Query 12

course Phy 232

6/27 10

Question: **** Query introductory set six, problems 11-14 **** given the length of a string how do we determine the wavelengths of the first few harmonics?

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Your Solution:

Harmonics 1, 2, and 3 of a string of a certain length will have 2, 3, and 4 nodes respectively. Therefore, the fundamental or first harmonic will have only half a wavelength over the entire length of the string. With each harmonic, the wavelengths contained in the string increase by one half. Therefore, the wavelength is easily found by multiplying the length of the string by the inverse of the number of wavelengths over the distance of the string.

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Given Solution:

** As wavelength decreases you can fit more half-waves onto the string. You can fit one half-wave, or 2 half-waves, or 3, etc..

So you get

1 half-wavelength = string length, or wavelength = 2 * string length; using `lambda to stand for wavelength and L for string length this would be

1 * 1/2 `lambda = L so `lambda = 2 L.

For 2 wavelengths fit into the string you get

2 * 1/2 `lambda = L so `lambda = L.

For 3 wavelengths you get

3 * 1/2 `lambda = L so `lambda = 2/3 L; etc.

Your wavelengths are therefore 2L, L, 2/3 L, 1/2 L, etc..

FOR A STRING FREE AT ONE END:

The wavelengths of the first few harmonics are found by the node - antinode distance between the ends. The node-antinode distance corresponds to 1/4 wavelength, so the wavelength is 4 times the length of the string.

The second harmonic is from node to antinode to node to antinode, or 3/4 of a wavelength. So 3/4 of this wavelength is equal to the length of the string, and the wavelength is therefore 4/3 the length of the string.

The third and fourth harmonics would therefore be 5/4 and 7/4 the length of the string, respectively. **

STUDENT QUESTION (instructor comments in bold)

In the explanation, I don’t understand why the wavelengths were halved [L = 1 * 1/2(‘lambda)].

As indicated in the given solution, you can fit an even number of half-wavelengths onto a string fixed at both ends.

• If you have a single half-wavelength, then the length of the string is 1/2 wavelength; hence L = 1 * (1/2 lambda).

• If you have two half-wavelengths, then the length of the string is 2 * 1/2 wavelength; hence L = 2 * (1/2 lambda).

• etc.

I get the explanation at the bottom were the 1st harmonic is 1/4 the wavelength and the 2nd is 3/4 the wavelength, etc….. but where does that come into play when determining the actual wavelength. I can’t tell if both of the explanations say the same things, or if it’s a 2-part explanation.

I believe you are referring to the solution for a string which is free at one end. For the string free at one end, the first harmonic isn't 1/4 of the wavelength. The first harmonic has a wavelength, which is related to the length of the string.

• For the first harmonic there is a single node, at one end, and a single antinode, at the other. The length of the string is therefore a single node-antinode distance. Since the node-antinode distance is 1/4 of the wavelength, the length of the string is 1/4 wavelength. (It would follow that the wavelength is 4 times the length of the string).

• For the second harmonic three node-antinode distances are spread along the wave, so the wavelength is 4/3 the length of the string, as indicated in the given solution.

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Question: **** Given the wavelengths of the first few harmonics and the velocity of a wave disturbance in the string, how do we determine the frequencies of the first few harmonics?

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Your Solution:

The frequencies are determined by dividing the velocity of the wave by the wavelength. This gives the frequency because it divides up the distance traveled by the wave in a second into the number of wavelengths traveled over that time.

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Given Solution:

** The frequency is the number of crests passing per unit of time.

We can imagine a 1-second chunk of the wave divided into segments each equal to the wavelength. The number of peaks is equal to the length of the entire chunk divided by the length of a 1-wavelength segment. This is the number of peaks passing per second.

So frequency is equal to the wave velocity divided by the wavelength. **

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Question: **** Given the tension and mass density of a string how do we determine the velocity of the wave in the string?

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Your Solution:

The velocity is equal to the square root of the tension divided by the mass density.

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Given Solution:

** We divide tension by mass per unit length:

v = sqrt ( tension / (mass/length) ). **

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Question: **** gen phy explain in your own words the meaning of the principal of superposition

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Your Solution:

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Given Solution:

** the principle of superposition tells us that when two different waveforms meet, or are present in a medium, the displacements of the two waveforms are added at each point to create the waveform that will be seen. **

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Question: **** gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence?

