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
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?
Your Solution:
Confidence Rating:
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. **
Self-critique (if necessary):
Self-critique Rating:
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?
Your solution:
Confidence Rating:
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. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`q **** Given the tension and mass density of a
string how do we determine the velocity of the wave in the string?
Your solution:
Confidence Rating:
Given Solution:
`a** We divide tension by mass per unit length:
v = sqrt ( tension / (mass/length) ). **
Self-critique (if necessary):
Self-critique Rating:
Question:
`q**** gen phy explain in your own words the meaning of the
principal of superposition
Your solution:
Confidence Rating:
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.
**
Self-critique (if necessary):
Self-critique Rating:
Question:
`q **** gen phy
what does it mean to say that the angle of reflection is equal to the
angle of incidence?
Your solution:
Confidence Rating:
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
**
Your solution:
Confidence Rating:
Self-critique Rating: