 2003