{"id":1183,"date":"2026-01-24T16:18:35","date_gmt":"2026-01-24T20:18:35","guid":{"rendered":"https:\/\/fiziko.net\/?page_id=1183"},"modified":"2026-02-12T22:15:14","modified_gmt":"2026-02-13T02:15:14","slug":"ondas-transversais","status":"publish","type":"page","link":"https:\/\/fiziko.net\/?page_id=1183","title":{"rendered":"Ondas Transversais"},"content":{"rendered":"\n<p>Uma <strong>onda mec\u00e2nica <\/strong>\u00e9 uma <strong>perturba\u00e7\u00e3o<\/strong> <strong>energ\u00edtica<\/strong> que se desloca atrav\u00e9s de um  <strong>meio material<\/strong>, no qual a onda se propaga.<\/p>\n\n\n\n<p>Ondas n\u00e3o transportam materia, a energia se propaga de mol\u00e9cula a mol\u00e9cula ao longo do meio material por onde a perturba\u00e7\u00e3o energ\u00e9tica \u00e9 transmitida.<\/p>\n\n\n\n<p>A forma mais simples de uma onda \u00e9 a <strong>onda transversal<\/strong>, gerada por um Oscilador Hmarm\u00f4nico Simples (OHS), posicionado ao longo do eixo y, preso a uma corda com densidade de massa <math data-latex=\"\\mu = \\frac{dm}{dl}\"><semantics><mrow><mi>\u03bc<\/mi><mo>=<\/mo><mfrac><mrow><mi>d<\/mi><mi>m<\/mi><\/mrow><mrow><mi>d<\/mi><mi>l<\/mi><\/mrow><\/mfrac><\/mrow><annotation encoding=\"application\/x-tex\">\\mu = \\frac{dm}{dl}<\/annotation><\/semantics><\/math>, que sofre uma tens\u00e3o F, conforme verificamos na anima\u00e7\u00e3o abaixo<\/p>\n\n\n\n<figure class=\"wp-block-video\"><video height=\"1388\" style=\"aspect-ratio: 1720 \/ 1388;\" width=\"1720\" controls src=\"https:\/\/fiziko.net\/wp-content\/uploads\/2026\/01\/Ondas-01-2.mp4\"><\/video><\/figure>\n\n\n\n<p>O ponto azul descreve o movimento do OHS que, como sabemos, \u00e9 descrito pela equa\u00e7\u00e3o a seguir<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>y(t)=A\\sin{(wt+\\phi)}<\/pre><\/div>\n\n\n\n<p>No entanto, como podemos verificar na anima\u00e7\u00e3o, todos os pontos da onda, ao longo do eixo x, movem-se tamb\u00e9m na vertical, ponto vermelho em destaque, logo a fun\u00e7\u00e3o que descreve a onda transversal \u00e9 do tipo<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>y(x,t)=A\\sin(\\alpha x-\\omega t+\\phi)<\/pre><\/div>\n\n\n\n<p>Devemos determinar o valor de <math data-latex=\"\\alpha\"><semantics><mi>\u03b1<\/mi><annotation encoding=\"application\/x-tex\">\\alpha<\/annotation><\/semantics><\/math>, sabendo que todo o argumento da fin\u00e7\u00e3o seno \u00e9 adimensional,<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\begin{cases}\n   \\displaystyle {\\omega = \\frac{2\\pi}{T}= 2\\pi f }, &amp; \\quad \\text{frequ\u00eancia angular} \\\\\n   \\displaystyle {\\alpha = k = \\frac{2\\pi}{\\lambda}}, &amp; \\quad \\text{n\u00famero de onda}\n\\end{cases}<\/pre><\/div>\n\n\n\n<p>podemos demostrar a equa\u00e7\u00e3o do n\u00famero de onda, considerando que a fase da fun\u00e7\u00e3o seno igual a zero e obtendo as rela\u00e7\u00f5es por meio das defini\u00e7\u00f5es a seguir<\/p>\n\n\n\n<p>Com que velocidade a energia se propaga ao longo do meio material? para responder essa pergunta basta considerar que a onda levar\u00e1 o tempo T (per\u00edodo) para concluir um ciclo e que a dist\u00e2ncia que o pulso de energia se movimentou \u00e9 igual ao comprimento de onda <math data-latex=\"\\lambda\"><semantics><mi>\u03bb<\/mi><annotation encoding=\"application\/x-tex\">\\lambda<\/annotation><\/semantics><\/math> (a dist\u00e2ncia de dois picos consecutivos ou a dist\u00e2ncia de dois vales consecutivos), logo<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>v=\\frac{\\Delta x}{\\Delta t}=\\frac{\\lambda}{T}=\\lambda f<\/pre><\/div>\n\n\n\n<p>assim,<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\alpha x-\\omega t=0 \\\\\n.