Assignments 9 10

course Phy 242

̩ʡT㟑assignment #009

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Physics II

09-28-2006

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20:37:50

Query introductory set 6, problems 1-10

explain how we know that the velocity of a periodic wave is equal to the product of its wavelength and frequency

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RESPONSE -->

We know that frequency is the amount of cycles of a wave per second. The wavelength is the distance between the crests of two seperate cycles. So we now know that the amount of cycles * wavelength will give us the distance traveled. Then we are given that this all occurs in one second. So naturally we know 'dx/'dt = velocity, and thus we know the velocity of the periodic wave.

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20:37:57

** we know how many wavelength segments will pass every second, and we know the length of each, so that multiplying the two gives us the velocity with which they must be passing **

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ok

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20:41:17

explain how we can reason out that the period of a periodic wave is equal to its wavelength divided by its velocity

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RESPONSE -->

If we divide a wavelength (m) by the velocity of a periodic wave (m/s) we get an answer that only has units of seconds. This amount of time is the period or the amount of peaks that pass in one second.

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20:41:30

** If we know how far it is between peaks (wavelength) and how fast the wavetrain is passing (velocity) we can divide the distance between peaks by the velocity to see how much time passes between peaks at a given point. That is, period is wavelength / velocity. **

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RESPONSE -->

ok

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20:51:51

explain why the equation of motion at a position x along a sinusoidal wave is A sin( `omega t - x / v) if the equation of motion at the x = 0 position is A sin(`omega t)

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RESPONSE -->

The quation of motion for a sinusoidla was is:

y = A sin('omega*t)

because, the wave is transverse wave and thus its medium is moving perpindicular to the direction of the wave. This means that the wave moves in the y and z directions only to be perpindicular to the x-axis. Since the wave is a sine wave it undergoes SHM but must take the amplitude into account. The 'omega is equal to 2'pi*frequency.

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20:52:07

** the key is the time delay. Time for the disturbance to get from x = 0 to position x is x / v. What happens at the new position is delayed by time x/v, so what happens there at clock time t happened at x=0 when clock time was t = x/v.

In more detail: If x is the distance down the wave then x / v is the time it takes the wave to travel that distance. What happens at time t at position x is what happened at time t - x/v at position x=0.

That expression should be y = sin(`omega * (t - x / v)).

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The sine function goes from -1 to 0 to 1 to 0 to -1 to 0 to 1 to 0 ..., one cycle after another. In harmonic waves the motion of a point on the wave (think of the motion of a black mark on a white rope with vertical pulses traveling down the rope) will go thru this sort of motion (down, middle, up, middle, down, etc.) as repeated pulses pass.

If I'm creating the pulses at my end, and that black mark is some distance x down in rope, then what you see at the black mark is what I did at time x/v earlier. **

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RESPONSE -->

Understood.

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20:57:31

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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RESPONSE -->

Given the length of the string one can assume that the wave will have at least two nodes at each end of the string. The first few harmonics will then have 2, 3, and 4 nodes in the length of the string. so they have respectly 1/2, 2/2, and 3/2 the wavelength themselves. So this gives you:

harmonic 1 = L/(1/2) = 2L

harmonic 2 = L/(2/2) = L

harmonic 3 = L/(3/2) = (2/3)L

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20:57:38

** 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.

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

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RESPONSE -->

ok

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21:02:05

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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RESPONSE -->

Knowing the wavelengths and the velocity we can determine that during a certain amoung of time (s) a certain distance ('dx) is traveled. We can now take that distance and divide by the wave length. This answer gives us the amount of distance of each wavelength. Which is the frequency.

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21:02:14

** 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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RESPONSE -->

Understood.

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21:04:13

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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RESPONSE -->

The velocity of a wave in a string given the mass density of the string and the tension is determined using the equation:

v = 'sqrt(tension/mass per unit length)

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21:04:18

** We divide tension by mass per unit length and take the square root:

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

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RESPONSE -->

ok

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21:04:24

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

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21:04:26

** 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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21:04:28

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

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21:04:30

** 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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؏PM౲H֢kXMҸ

assignment #010

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Physics II

09-28-2006

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21:05:21

**** 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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RESPONSE -->

This question was posed in the previous assignment.

