Assignment 12

course Phy 202

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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:

The length of a string is equal to ½ wavelength, so 2* the string length is equal to a whole wavelength. So we need to multiply the wavelength by 2 in order to get the string length, or divide the string length in half.

For the 2nd harmonic we have (2/2)*2 = 1*the string length

For the 3rd harmonic we have (3/2)*2= 2/3* string length

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Question: `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):

ok

Self-critique Rating:

3

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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= velocity/wavelength

Frequency is the number of peaks or cycles that pass a certain spot in a wavelength.

We take the velocity and divide it by the wavelength found for each separate harmonic. This will give us each frequency.

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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):

ok

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:

V= sqrt(tension/mass per unit length)

We divide the tension by the quotient of the mass divided by the length of the wave to find the velocity.

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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):

ok

Self-critique Rating:

3

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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:

This principle basically says that when two wavelengths meet at some point, the displacement that results in the sum of the two wavelengths sizes separately.

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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):

ok

Self-critique Rating:3

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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:

This law means that the angle of incidence, or the angle at which the ray hits the surface is the same as the angle at which the ray is reflected.

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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 **

Self Critique (if necessary):

Ok

Self Critique rating:

3

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