\\\\\n\\alpha = \\frac{\\omega t}{x}=\\frac{2\\pi}{T}\\frac{t}{x}=\\frac{2\\pi}{T}\\frac{1}{v}=\\frac{2\\pi}{T}\\frac{T}{\\lambda} \\\\\n.\\\\\n\\alpha=k=\\frac{2\\pi}{\\lambda}<\/pre><\/div>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full\"><img fetchpriority=\"high\" decoding=\"async\" width=\"624\" height=\"645\" src=\"http:\/\/fiziko.net\/wp-content\/uploads\/2026\/01\/onda-001.jpg\" alt=\"\" class=\"wp-image-1199\" srcset=\"https:\/\/fiziko.net\/wp-content\/uploads\/2026\/01\/onda-001.jpg 624w, https:\/\/fiziko.net\/wp-content\/uploads\/2026\/01\/onda-001-290x300.jpg 290w\" sizes=\"(max-width: 624px) 100vw, 624px\" \/><\/figure>\n<\/div>\n\n\n<p>Vamos considerar que a onda transversal ser\u00e1 produzida em uma corda de comptimento L, tensionada pela for\u00e7a F e com densidade de massa linear dada por <math data-latex=\"\\mu=\\frac{dm}{dl}=\\frac{\\delta m}{\\delta l}\"><semantics><mrow><mi>\u03bc<\/mi><mo>=<\/mo><mfrac><mrow><mi>d<\/mi><mi>m<\/mi><\/mrow><mrow><mi>d<\/mi><mi>l<\/mi><\/mrow><\/mfrac><mo>=<\/mo><mfrac><mrow><mi>\u03b4<\/mi><mi>m<\/mi><\/mrow><mrow><mi>\u03b4<\/mi><mi>l<\/mi><\/mrow><\/mfrac><\/mrow><annotation encoding=\"application\/x-tex\">\\mu=\\frac{dm}{dl}=\\frac{\\delta m}{\\delta l}<\/annotation><\/semantics><\/math>, veja figura<\/p>\n\n\n\n<p>A for\u00e7a resultante exercida sobre essa por\u00e7\u00e3o da corda ser\u00e1<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\begin{cases}\n   F_{2x} - F_{1x} = \\delta m. a_{x}, &amp;  \\quad a_{x}=0\\\\\n   F_{2y} - F_{1y} = \\delta m. a_{y}, &amp; \\quad a_{y}=\\frac{\\partial^2y}{\\partial t^2}\n\\end{cases}<\/pre><\/div>\n\n\n\n<p>como a por\u00e7\u00e3o de massa <math data-latex=\"\\delta m\"><semantics><mrow><mi>\u03b4<\/mi><mi>m<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">\\delta m<\/annotation><\/semantics><\/math> n\u00e3o se move ao longo do eixo x, tem-se que <math data-latex=\"F_{1x}=F_{2x}=F\"><semantics><mrow><msub><mi>F<\/mi><mrow><mn>1<\/mn><mi>x<\/mi><\/mrow><\/msub><mo>=<\/mo><msub><mi>F<\/mi><mrow><mn>2<\/mn><mi>x<\/mi><\/mrow><\/msub><mo>=<\/mo><mi>F<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">F_{1x}=F_{2x}=F<\/annotation><\/semantics><\/math>, considerando-se que <math data-latex=\"\\cos \\theta \\approx1\"><semantics><mrow><mrow><mi>cos<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mi>\u03b8<\/mi><mo>\u2248<\/mo><mn>1<\/mn><\/mrow><annotation encoding=\"application\/x-tex\">\\cos \\theta \\approx1<\/annotation><\/semantics><\/math> e <math data-latex=\"\\sin \\theta\\approx \\tan \\theta \\approx \\frac{\\partial y}{\\partial