#$&* course Phy 122 6/25 11pm 011. `Query 11
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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: I understand the first part of the explanation with a string held by both ends by using the equation in the class notes. But I am confused if there is another equation for when the string is only held by one hand and is free on one end? ???Is there an equation I can follow when the string is free on one end??? Your Self-Critique Rating: 2
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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: This is a much simpler way of thinking about it but I think the equation I found in my notes would find the same answer. Your Self-Critique Rating: 3 ********************************************* 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: According to my class notes you could find the velocity of a wave in a uniform string using this equation: v = `sqrt( T / `mu ) where T is the tension and ‘mu is the mass per unit length. 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 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!