Chapter 16
Waves
A wave is the motion of a disturbance in a medium. The medium for ocean waves is water, for example. When a string, fixed at both ends, is given a vertical hit by a stick, a dent appears in it that travels along the string. When it reaches an end point, it reflects and inverts and travels toward the other end. Fig. 1 shows the motion of a single disturbance.
Figure 1
If you hold end A of the string (Fig. 2) and try to give it a continuous upanddown motion, with a little adjustment of the pace of oscillations, you can make at least the following waveforms:
Figure 2
Each wave travels from A to B and reflects at B. When each reflected wave reaches point A, it reflects again and the process repeats. Of course, the up and down motion of hand keeps putting energy into the system by constantly generating waves that are in phase with the returned waves creating the above waveforms. Although the waves appear to be standing as they are called "standing waves," they are actually traveling backandforth along the string. The subject of waves is lengthy, complicated, and mathematically very involved. The above is enough to give you an idea.
Types of Waves:
There are two classifications: one classification of waves is: mechanical and electromagnetic.
Mechanical waves require matter for their transmission. Sound waves, ocean waves, and waves on a guitar string are examples. Air, water, and metal string are their media (matter), respectively.
Electromagnetic waves travel both in vacuum and matter. If light (an electromagnetic wave itself) could not travel in vacuum, we would not see the Sun. Radio waves, ultraviolet waves, and infrared waves are all electromagnetic waves and travel in vacuum.
Waves are also classified as transverse and longitudinal (See Fig. 3).
For a transverse wave the disturbance direction is perpendicular to the propagation direction. Water waves are transverse. Waves on guitar strings are also transverse.
For a longitudinal wave the disturbance direction is parallel to the propagation direction. Waves on a slinky as well as sound waves are longitudinal.
Figure 3
Frequency ( f ):
The frequency f of a wave is the number of full waveforms generated per second. This is the same as the number of repetitions per second or the number of oscillations per second. The SI unit for frequency is (1/s), or (s^{1}), called "Hertz (Hz)."
Period ( T ):
Period T is the number of seconds per waveform, or the number of seconds per oscillation. It is clear that frequency and period are reciprocals.
T = 1/f
Also, recall the useful relation between frequency f and the angular speed ω :
ω = 2π f.
ω is the number of radians per second, but f is the number of turns or oscillations per second. Each turn is 2π radians.
Wavelength ( λ ):
Wavelength λ is the distance between two successive points on a wave that are in the same state of oscillation. Points A and B in Fig. 4 are the nearest or successive points that are both the same amount passed the maximum and therefore in the same state of oscillation.
Figure 4
Wavelength may also be defined as the distance between a peak to the next, or the distance between a trough to the next, as shown above.
Also pay attention to Fig. 5. It shows that the distance between any node and the antinode next to it is λ/4.
Figure 5
Wave Speed ( v ):
The wave speed is the distance a wave travels per second. A wave source with frequency f generates f wavelengths per second and since each wavelength is λ units long; therefore, the wave speed formula is:
v = f λ.
Example 1: The speed of sound waves at STP condition is 331m/s. Calculate the wavelength of a sound wave with a frequency of 1324Hz at STP.
Solution: v = f λ ; λ = v/f ; λ = (331m/s)/(1324/s) = 0.250m.
The Vibrating String A stretched string fixed at both ends and brought into oscillation forms a "vibrating string." An example is a violin string on which waves keep traveling backandforth between its ends. If a violin string is observed closely or by a magnifying glass, at times it appears as shown on the right. The higher the pitch of the note it plays, the higher the frequency of oscillations and the shorter the wavelength or the sinewaves that appear along its length. The waveforms appear to be stationary, but in reality they are not. They are called "standing waves." 
Figure 6 Nodes are points of zero oscillation and antinodes are points of maximum oscillation as shown in Fig. 6. 
Now, look at the following example:
Example
2: In a 60.0cm long
violin string, three antinodes are observed. Find the
wavelength of the waves on it. Solution: Each loop has a length of 60.0cm /3 = 20.0cm. Each wavelength (a full sine wave) contains two of such loops; therefore, 
Fig. 7 λ = 40.0cm. 
Speed of Waves in a Stretched String:
The speed of waves in a stretched string depends on the tension F in the string as well as the mass per unit length μ of the string as explained below: The more a string is stretched, the faster waves travel in it. 
The formula that relates tension F in the string and the waves speed v is:

Proof:
If we model the peak of a wave as it passes through the medium (the string) at speed v as shown in Fig. 8, we may think that the peak segment is under a tensile force F that pulls it in opposite directions. The hump can be looked at as a portion of a circle from A to B with its center at C. The hump is being pulled down by a force of magnitude 2Fsinq . This pulling down force passes through the center and therefore acts as a centripetal force for the segment that is equal to Mv^{2}/R ; therefore, 2Fsinq = Mv^{2}/R. For small angles and in radians, sinq = q . The formula becomes:
2Fq = Mv^{2}/R (1) where M = the mass of the string segment.