x}\"><semantics><mrow><mrow><mi>sin<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mi>\u03b8<\/mi><mo>\u2248<\/mo><mrow><mi>tan<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mi>\u03b8<\/mi><mo>\u2248<\/mo><mfrac><mrow><mi>\u2202<\/mi><mi>y<\/mi><\/mrow><mrow><mi>\u2202<\/mi><mi>x<\/mi><\/mrow><\/mfrac><\/mrow><annotation encoding=\"application\/x-tex\">\\sin \\theta\\approx \\tan \\theta \\approx \\frac{\\partial y}{\\partial x}<\/annotation><\/semantics><\/math>, alem de <math data-latex=\"\\Delta \\theta \"><semantics><mrow><mrow><mi mathvariant=\"normal\">\u0394<\/mi><\/mrow><mi>\u03b8<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">\\Delta \\theta <\/annotation><\/semantics><\/math>, <math data-latex=\"\\Delta x\"><semantics><mrow><mrow><mi mathvariant=\"normal\">\u0394<\/mi><\/mrow><mi>x<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">\\Delta x<\/annotation><\/semantics><\/math> e <math data-latex=\"\\Delta y\"><semantics><mrow><mrow><mi mathvariant=\"normal\">\u0394<\/mi><\/mrow><mi>y<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">\\Delta y<\/annotation><\/semantics><\/math> muito pequenos.<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\begin{cases}\n   \\displaystyle F_{1y}=F \\sin(\\theta) \\simeq F \\tan(\\theta)=F \\left. \\frac{\\partial y}{\\partial x} \\right|_{\\theta} \\\\\n    \\displaystyle F_{2y}=F \\sin(\\theta+\\Delta \\theta) \\simeq F \\tan(\\theta+\\Delta \\theta)=F \\left. \\frac{\\partial y}{\\partial x} \\right|_{\\theta+\\Delta \\theta}\n\\end{cases}<\/pre><\/div>\n\n\n\n<p>tem-se que<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>F \\left. \\frac{\\partial y}{\\partial x} \\right|_{\\theta+\\Delta \\theta}-F \\left. \\frac{\\partial y}{\\partial x} \\right|_{\\theta} = \\delta m \\frac{\\partial^2y}{\\partial t^2} \\\\ .\\\\\nF \\left. \\frac{\\partial y}{\\partial x} \\right|_{\\theta+\\Delta \\theta}-F \\left. \\frac{\\partial y}{\\partial x} \\right|_{\\theta} = \\mu \\Delta x \\frac{\\partial^2y}{\\partial t^2} \\\\ .\\\\\n\\frac{ \\left. \\frac{\\partial y}{\\partial x} \\right|_{\\theta+\\Delta \\theta}- \\left. \\frac{\\partial y}{\\partial x} \\right|_{\\theta}}{\\Delta x} = \\frac{\\mu}{F} \\frac{\\partial^2y}{\\partial t^2}<\/pre><\/div>\n\n\n\n<p>se tomarmos o limite de <math data-latex=\"\\Delta x \\to 0\"><semantics><mrow><mrow><mi mathvariant=\"normal\">\u0394<\/mi><\/mrow><mi>x<\/mi><mo>\u2192<\/mo><mn>0<\/mn><\/mrow><annotation encoding=\"application\/x-tex\">\\Delta x \\to 0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\frac{\\partial^2y}{\\partial x^2}=\\frac{\\mu}{F}\\frac{\\partial^2y}{\\partial t^2}<\/pre><\/div>\n\n\n\n<p>dizemos que a equa\u00e7\u00e3o abaixo \u00e9 a <strong>fun\u00e7\u00e3o da onda transversal<\/strong><\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>y(x,t)=A\\sin(kx-\\omega t+\\phi)<\/pre><\/div>\n\n\n\n<p>a equa\u00e7\u00e3o acima \u00e9 chamada de <strong>equa\u00e7\u00e3o de movimento da onda transversal<\/strong>, aplicando as derivadas parciais tem-se<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\begin{cases}\n   \\displaystyle\\frac{\\partial^2y}{\\partial t^2}=-A \\omega^2\\sin(kx-\\omega t + \\phi) \\\\ .