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Your Solution:

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Given Solution:

** angle of incidence with a surface is the angle with the perpendicular to that surface; when a ray comes in at a given angle of incidence it reflects at an equal angle on the other side of that perpendicular **

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Question: query univ phy problem 15.50 11th edition 15.48 (19.32 10th edition) y(x,t) = .75 cm sin[ `pi ( 250 s^-1 t + .4 cm^-1 x) ] What are the amplitude, period, frequency, wavelength and speed of propagation?

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Your Solution:

The amplitude is .75.

The wavelength is 2 pi / .4 pi = 5 cm

Frequency is 250 pi / 2 pi = 125 Hz

Period is 1 / 125 = .008 seconds

Speed of propagation is 125 * 5 = 625 cm /s

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Given Solution:

** y(x, t) = A sin( omega * t + k * x), where amplitude is A, frequency is omega / (2 pi), wavelength is 2 pi / k and velocity of propagation is frequency * wavelength. Period is the reciprocal of frequency.

For A = .75 cm, omega = 250 pi s^-1, k = .4 pi cm^-1 we have

A=.750 cm

frequency is f = 250 pi s^-1 / (2 pi) = 125 s^-1 = 125 Hz.

period is T = 1/f = 1 / (125 s^-1) = .008 s

wavelength is lambda = (2 pi / (.4 * pi cm^-1)) = 5 cm

speed of propagation is v = frequency * wavelength = 125 Hz * 5 cm = 625 cm/s.

Note that v = freq * wavelength = omega / (2 pi) * ( 2 pi ) / k = omega / k. **

STUDENT COMMENT:

2*pi is one full cycle, but since the function is cos, everything is multiplied by pi. So does this mean that the cos function only represents a half a cycle?

It's late and I might well be missing something, but I don't think the cosine is mentioned in this problem. Let me know if I'm wrong.
I believe that only the sine function is mentioned. However there isn't much difference between the two (only a phase difference of pi/2) and everything below would apply to either:
pi is there because of the periodicity of the sine and cosine functions, as described below.
A sine or cosine function completes a full cycle as its argument changes by 2 pi.
The argument might be a function of clock time, or of position.
If the argument is of the form omega * t, then a period is completed every time omega * t changes by 2 pi. This occurs when t changes by 2 pi / omega. So the period of the function is 2 pi / omega.
If the argument is of the form k x, then it changes by 2 pi every time x changes by 2 pi / k, so the wavelength of the function is 2 pi / k.

STUDENT COMMENT

the book has this as position = Acos(k*x+ -omega*t)

and velocity omega*A*sin(k*x-omega*t)

this is no big deal, they mean the same as the student sort of mentions, that the sine is 2pi shifted which is in the wave number.

INSTRUCTOR RESPONSE

It doesn't make a lot of difference. The sine or cosine is a valid function to use, and whether you use k x - omega t, kx + omega t, -kx + omega t or -kx - omega t you get the equation of a traveling harmonic wave that satisfied the wave equation y_tt = 1/c * y_xx, where for example _tt represents second derivative with respect to t, and c represents the propagation velocity of the wave.

The general form of a traveling harmonic wave can be taken as

y(x, t) = A cos(k x - omega t + phi)

where phi is the initial phase. Any initial phase is possible. If phi = -pi/2 then since sin(theta) = cos(theta - pi/2) the equation could be

y(x, t) = A sin(kx - omega t).

If phi = 3 pi / 2 then since sin(theta) = - sin(theta + pi) = sin(-theta) we have

y(x, t) = A sin(omega t - k x).

Recall that the graph of y = f( x - h) is identical in shape to the graph of y = f(x), but translated h units in the horizontal direction.

Now kx - omega t can be written as k ( x - omega / k * t), and f(k ( x - omega / k * t) ) can therefore be seen as a horizontal translation of the graph of y = f(kx), the amount of the translation being omega / k * t . This translation is positive, so as t increases the horizontal translation increases, and the graph progresses to the right. The rate of progression with respect to t is omega / k, which is therefore the speed of propagation.

Thus the argument k x - omega t guarantees that the function y = f( k x - omega t) represents an unchanging shape that moves to the right with speed omega / k.

If f is a sine or cosine function, or any superposition of sine and cosine functions, we have a harmonic wave.

