Query 12

#$&*

course Phy 202

2/23 1:37pm

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.

012. `Query 10

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Question: `q**** Query introductory set six, problems 11-13 **** given the length of a string how do we

determine the wavelengths of the first few harmonics?

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Your Solution:

n/2*'lambda = L

lambda = 2L/n

I don't understand the string attached at one end. I can only find one thing about it in the book, p310

in 6th edition.

Looks like length n/4*'lambda = length of string

@&

If a string is attached at one end, with the other end free, then the free end is an antinode. This determines the possible series of nodes and antinodes.

*@

confidence rating #$&*:

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Given Solution: `q** 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 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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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q**** 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:

f = v/lambda = v/(2L/n) = (nv)/(2L)

confidence rating #$&*:

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Given Solution:

`a** 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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Self-critique (if necessary):

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Question: `q **** 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:

F_T = m/L*v^2 so v = sqrt(F_T*L/m)

confidence rating #$&*:

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Given Solution:

`a** We divide tension by mass per unit length:

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

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Self-critique (if necessary):

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

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Your solution:

When you have 2 waves coming towards each other you add their displacements together

confidence rating #$&*:

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Given Solution:

`a** 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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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q **** 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:

The angle of incidence is the angle the ray makes with a line perpendicular to the surface it's being

reflected. The angle of reflection is the angle a reflected ray makes with the perpendicular of the

reflected surface. To saythey are equal means the angle coming in equals the angle going out.

confidence rating #$&*:

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Given Solution:

`a** 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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Your solution:

confidence rating #$&*:

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Self-critique Rating:"

Self-critique (if necessary):

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Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Your work looks good. See my notes. Let me know if you have any questions. &#