Query 12

#$&*

course Phy 122

2/21 1557hrs

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:

One complete wave in a standing wave consist of 2 complete loops Length (L) = Lambda, if you were to get one complete wave out of the length of string. If you got one loop out of the string, that would be half of a wave so, Length (L) = ½ Lambda.

I believe the wave is harmonic if the force needed to bring it back to equilibrium = the force of the displacement. So if you doubled the length of the string, you would get double the amount of waves. From above, if Length (L) = ½ Lambda, then 2(L) = Lambda, and so on, adding one loop per Length (L) of sring. So 3(L) = 1.5Lambda.

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): I think we are talking about the same thing but, if the wave decreases then the wavelength would be damped I believe. This would mean that the tension on the string is slowly overcoming the force of original displacement.

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Self-critique Rating:2

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Damping would affect the amplitude, not the wavelength. The frequency of the disturbance wouldn't change, nor would the speed of propagation of disturbances, so that wavelength would be unaffected. Energy can be lost, though, and energy depends on the amplitude of the oscillations.
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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:

Frequency is the number of high points (in the wave) per a unit of time. The velocity would be how fast that peak travels to the next peak. So frequency would = number of crests/time allotment. And velocity would be how fast that point (on the crest) takes to crest again.

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): This explanation is a lot easier to understand

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Self-critique Rating:3

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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:

Velocity is the squareroot of (tension/ (mass/length))

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):3

Self-critique (if necessary):

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Self-critique rating:

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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:

Velocity is the squareroot of (tension/ (mass/length))

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):3

Self-critique (if necessary):

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Self-critique rating:

#*&!

Query 12

@& Damping would affect the amplitude, not the wavelength. The frequency of the disturbance wouldn't change, nor would the speed of propagation of disturbances, so that wavelength would be unaffected. Energy can be lost, though, and energy depends on the amplitude of the oscillations. *@

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

@&
Damping would affect the amplitude, not the wavelength. The frequency of the disturbance wouldn't change, nor would the speed of propagation of disturbances, so that wavelength would be unaffected. Energy can be lost, though, and energy depends on the amplitude of the oscillations.
*@