Another interpretation is that k x - omega t = -omega ( t - k / omega * x). In this case k / omega * x is regarded as the time required for the wave to propagate from position 0 to position x. The function f(-omega * t) that describes the motion of the point x = 0 in time is translated in space, so that the same motion occurs at position x after a 'time lag' of k / omega * x.

Similar analysis shows that a function of the form y = f( omega t + k x) represents a wave traveling in the negative x direction.

You'll probably have more questions, which I'll welcome.

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Question: **** Describe your sketch for t = 0 and state how the shapes differ at t = .0005 and t = .0010.

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Your Solution:

At t=0, the graph is one of a typical sine function with an amplitude of .75 and a wavelength of 5. At t=.0005 and t=.001, the graphs are still sine functions with the same amplitude and wavelength as at t=0. However, in each case, the graph is shifted by a certain amount in the x direction.

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Given Solution:

** Basic precalculus: For any function f(x) the graph of f(x-h) is translated `dx = h units in the x direction from the graph of y = f(x).

The graph of y = sin(k * x - omega * t) = sin(k * ( x - omega / k * t) ) is translated thru displacement `dx = omega / k * t relative to the graph of sin(k x).

At t=0, omega * t is zero and we have the original graph of y = .75 cm * sin( k x). The graph of y vs. x forms a sine curve with period 2 pi / k, in this case 2 pi / (pi * .4 cm^-1) = 5 cm which is the wavelength. A complete cycle occurs between x = 0 and x = 5 cm, with zeros at x = 0 cm, 2.5 cm and 5 cm, peak at x = 1.25 cm and 'valley' at x = 3.75 cm.

At t=.0005, we are graphing y = .75 cm * sin( k x + .0005 omega), shifted -.0005 * omega / k = -.313 cm in the x direction. The sine wave of the t=0 function y = .75 cm * sin(kx) is shifted -.313 cm, or .313 cm left so now the zeros are at -.313 cm and every 2.5 cm to the right of that, with the peak shifted by -.313 cm to x = .937 cm.

At t=.0010, we are graphing y = .75 cm * sin( k x + .0010 omega), shifted -.0010 * omega / k = -.625 cm in the x direction. The sine wave of the t = 0 function y = .75 cm * sin(kx) is shifted -.625 cm, or .625 cm left so now the zeros are at -.625 cm and every 2.5 cm to the right of that, with the peak shifted by -.625 cm to x = +.625 cm.

The sequence of graphs clearly shows the motion of the wave to the left at 625 cm / s. **

Your Self-Critique: In my solution, I did not address where the zeros of the graphs occurred and I failed to realize that the graphs would be shifted by the amount omega / k * t. However, I now understand this aspect of the problem.

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Question:

**** If mass / unit length is .500 kg / m what is the tension?

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Your Solution:

Velocity is equal to the square root of the tension divided by the mass density. Therefore, tension is equal to (6.25 m/s)^2 * .5 kg / m = 19.5 N.

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Given Solution:

** Velocity of propagation is

v = sqrt(T/ (m/L) ). Solving for T:

v^2 = T/ (m/L)

v^2*m/L = T

T = (6.25 m/s)^2 * 0.5 kg/m so

T = 19.5 kg m/s^2 = 19.5 N approx. **

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Question: **** What is the average power?

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Your Solution:

Average power is equal to .5 * sqrt( m / L * F) * omega^2 * amplitude^2. Therefore, the average power equals .5 * sqrt(.5 kg/m * 19.5 N) * (250 pi s^-1)^2 * (.0075 m)^2 = 54.2 watts.

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Given Solution:

** The text gives the equation Pav = 1/2 sqrt( m / L * F) * omega^2 * A^2 for the average power transferred by a traveling wave.

Substituting m/L, tension F, angular frequency omeage and amplitude A into this equation we obtain

Pav = 1/2 sqrt ( .500 kg/m * 19.5 N) * (250 pi s^-1)^2 * (.0075 m)^2 =

.5 sqrt(9.8 kg^2 m / (s^2 m) ) * 62500 pi^2 s^-2 * .000054 m^2 =

.5 * 3.2 kg/s * 6.25 * 10^4 pi^2 s^-2 * 5.4 * 10^-5 m^2 =

5.5 kg m^2 s^-3 = 5.5 watts, approx..

The arithmetic here was done mentally so double-check it. The procedure itself is correct. **

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