Assingment 11

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course PHY 232

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htmYour solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem

along with a statement of what you do or do not understand about it. This response should be

given, based on the work you did in completing the assignment, before you look at the given

solution.

011. `Query 10

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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:

When given the length of a string we can determine the wavelength as shown below

lambda=2L Where lambda is the wavelength and L is the length

so for 2waves we can say

2(1/2)lambda= L

lambda=L

While for 3waves

3(1/2)lambda=L

lambda=2/3L

confidence rating #$&*:

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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.

Your Self-Critique:

The given solution is lengthy but explains better how to solve the problem.

Your Self-Critique Rating:ok

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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:

Since we already given the wavelength and the velocity its easier to rearrange the wavelength

equation and solve for frequecy as f=v/lambda

confidence rating #$&*:

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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. **

Your Self-Critique:

The given solution has a different approach but same idea.

Your Self-Critique Rating:ok

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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:

To obtain the velocity of the string we divide the tension by mass density and then obtain the

sqrt

confidence rating #$&*:

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Given Solution:

** We divide tension by mass per unit length:

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

Your Self-Critique:

The given solution is similar to mine.

Your Self-Critique Rating:Ok

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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:

Superposition is two different waveforms that meet and create a waveform that can be seen

confidence rating #$&*:

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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. **

Your Self-Critique:

The given solution is a good definition of the superposition.

Your Self-Critique Rating:OK

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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?

Your Solution

Since the angle of incidence is perpendicular to the surface and the reflection will be on the

opposite side as well thats they are equal.

confidence rating #$&*:

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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 **

Your Self-Critique:

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.

Your Self-Critique:

The given solution is well detailed and explains the process much better.

Your Self-Critique Rating:Ok

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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:

From the equation given above we were given that y(x,t)=0.75cm sin[pi(250s^-1 t+0.4cm^-1 x)]

At t=0 when substituted to the equation above gives a graph of y vs x which is a sine curve

At t=0.0005 the only difference is that the peak is shifted

At t=0.0010 again the peak is slightly shifted on this too

confidence rating #$&*:

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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. **

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Your Solution:

Since we know v=sqrt(T/m) then we can rearrange the equation and solve for T=v^2*m

Therefore T=(6.25m/s)^2(0.500kg/m)=19.5N

confidence rating #$&*:

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Given Solution:

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

Your Self-Critique:

The given solution has a similar idea.

Your Self-Critique Rating:ok

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Question: ** 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:

Since Pav=1/2sqrt(m/L*F)*omega^2*A^2 then we can substitute the data we already have from the

previous question and solve the problem as shown below;

0.5sqrt(0.5kg/m *19.5)*(250pi s^-1)^2(0.0075m)^2=54.2 Watts

confidence rating #$&*:

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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 =

54 kg m^2 s^-3 = 54 watts, approx..

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

If your solution to stated problem does not match the given solution, you should self-critique per

instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-

phrase interpretation of the problem along with a statement of what you do or do not understand

about it. This response should be given, based on the work you did in completing the assignment,

before you look at the given solution.

Your Self-Critique:

The given solution uses a similar approach to my solution based by the book formula

Your Self-Critique Rating:Ok

QUESTION FROM STUDENT

My question is on the following question I was trying during a practice test:

Analyze the pressure vs. volume of a 'bottle engine' consisting of 8 liters of an ideal gas as it

operates between minimum

temperature 200 Celsius and maximum temperature 360 Celsius, pumping water to half the maximum

possible height.

Sketch a pressure vs. volume graph from the original state to the maximum-temperature state and

use the graph to

determine the useful work done by the expansion. Then, assuming a diatomic gas, determine the

thermal energy required

to perform the work and the resulting practical efficiency of the process.