\\\\\n    \\displaystyle\\frac{\\partial^2y}{\\partial x^2}=-A k^2\\sin(kx-\\omega t + \\phi) \n\\end{cases}<\/pre><\/div>\n\n\n\n<p>igualando as derivadas, temos<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\frac{\\partial^2y}{\\partial x^2}=\\frac{k^2}{\\omega^2}\\frac{\\partial^2y}{\\partial t^2}<\/pre><\/div>\n\n\n\n<p>Como <math data-latex=\"v=\\lambda f=\\frac{2\\pi f\\lambda}{2\\pi}=\\frac{\\omega}{k}\"><semantics><mrow><mi>v<\/mi><mo>=<\/mo><mi>\u03bb<\/mi><mi>f<\/mi><mo>=<\/mo><mfrac><mrow><mn>2<\/mn><mi>\u03c0<\/mi><mi>f<\/mi><mi>\u03bb<\/mi><\/mrow><mrow><mn>2<\/mn><mi>\u03c0<\/mi><\/mrow><\/mfrac><mo>=<\/mo><mfrac><mi>\u03c9<\/mi><mi>k<\/mi><\/mfrac><\/mrow><annotation encoding=\"application\/x-tex\">v=\\lambda f=\\frac{2\\pi f\\lambda}{2\\pi}=\\frac{\\omega}{k}<\/annotation><\/semantics><\/math>, temos que<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\frac{1}{v^2}=\\frac{k^2}{\\omega^2}=\\frac{\\mu}{F} \\\\ .\n\\\\\nv=\\sqrt{\\frac{F}{\\mu}}<\/pre><\/div>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Pot\u00eancia Transportada por uma Onda Transversal<\/strong><\/h2>\n\n\n\n<p>Um oscilador que produz um OHS realiza trabalho e transmite energia \u00e0 corda, que passa a oscilar. Essa energia, obviamente, n\u00e3o fica armazenada em um ponto da corda, mas, sim, propaga-se com a onda. A <strong>for\u00e7a transversal restauradora aplicada<\/strong> <math data-latex=\"F_{y}\"><semantics><msub><mi>F<\/mi><mi>y<\/mi><\/msub><annotation encoding=\"application\/x-tex\">F_{y}<\/annotation><\/semantics><\/math>, no ponto x, a um elemento dx da corda sob tens\u00e3o F , \u00e9<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>F_{y}=-F \\frac{\\partial y}{\\partial x}<\/pre><\/div>\n\n\n\n<p>a <strong>Pot\u00eancia Instant\u00e2nea<\/strong> \u00e9 definida pela equa\u00e7\u00e3o abaixo, com <math data-latex=\"v_{y}=\\frac{\\partial y}{\\partial t}\"><semantics><mrow><msub><mi>v<\/mi><mi>y<\/mi><\/msub><mo>=<\/mo><mfrac><mrow><mi>\u2202<\/mi><mi>y<\/mi><\/mrow><mrow><mi>\u2202<\/mi><mi>t<\/mi><\/mrow><\/mfrac><\/mrow><annotation encoding=\"application\/x-tex\">v_{y}=\\frac{\\partial y}{\\partial t}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>P_{ot}(x,t)=F_{y}v_{y}=-F \\frac{\\partial y}{\\partial x}\\frac{\\partial y}{\\partial t}<\/pre><\/div>\n\n\n\n<p>logo,<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>P_{ot}(x,t)=FA^2k\\omega\\cos(kx-\\omega t+\\phi)\\cos(kx-\\omega t+\\phi) \\\\ .\n\\\\\nP_{ot}(x,t)=FA^2k\\omega\\cos^2(kx-\\omega t+\\phi)<\/pre><\/div>\n\n\n\n<p>A m\u00e9dia temporal do <math data-latex=\"\\cos^2(kx-\\omega t+\\phi)=\\frac{1}{2}\"><semantics><mrow><msup><mi>cos<\/mi><mn>2<\/mn><\/msup><mo>\u2061<\/mo><mo form=\"prefix\" stretchy=\"false\">(<\/mo><mi>k<\/mi><mi>x<\/mi><mo>\u2212<\/mo><mi>\u03c9<\/mi><mi>t<\/mi><mo>+<\/mo><mi>\u03d5<\/mi><mo form=\"postfix\" stretchy=\"false\">)<\/mo><mo>=<\/mo><mfrac><mn>1<\/mn><mn>2<\/mn><\/mfrac><\/mrow><annotation encoding=\"application\/x-tex\">\\cos^2(kx-\\omega t+\\phi)=\\frac{1}{2}<\/annotation><\/semantics><\/math>, assim, a <strong>pot\u00eancia m\u00e9dia<\/strong> ser\u00e1<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\bar{P}_{ot}=\\frac{1}{2}\\omega k F A^2=\\frac{1}{2}\\mu v \\omega^2A^2<\/pre><\/div>\n\n\n\n<h2 class=\"wp-block-heading\">Intensidade da onda<\/h2>\n\n\n\n<p>A<strong>\u00a0intensidade <math