If we calculate mass M of the hump, it results in M = 2μRθ. This is because the length of the hump is 2Rθ and μ is the mass per unit length of the string. In other words, μ= mass /length. Eqn. (1) takes the form: 2Fq = 2μRθ v^{2}/R (2) Solving for v results in:


Figure 8
Answer the following:
What is F and what is v? Why are the angles marked as q equal? What happens to Fcosq ? What is the total downward force that is trying to bring the string to normal as the wave passes through? How do you calculate the arclength AB and its mass M?
Example 3: A 120cm guitar string is under a tension of 400N. The mass of the string is 0.480 grams. Calculate (a) the mass per unit length of the string and (b) the speed of waves in it. (c) In a diagram show the number of (1/2)λ's that appear in this string if it is oscillating at a frequency of 2083Hz.
Solution: (a) μ = M/L ; μ = 0.480x10^{3}kg/1.20m = 4.00x10^{4} kg/m.
(b) v = (F/μ)^{1/2} ; v = [400N/4.00x10^{4}kg/m]^{1/2} = 1000 m/s.
(c) v = f λ ; λ = v/f ; λ = (1000m/s)/(2083/s) = 0.480m.
(1/2)λ = 0.480m/2 = 0.240m.
The number of (λ/2)'s that fit in this string length of 120cm is
1.20m/0.240m = 5.00, as shown in Fig. 8.
Figure 9
A Good Link to Try: http://surendranath.tripod.com/Applets.html .
Traveling Harmonic Waves:
We are interested in finding a formula that calculates the yvalue at any point in a one dimensional medium as harmonic waves travel in it at speed v. This means that at any point x and at any instant t, we want y(x,t). For harmonic waves, such equation has the general form:
y(x,t) = A sin(kx  ωt + φ)
k is called the wave number and its unit in SI is m^{1}. The above equation is for one dimensional harmonic waves traveling along the +x axis. If the waves are moving along the x axis, the appropriate equation is:
y(x,t) = A sin(kx + ωt + φ).
If y = 0 at t = 0 and x = 0, then φ = 0. It is important to distinguish between the wave propagation velocity v (along the xaxis) and the medium's particles velocity v_{y} (along the yaxis) as transverse waves pass by the particles of the medium. The wave propagation velocity is V = f λ , but the particles velocity in the ydirection is v_{y }= ∂y/∂t. The symbol "∂ " denotes "partial derivative."
Example 4: The equation of certain traveling waves is
y(x,t) = 0.0450 sin(25.12x  37.68t  0.523)
where x and y are in meters, and t in seconds. Determine the following: (a) Amplitude, (b) wave number, (c) wavelength, (d) angular frequency, (e) frequency, (f) phase angle, (g) the wave propagation speed, (h) the expression for the medium's particles velocity as the waves pass by them, and (i) the velocity of a particle that is at x = 3.50m from the origin at t = 21.0s.
Solution: Comparing this equation with the general form, results in
(a) A = 0.0450m ; (b) k = 25.12m^{1} ; (c) λ = (2π/k) = 0.250m
(d) ω = 37.68rd/s ; (e) f = ω/(2π) = 6.00Hz ; (f) φ = 0.523 rd ;
(g) v = f λ =1.50m/s; (h) v_{y} = ∂y/∂t = 0.045(37.68) cos (25.12x  37.68t  0.523)
(i) v_{y}(3.5m, 21s) = 0.045(37.68) cos (25.12*3.537.68*210.523) = 1.67m/s.
Standing Harmonic Waves:
When two harmonic waves of equal frequency and amplitude travel through a medium in opposite directions, they combine and the result can form standing waves.
If the equation of the wave going to the right is y_{1} = A sin(kx  ωt) and that of the one going to the left is y_{2} = A sin(kx + ωt), we may add the two to obtain the equation of the combination wave as
y(x,t) = Asin(kx  ωt) + A sin(kx + ωt)
Using the trigonometric identity: sin a + sin b = 2sin[(a + b)/2] cos [(a  b)/2],
we get: y(x,t) = 2Acos(ωt) ∙ sin(kx).
In this equation, sin(kx) determines the shape of the standing wave and
2Acos(ωt) determines how its amplitude varies with time.
If at t = 0, the red wave that is going to the right is in phase with the blue wave that is going to the left, they add up constructively and the amplitude of the sum (the gray) is twice each amplitude as shown at t = 0. At t = T/4 the become completely out of phase and cancel each others effect to a sum of zero. This is only for a very brief instant (the gray becomes a straight line). At t = T/2, the opposite of t = 0 occurs. The two humps become troughs and the single trough becomes s hump. In another half a period the shape of the gray (or the sum) become like the one at t = 0. Confirm the above explanation by looking at the different figures shown on the right (Fig. 10). I 
Figure 10
Resonant Standing Waves in A String:
In a medium with limited boundary such as a string fixed at both ends, standing waves can only be formed for a set of discrete frequencies or wavelengths.