I understand the pressure vs volume graph. Does this question basically mean that I should find

d'Q(v) and d'Q(p) and if

so, what temperatures do I use for the temperature change. Then once i find d'Q, how do i find

work. It was stated that

it is the area under the curve, but is this the same as the equation d'Q(p)-d'Q(v). Also

efficiency is found by taking Max

temp-min temp/max temp, so I know how to do that, but why would this change with the amount of

energy needed to

perform work. I am very confused on this problem.

INSTRUCTOR RESPONSE

Your questions are well posed and very relevant.

Note also that the Bottle Engine is addressed fairly extensively in Class Notes between #08 and

#12, and is the subject of the two video experiments to be viewed as part of Assignment 11.

For the situation in question the maximum pressure possible, operating the system between 200 C

and 360 C, is about

T_max / T_min * P_min = 623 K / (473 K) * 100 kPa = 132 kPa.

This would allow us to support a column of water which exerts a pressure of 32 kPa. This column

would be about 3.2 meters high (easily found using Bernoulli's equation).

To raise water to half this height would require a temperature of about 280 C, after which the gas

would expand by factor

623 K / (553 K) = 1.14.

The volume would change by this factor by displacing an equal volume of water, which would be

raised to the 1.6 meter height.

The temperature of the gas would be raised from 200 C to 280 C at constant volume, then from 280 C

to 360 C at constant pressure.

The efficiency we calculate here is the 'practical efficiency', which is the ratio of the

mechanical work done to the thermal energy added to the system. The mechanical work is the

raising of the water, so is equal to the PE change of the water which is raised to the 1.6 m

height. The thermal energy added is the energy required to heat the gas, first at constant volume

then at constant pressure."

Self-critique (if necessary):

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Self-critique rating:

QUESTION FROM STUDENT

My question is on the following question I was trying during a practice test:

Analyze the pressure vs. volume of a 'bottle engine' consisting of 8 liters of an ideal gas as it

operates between minimum

temperature 200 Celsius and maximum temperature 360 Celsius, pumping water to half the maximum

possible height.

Sketch a pressure vs. volume graph from the original state to the maximum-temperature state and

use the graph to

determine the useful work done by the expansion. Then, assuming a diatomic gas, determine the

thermal energy required

to perform the work and the resulting practical efficiency of the process.

I understand the pressure vs volume graph. Does this question basically mean that I should find

d'Q(v) and d'Q(p) and if

so, what temperatures do I use for the temperature change. Then once i find d'Q, how do i find

work. It was stated that

it is the area under the curve, but is this the same as the equation d'Q(p)-d'Q(v). Also

efficiency is found by taking Max

temp-min temp/max temp, so I know how to do that, but why would this change with the amount of

energy needed to

perform work. I am very confused on this problem.

INSTRUCTOR RESPONSE

Your questions are well posed and very relevant.

Note also that the Bottle Engine is addressed fairly extensively in Class Notes between #08 and

#12, and is the subject of the two video experiments to be viewed as part of Assignment 11.

For the situation in question the maximum pressure possible, operating the system between 200 C

and 360 C, is about

T_max / T_min * P_min = 623 K / (473 K) * 100 kPa = 132 kPa.

This would allow us to support a column of water which exerts a pressure of 32 kPa. This column

would be about 3.2 meters high (easily found using Bernoulli's equation).

To raise water to half this height would require a temperature of about 280 C, after which the gas

would expand by factor

623 K / (553 K) = 1.14.

The volume would change by this factor by displacing an equal volume of water, which would be

raised to the 1.6 meter height.

The temperature of the gas would be raised from 200 C to 280 C at constant volume, then from 280 C

to 360 C at constant pressure.

The efficiency we calculate here is the 'practical efficiency', which is the ratio of the

mechanical work done to the thermal energy added to the system. The mechanical work is the

raising of the water, so is equal to the PE change of the water which is raised to the 1.6 m

height. The thermal energy added is the energy required to heat the gas, first at constant volume

then at constant pressure."

Self-critique (if necessary):

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Self-critique rating:

#*&!

Assingment 11

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