data-latex=\"I\"><semantics><mi>I<\/mi><annotation encoding=\"application\/x-tex\">I<\/annotation><\/semantics><\/math> <\/strong>de uma onda \u00e9 uma\u00a0grandeza f\u00edsica que descreve a quantidade de energia que a onda transporta por unidade de \u00e1rea, em um intervalo de tempo, atrav\u00e9s de uma superf\u00edcie perpendicular \u00e0 dire\u00e7\u00e3o de propaga\u00e7\u00e3o, portanto tem-se<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>I=\\frac{\\bar{P}_{ot}}{A_{T}}<\/pre><\/div>\n\n\n\n<p>onde, <math data-latex=\"\\bar{P}_{ot}\"><semantics><msub><mover><mi>P<\/mi><mo stretchy=\"false\" class=\"tml-capshift\" style=\"math-style:normal;math-depth:0;\">\u203e<\/mo><\/mover><mrow><mi>o<\/mi><mi>t<\/mi><\/mrow><\/msub><annotation encoding=\"application\/x-tex\">\\bar{P}_{ot}<\/annotation><\/semantics><\/math> \u00e9 a pot\u00eancia m\u00e9dia da onda e <math data-latex=\"A_{T}\"><semantics><msub><mi>A<\/mi><mi>T<\/mi><\/msub><annotation encoding=\"application\/x-tex\">A_{T}<\/annotation><\/semantics><\/math> \u00e9 a \u00e1rea perpendicular \u00e0 dire\u00e7\u00e3o de propaga\u00e7\u00e3o da onda.<\/p>\n\n\n\n<p>Como a pot\u00eancia m\u00e9dia \u00e9 dada por<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\bar{P}_{ot}=\\frac{1}{2}\\omega k F A^2=\\frac{1}{2}\\mu v \\omega^2A^2<\/pre><\/div>\n\n\n\n<p>vamos escrever a densidade linear <math data-latex=\"\\mu\"><semantics><mi>\u03bc<\/mi><annotation encoding=\"application\/x-tex\">\\mu<\/annotation><\/semantics><\/math> com rela\u00e7\u00e3o a densidade volum\u00e9trica da corda <math data-latex=\"\\rho\"><semantics><mi>\u03c1<\/mi><annotation encoding=\"application\/x-tex\">\\rho<\/annotation><\/semantics><\/math>, assim<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\mu=\\frac{m}{l} \\cdot \\frac{A_{T}}{A_{T}}=\\frac{m}{lA_{T}}A_{T}=\\rho A_{T}<\/pre><\/div>\n\n\n\n<p>logo, podemos reescrever a equa\u00e7\u00e3o da pot\u00eancia m\u00e9dia na forma<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\bar{P}_{ot}=\\frac{1}{2}\\rho A_{T} v \\omega^2A^2<\/pre><\/div>\n\n\n\n<p>obtendo-se, para a intensidade da onda transversal na corda, com densidade volum\u00e9trica da corda <math data-latex=\"\\rho\"><semantics><mi>\u03c1<\/mi><annotation encoding=\"application\/x-tex\">\\rho<\/annotation><\/semantics><\/math> e \u00e1rea da se\u00e7\u00e3o transversal da corda <math data-latex=\"A_{T}\"><semantics><msub><mi>A<\/mi><mi>T<\/mi><\/msub><annotation encoding=\"application\/x-tex\">A_{T}<\/annotation><\/semantics><\/math>, como<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>I=\\frac{\\bar{P}_{ot}}{A_{T}}=\\frac{1}{2}\\rho v \\omega^2A^2<\/pre><\/div>\n\n\n\n<h2 class=\"wp-block-heading\">Interfer\u00eancia de Ondas e o Princ\u00edpio da Superposi\u00e7\u00e3o<\/h2>\n\n\n\n<p>Da mesma forma que acontece com as for\u00e7as que atuam sobre um corpo de massa m, o princ\u00edpio da superposi\u00e7\u00e3o \u00e9 aplicado ao estudo das ondas. A soma vetorial de todas as for\u00e7as (<strong>princ\u00edpio da superposi\u00e7\u00e3o<\/strong>) atuando sobre um corpo de massa m \u00e9 igual a for\u00e7a resultante, nas ondas a soma das fun\u00e7\u00f5es de onda (<strong>princ\u00edpio da superposi\u00e7\u00e3o<\/strong>) ocasionam o fen\u00f4meno de interfer\u00eancia das ondas, logo, podemos dizer que o <strong>princ\u00edpio de superposi\u00e7\u00e3o<\/strong>, tamb\u00e9m conhecido como&nbsp;<strong>propriedade de superposi\u00e7\u00e3o<\/strong>, afirma que, para todos os&nbsp;sistemas lineares, a resposta l\u00edquida causada por dois ou mais est\u00edmulos \u00e9 a soma das respostas que teriam sido causadas por cada est\u00edmulo individualmente.