If you hold one end of a rope, say 19ft long, and tie the other end of it to a wall 16ft away from you, there will be a slack of 3ft in it allowing you to swing it up and down and make waves. By adjusting the frequency of the oscillatory motion you give to the end you are holding, you can generate a sequence of waves in the rope that will have an integer number of loops in it. For a frequency f, there is a corresponding wavelength λ such that V = f λ .
It is very clear from this equation that, since the
waves
speed, V ,
in a given medium is constant, the product f λ is
also constant.
This means that if you increase frequency f, the wavelength λ of the waves in the rope has to decrease. Of course, for resonance, the values of such frequencies, as was mentioned, are discrete, and so are their corresponding wavelengths. All you need to do is to adjust your hand's oscillations for each case to observe a full number of loops in the rope between you and the wall. It is also clear from Example 2 that each loop is one half of the wavelength in each case. When the entire length of the rope accommodates one loop only, it is called the fundamental frequency and that is the lowest possible frequency. The subsequent 2loop, 3loop, 4loop, and ... cases are called the 2nd, 3rd, 4th, and .... harmonics of that fundamental frequency as shown on the right (Fig. 11). 

Figure 11
From the above figures, at resonance, the length L of the string is related to the number of loops or λ/2 as follows:
Example 5: Find the frequency of the 4th harmonic waves on a violin string that is 48.0cm long with a mass of 0.300grams and is under a tension of 4.00N.
Solution: Using the above formula,
f_{4} = (4/0.96m) ∙ [4.00N/(0.000300kg/0.480m)]^{1/2} = 333Hz (Verify).
The Wave Equation:
The onedimensional wave equation for mechanical waves applied to traveling waves has the following form:
where v is the speed of waves in the medium such that v = ω/k. 
The solution to this equation is y(x,t) = A sin( kx  ωt + φ) .
Example 6: Show that the equation y(x,t) = A sin(kx  ωt + φ) satisfies the wave equation.
Solution: Take the appropriate partial derivatives and verify by substitution.
Energy Transport on a String:
As a wave travels along a string, it transports energy by being flexed point by point, or dx by dx. By dx , we mean differential length. It is easy to calculate the K.E. and P.E. of a differential element as shown in Fig. 12. Let dl be a differential segment of the string. Corresponding to this dl , there is a dx and a dy as shown.
Note that dl is a stretched dx but has the same mass as dx.
The conclusion is that the power transmission by a wave on a string is proportional to the squares of angular speed and amplitude and linearly proportional to the wave speed V in the string.
Example 7: A 1.00mlong string has a mass of 2.5 grams and is forced to oscillate at 400Hz while under a tensile force of 49N. If the maximum displacement of the string in the direction perpendicular to the waves propagation is 8.00mm, find its average power transmission.
Solution: We need to apply the formula P_{avg} = 0.5μ(ωA)^{2}v.
First μ = M/L, ω = 2πf, A, and V = (F/μ)^{0.5} must be calculated.
μ = 2.5x10^{3} kg/m, ω = 2512 rd/s, V = 140. m/s, A = 4.00x10^{3}m , and finally, P_{avg} = 17.7 watts. Verify all calculations.
Test Yourself 1:
1) A wave is the motion of (a) a particle along a straight line in a backandforth manner (b) a disturbance in a medium (c) an electric disturbance in vacuum (d) both b & c. click here.
2) A mechanical wave (a) can travel in vacuum (b) requires matter for its transmission (c) both a & b.
3) An electromagnetic wave (a) can travel in vacuum (b) can travel in matter (c) both a & b.
4) A longitudinal wave travels (a) perpendicular to the disturbance direction (b) parallel to the disturbance direction (c) in one direction only. click here.
5) A transverse wave (a) travels perpendicular to the disturbance direction (b) can also travel parallel to the disturbance direction (c) travels sidewise.
6) A stick that gives a downward hit to a horizontally stretched string, generates a trough that travels along the string. When the moving trough reaches a fixed end, it returns not as a trough, but a hump. The reason is that (a) the moving disturbance is not capable of pulling that end point down (b) conservation of momentum requires the string to be pulled upward by the fixed point and hence the wave reflects (c) the conservation of gravitational potential energy must be met. (d) both a & b. click here.
7) Wavelength is (a) the distance between two crests (b) the distance between a crest and the next one (c) the distance between a trough and the next one (d) b & c.
8) The frequency of a wave is the number per second of (a) full wavelengths generated by a source (b) full waves passing by a point (c) a & b. click here.
