#$&* course PHY 202 3/31 about 8:30 am 011. `Query 10
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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:OK Your Self-Critique Rating:OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Frequency is the number of cycles per one second. Velocity is m/s and wavelength is m. Dividing velocity by ‘lambda will result in a per second units. So 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:OK Your Self-Critique Rating:OK ********************************************* Question: **** Given the tension and mass density of a string how do we determine the velocity of the wave in the string? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Velocity is in units of m/s. T is in N, or kg*m / s^2. ‘mu is kg/m. dividing T/mass per unit length, ‘mu, will give units of kg*m*m / (s^2*kg), simplified to m^2/s^2. Take the square root and you get m/s. v = ‘sqrt(T/’mu) or v = ‘sqrt(T / (m/L)). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** We divide tension by mass per unit length: v = sqrt ( tension / (mass/length) ). ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: **** gen phy explain in your own words the meaning of the principal of superposition YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: If two waves arrive at the same point, each wave corresponds to a force, and forces can be added. So if two waves combine, the amplitudes can be added together. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:OK Your Self-Critique Rating:OK ********************************************* Question: **** gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: It means that a wave that is reflected on a surface will be reflected at the same angle measured perpendicular to the point of the surface in which it reflects. 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:OK Your Self-Critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: