{"id":9021,"date":"2023-08-31T08:15:44","date_gmt":"2023-08-31T13:15:44","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=9021"},"modified":"2025-12-13T06:54:57","modified_gmt":"2025-12-13T11:54:57","slug":"2eva2023paoi_t2-pendulo-vertical-amortiguado","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-2eva30\/2eva2023paoi_t2-pendulo-vertical-amortiguado\/","title":{"rendered":"2Eva2023PAOI_T2 EDO P\u00e9ndulo vertical amortiguado"},"content":{"rendered":"\n<h2 class=\"wp-block-heading\">2da Evaluaci\u00f3n 2023-2024 PAO I. 29\/Agosto\/2023<\/h2>\n\n\n\n<p><strong>Tema 2<\/strong> (35 puntos)&nbsp;Una mejor aproximaci\u00f3n a un p\u00e9ndulo oscilante con un \u00e1ngulo \u03b8 m\u00e1s amplio y con un coeficiente de amortiguamiento \u03bc se expresa con una ecuaci\u00f3n diferencial ordinaria de segundo orden.<\/p>\n\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\">\\frac{d^2 \\theta}{dt^2} = -\\mu \\frac{d\\theta}{ dt}-\\frac{g}{L}\\sin (\\theta) <\/span>\n\n\n\n<figure class=\"wp-block-image alignright size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"176\" height=\"268\" src=\"http:\/\/blog.espol.edu.ec\/algoritmos101\/files\/2023\/08\/penduloAmortuguado02.png\" alt=\"p\u00e9ndulo Amortiguado\" class=\"wp-image-17524\" \/><\/figure>\n\n\n\n<p>g = 9.81 m\/s<sup>2<\/sup><\/p>\n\n\n\n<p>L = 2 m<\/p>\n\n\n\n<p>\u03b8(0) = \u03c0\/4 rad<\/p>\n\n\n\n<p>\u03b8' (0) = 0 rad\/s<\/p>\n\n\n\n<p>El p\u00e9ndulo se suelta desde el reposo, desde un \u00e1ngulo de \u03c0\/4 respecto al eje vertical. El coeficiente de amortiguamiento \u03bc=0.5 es proporcional a la velocidad angular.<\/p>\n\n\n\n<p>a. Realice el planteamiento del ejercicio usando Runge-Kutta de 2do Orden<\/p>\n\n\n\n<p>b. Desarrolle tres iteraciones para \u03b8(t) con tama\u00f1o de paso h=0.2<\/p>\n\n\n\n<p>c. Usando el algoritmo, aproxime la soluci\u00f3n entre t=0 a t=10 s, adjunte sus resultados en la evaluaci\u00f3n.<\/p>\n\n\n\n<p>d. Realice una observaci\u00f3n sobre el movimiento estimado del p\u00e9ndulo a lo largo del tiempo.<\/p>\n\n\n\n<p><strong>R\u00fabrica<\/strong>: literal a (5 puntos), literal b (15 puntos), literal c (10 puntos), literal d (5 puntos)<\/p>\n\n\n\n<p><strong>Referencia<\/strong>: <a href=\"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-eval\/mn-2e20\/2eva2019ti_t2-pendulo-vertical\/\" data-type=\"post\" data-id=\"4089\">2Eva2019TI_T2 P\u00e9ndulo vertical<\/a><\/p>\n\n\n\n<p>Vista general de ecuaciones diferenciales I Cap\u00edtulo 1, 6min 54s. 3Blue1Brown 31-Marzo-2023.<\/p>\n\n\n\n<figure class=\"wp-block-embed is-type-video is-provider-youtube wp-block-embed-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<iframe loading=\"lazy\" title=\"Differential equations, a tourist&#039;s guide | DE1\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/p_di4Zn4wz4?start=413&feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>2da Evaluaci\u00f3n 2023-2024 PAO I. 29\/Agosto\/2023 Tema 2 (35 puntos)&nbsp;Una mejor aproximaci\u00f3n a un p\u00e9ndulo oscilante con un \u00e1ngulo \u03b8 m\u00e1s amplio y con un coeficiente de amortiguamiento \u03bc se expresa con una ecuaci\u00f3n diferencial ordinaria de segundo orden. g = 9.81 m\/s2 L = 2 m \u03b8(0) = \u03c0\/4 rad \u03b8' (0) = 0 [&hellip;]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"wp-custom-template-entrada-mn","format":"standard","meta":{"footnotes":""},"categories":[22],"tags":[56],"class_list":["post-9021","post","type-post","status-publish","format-standard","hentry","category-mn-2eva30","tag-edo"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/9021","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=9021"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/9021\/revisions"}],"predecessor-version":[{"id":17526,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/9021\/revisions\/17526"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=9021"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=9021"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=9021"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}