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21:05:32

** 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.

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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 first node corresponds to 1/4 wavelength. The second harmonic is from node to antinode to node to antinode, or 4/3. the third and fourth harmonics would therefore be 5/4 and 7/4 respectively. **

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RESPONSE -->

ok

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21:05:51

**** 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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RESPONSE -->

Question posed in the previous assignment.

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21:05:53

** 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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21:06:07

**** 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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RESPONSE -->

Question posed in the last assignment.

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21:06:09

** We divide tension by mass per unit length:

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

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RESPONSE -->

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21:06:13

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

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RESPONSE -->

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21:06:15

** 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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RESPONSE -->

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21:06:17

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

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RESPONSE -->

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21:06:21

** 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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21:40:19

query univ phy problem 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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RESPONSE -->

Given the equation:

y(x,t) = .75 cm * cos {'pi [(x/0.4 cm) + (t/250 s)]}

You are treating 250 s^-1 as (250 s)^(-1). Only the s is taken to the -1 power.

I note that the text is sometimes ambiguous on such things, implying parentheses in some cases (e.g., 40 m / 2 s means 40 m / (2 s), even though order of operations would say that 40 m / 2 s = 20 m * s) and not others. However there is no parenthesis here, and no parentheses implied.

The same holds for the expression .400 cm^-1.

Knowing the standard equation:

y (x,t) = A cos 2*'pi [(x/'lambda) - (t/T)]

The given equation contains only factor pi in the expression pi [(x * 0.4 cm^-1) + (250 t s^-1)]; the standard equation has 2 pi.

The given expression would have to be put into the form 2 pi [(x * 0.2 cm^-1) + (125 s^-1)]

Amplitude is given as 0.75 cm

Period is given as 250 s

Wavelength is given as 0.400 cm

Because of the above, period is 1/125 s = .008 s, in agreement with the given solution.

The wavelength would be 1 / (.2 cm^-1) = 5 cm, as in the given solution.

Frequency can be found using the T = 1/f equation and solving for the frequency this gives us a frequency of 0.004/s

The Speed of Propagation is found using V = f *wavelength, which equals 0.004/s * 0.004 m = 0.000016 m/s.

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21:40:59

** y(x, t) = A sin( omega * t + k * x), where amplitude is A, frequency is omega / (2 pi), wavelength is 2 pi / x 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. **

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RESPONSE -->

My book had a different equation. Can you see if my thinking is correct?

Watch out for the 2 pi vs. just pi.

Otherwise your only error was the order of operations and confusion over implied parentheses.

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21:43:47

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

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RESPONSE -->

I am not sure how to go about drawing a graph using the given information. I will be sure to print out the answer and study the procedure.

Presumably you have drawn the graph. You can give a convincing description of the important features in a couple of sentences.

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21:44:05

** 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. **

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RESPONSE -->

Understood.

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21:53:16

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

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RESPONSE -->

Since we know the Velocity as 1.6*10^-5m/s and the mass/unit length we can use the following equation and solve for the T in the wave.

V = 'sqrt(T/'mu)

1.6*10^-5 m/s = 'sqrt(T/0.0500kg/m)

(1.6*10^-5m/s)^2 = T / 0.0500kg/m

T = 2.56*10^-10m^2/s^2 * 0.0500kg/m

T = 1.28*10^-11 kg*m/s^2

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21:54:05

** 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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RESPONSE -->

I apparently did something wrong. I believe that the magnitude of the velocity I used was far from correct.

That was the problem, and it goes back the order of operations thing. Also check the mass per unit length to make sure you're right about the .0500 kg / m.

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21:59:29

**** What is the average power?

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RESPONSE -->

Pav = 0.5 * 'sqrt('mu*F)'omega^2*A^2

I know that we use this equation and should be able to simple take what we have thus far and plug it in or solve for a peice and then plug it in but since my answers are off already I will not get this correct.

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21:59:54

** 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 * 195 N) * (250 pi s^-1)^2 * (.0075 m)^2 =

.5 sqrt(98 kg^2 m / (s^2 m) ) * 62500 pi^2 s^-2 * .000054 m^2 =

.5 * 9.9 kg/s * 6.25 * 10^4 pi^2 s^-2 * 5.4 * 10^-5 m^2 =

17 kg m^2 s^-3 = 17 watts, approx..

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

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RESPONSE -->

Understood.

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"

Except for a couple of relatively minor errors, which however made a big difference in your results, your work looks good. Let me know if you have questions.