<\/p>\n\n\n\n<p>Tomemos duas ondas de mesma amplitude A, mesma frequ\u00eancia <math data-latex=\"\\omega\"><semantics><mi>\u03c9<\/mi><annotation encoding=\"application\/x-tex\">\\omega<\/annotation><\/semantics><\/math> e mesmo comprimento de onda <math data-latex=\"\\lambda\"><semantics><mi>\u03bb<\/mi><annotation encoding=\"application\/x-tex\">\\lambda<\/annotation><\/semantics><\/math>, mas com constantes de fase diferentes, dizemos ent\u00e3o que as ondas est\u00e3o defasadas<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\begin{cases}\n   y_{1}=A\\sin(kx-\\omega t+\\phi_{1}) \\\\\n   y_{2}=A\\sin(kx-\\omega t+\\phi_{2})\n\\end{cases}<\/pre><\/div>\n\n\n\n<p>\u00c9 poss\u00edvel estudarmos os efeitos quando as ondas se interferem, aplicando o princ\u00edpio da superposi\u00e7\u00e3o que nos d\u00e1<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>y=y_{1}+y_{2} \\\\ .\n\\\\\ny=A\\sin(kx-\\omega t+\\phi_{1})+A\\sin(kx-\\omega t+\\phi_{2})<\/pre><\/div>\n\n\n\n<p>utilizando a identidade <math data-latex=\"\\sin \\theta + \\sin \\gamma = 2\\sin {\\frac{\\theta+\\gamma}{2}}\\cos {\\frac{\\theta-\\gamma}{2}}\"><semantics><mrow><mrow><mi>sin<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mi>\u03b8<\/mi><mo>+<\/mo><mrow><mi>sin<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mi>\u03b3<\/mi><mo>=<\/mo><mn>2<\/mn><mrow><mspace width=\"0.1667em\"><\/mspace><mi>sin<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mfrac><mrow><mi>\u03b8<\/mi><mo>+<\/mo><mi>\u03b3<\/mi><\/mrow><mn>2<\/mn><\/mfrac><mrow><mspace width=\"0.1667em\"><\/mspace><mi>cos<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mfrac><mrow><mi>\u03b8<\/mi><mo>\u2212<\/mo><mi>\u03b3<\/mi><\/mrow><mn>2<\/mn><\/mfrac><\/mrow><annotation encoding=\"application\/x-tex\">\\sin \\theta + \\sin \\gamma = 2\\sin {\\frac{\\theta+\\gamma}{2}}\\cos {\\frac{\\theta-\\gamma}{2}}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n<p>fa\u00e7amos<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\begin{cases}\n   \\theta=kx-\\omega t+\\phi_{1} \\\\\n   \\gamma=kx-\\omega t+\\phi_{2}\n\\end{cases}<\/pre><\/div>\n\n\n\n<p>portanto,<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\begin{cases}\n   \\displaystyle\\frac{\\theta+\\gamma}{2}=kx-\\omega t + \\phi_{R}, &amp; \\text{onde }\\phi_{R}=\\displaystyle\\frac{\\phi_{1}+\\phi_{2}}{2} \\\\\n   \\displaystyle\\frac{\\theta-\\gamma}{2}=\\frac{\\Delta \\phi}{2}, &amp; \\text{onde }\\Delta\\phi=\\phi_{1}-\\phi_{2}\n\\end{cases}<\/pre><\/div>\n\n\n\n<p>dizemos que <math data-latex=\"\\phi_{R}\"><semantics><msub><mi>\u03d5<\/mi><mi>R<\/mi><\/msub><annotation encoding=\"application\/x-tex\">\\phi_{R}<\/annotation><\/semantics><\/math> \u00e9 a constante de fase da onda resultante e que <math