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 reverses and travels toward the other end. The following figure shows the motion of a single disturbance.
If you hold end A of the above string 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:
Each wave travels from A to B and gets reflected at B. When each reflected wave reaches point A, it gets reflected again and the process repeats. Of course the hand motion keeps putting energy into the system by constantly generating waves that are in phase with the returned waves creating the above waveforms. Such waves are called "standing waves." The subject of waves is lengthy and mathematically very involved. For now, the above is sufficient to give you an idea of wavelength and standing waves.
Types of Waves:
There are two classifications: one classification is: mechanical and electromagnetic.
Mechanical waves require matter for their transmission. Sound waves, ocean waves, and waves on a guitar string are examples..
Electromagnetic waves can travel both in vacuum and matter. If light could not travel in vacuum, we would not see the Sun. Light is an electromagnetic wave. Radio waves, Ultraviolet waves, and infrared waves are all electromagnetic waves.
Waves can also be classified as transverse and longitudinal. (See Figure)
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.
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 unit is (1/s), or (s ^{1}), or (Hz). If the repeated waveforms have a sinefunction shape, the waves are called harmonic waves.
Period ( T ):
Period 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 the 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 (cycles) per second. Each cycle 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. The distance from A to B, in the following figure, is equal to one wavelength ( λ ) because B is the first point after A that is in the same state of oscillation as A is.
Wavelength may also be defined as the distance between a peak to the next peak, or the distance between a trough to the next trough, as shown.
Wave Speed ( v ):
The wave speed is the distance a wave travels per second. Since each wave source generates ( f ) wavelengths per second and each wavelength is ( λ ) units of length long; therefore the wave speed formula is:
v = f λ.
Example 1: The speed of sound waves at STP conditions is 331 m/s. Calculate the wavelength of a sound wave which frequency is 1324 Hz at STP conditions.
Solution: Using v = f λ, & solving for λ, yields: λ = v / f ; λ = (331m/s) / (1324/s) = 0.250m
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A Vibrating String: A stretched string fixed at two points and brought into oscillations such as a violin string has waves on it that keep going backandforth between the two fixed points. If a string is observed closely or by a magnifying glass, at times it appears as shown: The higher the pitch of the note it is playing, the higher the frequency of oscillations and the shorter the wavelength or the sinewaves that appear on it. The waveforms appear to be stationary, but in reality they are not. They are called "standing waves." Nodes are points of the medium (the string) that at zero oscillation state and antinodes are points at maximum oscillation state.


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 has two of such loops; therefore, λ = 40.0cm

Speed of waves in a string depends on the tension in the string as well as the mass per unit length of it as explained below:
Speed of Waves in a Stretched String: The more a string is stretched, the faster waves travel in it. The formula that relates the tension ( F ) in the string and the wave speed v is:

Proof: The proof of this formula is as follows:
If we model the peak of a wave as it passes through the medium (the string) at speed v as shown, 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 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 and since sinq = q for small angles in radians, the formula becomes: 2Fq = Mv^{2}/R (1) where M = the mass of the string segment If we calculate the mass of the hump segment M, it results in M = 2μRθ. This is easy to understand because is the length of the hump is Rθ and let the mass per unit length of the string by defined as μ. In other words, μ= mass / length. Equation (1) takes the form 2Fq = 2μRθ v^{2}/R (2) and solving for v results in

Answer the following: What is F and what is v ? Why are the angles marked are equal? What happens to (Fcosq )'s ? What is the total downward force that is trying to bring the string to normal as the wave passes through? What is the length of arc AB that has a mass of M? 
Example 3: A 120.cm guitar string is under a tension of 400.N. 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)λs that appear in this string if it is oscillating at a frequency of 2083Hz. Note: With 1.20m in the denominator of the first line of solution, the rest of the problem is correct.
Solution: (a) μ = M / L ; μ = (0.480x10^{3}kg) / 1.20m = 4.00x10^{4} kg/m (Correct!)
(b) v = (F/μ)^{1/2} ; v = (400.N / 4.00x10^{4} kg/m)^{1/2} = 1000 m/s (3 sig. fig.)
(c) v = f λ ; λ = v / f ; λ = (1000.m/s) / (2083 /s) = 0.480m
(1/2)λ = 0.480m / 2 = 0.240m
In a length of 1.20m , 5.00 (λ/2)s do fit, as shown:
Traveling Harmonic Waves:
We are interested in finding a formula that calculates the yvalue of any point in a one dimensional medium as harmonic waves are traveling in that 1D medium 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 1D 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 particles of the medium. The wave propagation velocity is V = f λ , but the particles velocity in the ydirection is v_{y }= ∂y / ∂ t.
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 = 37.68 rd/s
(e) f = (ω / 2π) = 6.00Hz (f) = 0.523 rd (g) v = fλ = (6.00 Hz)(0.250m) = 1.50 m/s
(h) v_{y }= ∂y _{ /} ∂ t = 0.0450( 37.68) cos (25.12x  37.68t  0.523)
( i ) v_{y}(3.50m , 21.0s) = 0.0450( 37.68) cos (25.12*3.50  37.68*21.0  0.523) = 1.67 m/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 is a standing wave. If the equation of the wave going to the right is A sin(kx  ωt) and that of one going to the left is A sin(kx + ωt), we may add the two to obtain the equation of the combination wave (Gray) as y(x,t) = A sin(kx  ωt) + A sin(kx + ωt) Using the trigonometric identity: sin A+sin B = 2sin [(A+B)/2] .cos [(AB)/2], we get: y ( x, t ) = 2A cos(ωt) * sin kx In this equation, sinkx determines the shape of the standing wave and 2A cos(ωt) determines how its amplitude varies with time.

Resonant Standing Waves on A String:
If the medium in which
standing waves are formed is infinite, there is no restriction on the
wavelength or frequency of the waves. In case of a bound medium,
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 full loops in it. For any 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, and if you increase the frequency, the wavelength 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 is accommodating one loop only, it is called the fundamental frequency and that is the lowest possible frequency for that rope under that particular tension. The subsequent 2loop, 3loop, 4loop, and ... cases are called the 2nd, 3rd, 4th, and .... harmonics of that fundamental frequency. They are shown on the right.


From the above figure, 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} = 333 Hz (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 of course mean differential length. It is easy to calculate the K.E. and P.E. of a differential element as shown.
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 (Verify), ω = 2512 rd/s (Verify) , v = 140. m/s (Verify), A = 4.00x10^{3}m , and finally, P_{avg} = 17.7 watts
Chapter 16 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) a 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 point, 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 formula for wave speed, in general, 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 Metric 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 is playing 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) 1/8 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)
A snapshot of a traveling wave taken at t = 0.40s is shown on the right. 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.


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 (m) and t in (s). Find the (a) wavelength, (b) period, and (c) wave speed. Assume 3 sig. fig. 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 liner 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.