9) The wave speed formula is (a) V = λ /f (b) V = f λ (c) V = f /λ.
10) The power transmission by a wave on a string is proportional to the (a) amplitude (b) square root of the amplitude (c) square of the amplitude.
11) The power transmission by a wave on a string is proportional to the (a) frequency (b) square of frequency (c) square root of frequency. click here.
12) The energy transmission by a wave on a string is proportional to the (a) wave speed (b) square of wave speed (c) square root of wave speed.
13) In y(x.t) = A sin (kx  ωt) for a transverse wave, dy/dt gives us the (a) wave speed (b) wave acceleration (c) medium's particles speed.
14) For the resonance of a string fixed at both ends, it is possible to generate (a) all (b) only odd (c) only even multiples of λ/2 in its entire length.
15) If the tension in a stretched string is quadrupled, then a disturbance made in it travels (a) 4 times faster (b) 1/2 times slower (c) 2 times faster. click here.
16) If the tension in a stretched string is increased by a factor of 9, then a disturbance made in it travels (a) 3 times faster (b) 1/3 times slower (c) 9 times faster.
17) The speed of waves in a stretched string is proportional to (a)F, the tension (b)F^{1/2} (c)F^{2}. click here.
18) The quantity μ = M/L, mass of a string divided by its length, is called (a) mass per unit length (b) mass length (c) length per unit mass.
19) The SI unit for μ is (a) kg/m (b) slug/ft (c) gr/cm.
20) Mechanical waves travel faster in a string that is (a) thicker and therefore less flexible (b) thinner and therefore more flexible (c) neither a nor b.
21) The speed of waves in a stretched string is (a) directly proportional to μ (b) inversely proportional to μ (c) inversely proportional to 1/μ (d) directly proportional (1/μ)^{0. 5}. click here.
22) The tension in a guitar string is 576N and its 1.00m length has a mass of 0.100gram. The waves speed in this stretched string is (a) 2400m/s (b) 3600m/s (c) 1800m/s.
23) If two full wavelengths can be observed in this string, the wavelength of the waves is (a) 1.00m (b) 0.500m (c) 2.00m.
24) What frequency the guitar string in the previous questions does play for a wavelength of 0.500m? (a) 1200Hz (b) 400Hz (c) 4800Hz. click here.
25) The distance between a node and the next antinode on a wave is (a) (1/2)λ (b) (1/4)λ (c) (1/2)λ . click here.
Problems:
1) Find the corresponding wavelengths for each of the following radio waves: (a) The AM band ranging from 550kHz to 1600kHz. (b) The FM band ranging from 88MHz to 108MHz. The speed of E&M waves is 3.00x10^{8} m/s.
2)
The snapshot of a traveling wave taken at t = 0.40s is shown in Fig. 13. If the wavelength is 8.0cm and the amplitude 2.4cm, and at t = 0, crest C occurred at x = 0, write the equation of the wave.


Figure 13
3) For a tension of 36.0N in a string, waves travel at 42.0m/s, At what tension do waves travel at a speed of 21.0m/s?
4) Use the equation y(x.t) = A sin (kx  ωt) for traveling waves on a string, to find (a) the slope of the string at any position x and time t. (b) How is the maximum slope related to the wave speed and the maximum particle speed?
5) The equation of a traveling harmonic wave is y = 0.06sin( x/5  t ) where x and y are in meters and t in seconds. Find the (a) wavelength , (b) period, and (c) wave speed. Assume 3 sig. figs. on all numbers.
6) The amplitude of the standing waves on a stretched string is 3.00mm, and the distance between a node and its nearest antinode is 12.5 cm. If the linear mass density of the string is 3.60 grams per meter, and the tension in the string is 9.00N, write the equation of the standing waves in the string.
7) Two strings S_{1} and S_{2} that have the same linear mass density are under tensions F_{1} and F_{2} such that F_{1} = 2F_{2} , but have different lengths (L_{1} = 0.333L_{2}). Find the ratio of their fundamental frequencies.
8) A mechanical oscillator imparts 5.00 watts of power at a frequency of 60.0Hz to a thin metal wire of length 16.0m that weighs 0.4905N. For a tension of 48.0N in the wire, find the amplitude of the generated waves. g = 9.81m/s^{2}.
9) The tensile stress in a steel wire is 2.2x10^{8} Pa. The density of steel is 7.8 grams/cm^{3}. Find the speed of transverse waves in the wire.
10) The differential energy of the nth standing wave in a string that is fixed at both ends is dE=μ(ωA)^{2}dx where μ, ω, and A are the linear mass density, angular frequency, and amplitude of the waves, respectively. Calculate (a) the total energy of the wave along the entire length (from 0 to L ) of the string, and (b) show that the energy per loop of the standing wave is E = 2π^{2 }μ A^{2 }f v.