data-latex=\"\\Delta\\phi\"><semantics><mrow><mrow><mi mathvariant=\"normal\">\u0394<\/mi><\/mrow><mi>\u03d5<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">\\Delta\\phi<\/annotation><\/semantics><\/math> \u00e9 a diferen\u00e7a de fase das ondas, assim a forma da onda resultante \u00e9<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>y(x,t)=2A\\cos{\\frac{\\Delta\\phi}{2}}\\sin(kx-\\omega t+\\phi_{R})<\/pre><\/div>\n\n\n\n<p>nota-se que a nova amplitude resultante \u00e9<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>A_{R}=2A\\cos{\\frac{\\Delta\\phi}{2}}<\/pre><\/div>\n\n\n\n<p>A <strong>Interfer\u00eancia Construtiva<\/strong> \u00e9 determinada quando a amplitude resultante assume o seu maior valor, que ocorre quando <math data-latex=\"\\cos \\frac{\\Delta\\phi}{2}=\\pm 1\"><semantics><mrow><mrow><mi>cos<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mfrac><mrow><mrow><mi mathvariant=\"normal\">\u0394<\/mi><\/mrow><mi>\u03d5<\/mi><\/mrow><mn>2<\/mn><\/mfrac><mo>=<\/mo><mo form=\"prefix\" stretchy=\"false\">\u00b1<\/mo><mn>1<\/mn><\/mrow><annotation encoding=\"application\/x-tex\">\\cos \\frac{\\Delta\\phi}{2}=\\pm 1<\/annotation><\/semantics><\/math>, logo<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\frac{\\Delta\\phi}{2}=0,\\pi, 2\\pi,3\\pi\\dots \\\\.\n\\\\\n\\frac{\\Delta\\phi}{2}=n\\pi, \\quad\\text{com }n=0,1,2,3 \\dots\n<\/pre><\/div>\n\n\n\n<p>A <strong>Interfer\u00eancia Destrutiva<\/strong> ocorre quando <math data-latex=\"\\cos{\\frac{\\Delta\\phi}{2}}=0\"><semantics><mrow><mrow><mi>cos<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mfrac><mrow><mrow><mi mathvariant=\"normal\">\u0394<\/mi><\/mrow><mi>\u03d5<\/mi><\/mrow><mn>2<\/mn><\/mfrac><mo>=<\/mo><mn>0<\/mn><\/mrow><annotation encoding=\"application\/x-tex\">\\cos{\\frac{\\Delta\\phi}{2}}=0<\/annotation><\/semantics><\/math>, logo<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\frac{\\Delta\\phi}{2}=\\frac{\\pi}{2}, 3\\frac{\\pi}{2}, 5\\frac{\\pi}{2}  \\dots \\\\.\n\\\\\n{\\Delta\\phi}=m\\pi, \\quad\\text{com }m=1,3,5 \\dots<\/pre><\/div>\n\n\n\n<p>Testando a teoria em uma simula\u00e7\u00e3o no GeoGebra.<\/p>\n\n\n\n<iframe src=\"https:\/\/www.geogebra.org\/calculator\" width=\"100%\" height=\"500\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" allowfullscreen><\/iframe>\n\n\n\n<p>Consideremos o caso em que a corda est\u00e1 fixa nas duas extremidades, quando produzimos a onda na corda esta se propaga na dire\u00e7\u00e3o x positivo e \u00e9 refletida, defasando em <math data-latex=\"\\pi\"><semantics><mi>\u03c0<\/mi><annotation encoding=\"application\/x-tex\">\\pi<\/annotation><\/semantics><\/math>, e se movendo na dire\u00e7\u00e3o x negativo, na pr\u00e1tica tem-se<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\begin{cases}\n   y_{1}=A\\sin(kx-\\omega t) \\\\\n   y_{2}=A\\sin(kx+\\omega t)\n\\end{cases}<\/pre><\/div>\n\n\n\n<p>com a onda resultante dada por<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>y(x,t)=2A\\sin{(kx)}\\cos{(\\omega t)}<\/pre><\/div>\n\n\n\n<p>Onde a amplitude resultante \u00e9 dada por <\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>A_{R}=2A\\sin{kx}<\/pre><\/div>\n\n\n\n<p>A <strong>Interfer\u00eancia Construtiva<\/strong> \u00e9 determinada quando a amplitude resultante assume o seu maior valor, que ocorre quando <math data-latex=\"\\sin kx=\\pm 1\"><semantics><mrow><mrow><mi>sin<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mi>k<\/mi><mi>x<\/mi><mo>=<\/mo><mo form=\"prefix\" stretchy=\"false\">\u00b1<\/mo><mn>1<\/mn><\/mrow><annotation encoding=\"application\/x-tex\">\\sin kx=\\pm 1<\/annotation><\/semantics><\/math>, logo<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>kx=\\frac{\\pi}{2}, 3\\frac{\\pi}{2}, 5\\frac{\\pi}{2} \\dots \\\\.\n\\\\\n\\frac{2\\pi}{\\lambda}x=m\\frac{\\pi}{2}, \\quad\\text{com }m=1,3,5 \\dots \\\\ .\n\\\\\nx=m\\frac{\\lambda}{4}, \\quad\\text{com }m=1,3,5 \\dots <\/pre><\/div>\n\n\n\n<p>A equa\u00e7\u00e3o acima localiza a posi\u00e7\u00e3o dos <strong>antin\u00f3s<\/strong>. Se tomarmos a dist\u00e2ncia de dois antin\u00f3s consecutivos teremos<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>x_{m+2}-x_{m}=(m+2)\\frac{\\lambda}{4}-m\\frac{\\lambda}{4} \\\\.\n\\\\\nx_{m+2}-x_{m}=\\frac{\\lambda}{2}<\/pre><\/div>\n\n\n\n<p>A <strong>Interfer\u00eancia Destrutiva<\/strong> ocorre quando <math data-latex=\"\\sin kx=0\"><semantics><mrow><mrow><mi>sin<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mi>k<\/mi><mi>x<\/mi><mo>=<\/mo><mn>0<\/mn><\/mrow><annotation encoding=\"application\/x-tex\">\\sin kx=0<\/annotation><\/semantics><\/math>, logo<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>kx=0,\\pi,2\\pi,3\\pi \\dots \\\\.\n\\\\\n\\frac{2\\pi}{\\lambda}x=n\\pi, \\quad \\text{com }n=0,1,2,3 \\dots \\\\.\n\\\\\nx=n\\frac{\\lambda}{2}, \\quad \\text{com }n=0,1,2,3 \\dots<\/pre><\/div>\n\n\n\n<p>A equa\u00e7\u00e3o acima localiza a posi\u00e7\u00e3o dos <strong>n\u00f3s<\/strong>. Esta forma de onda \u00e9 chamada de <strong>onda estacion\u00e1ria<\/strong>, pois o ponto de interfer\u00eancia destrutiva \u00e9 localizado em pontos fixos. Se tomarmos a dist\u00e2ncia de dois antin\u00f3s consecutivos teremos<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>x_{n+1}-x_{n}=(n+1)\\frac{\\lambda}{2}-n\\frac{\\lambda}{2} \\\\.\n\\\\\nx_{n+1}-x_{n}=\\frac{\\lambda}{2}<\/pre><\/div>\n\n\n\n<p>Conclui-se que a dist\u00e2ncia entre dois n\u00f3s consecutivos \u00e9 igual a meio comprimento de onda, desta forma, podemos obter um conjunto de <strong>frequ\u00eancias de resson\u00e2ncia<\/strong> chamadas de <strong>modos normais de vibra\u00e7\u00e3o<\/strong>, a onda estacion\u00e1ria s\u00f3 se forma nestes valores de frequ\u00eancia, veja a figura abaixo<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full\"><img decoding=\"async\" width=\"430\" height=\"422\" src=\"https:\/\/fiziko.net\/wp-content\/uploads\/2026\/02\/onda-002.jpg\" alt=\"\" class=\"wp-image-1215\" srcset=\"https:\/\/fiziko.net\/wp-content\/uploads\/2026\/02\/onda-002.jpg 430w, https:\/\/fiziko.net\/wp-content\/uploads\/2026\/02\/onda-002-300x294.jpg 300w\" sizes=\"(max-width: 430px) 100vw, 430px\" \/><\/figure>\n<\/div>\n\n\n<p>O primeiro modo de vibra\u00e7\u00e3o n=1 \u00e9 chamado de <strong>primeiro harm\u00f4nico<\/strong> ou <strong>harm\u00f4nico fundamental<\/strong>, todos os outros padr\u00f5es s\u00e3o m\u00faltiplos inteiros deste, assim, quando <math data-latex=\"n=2\"><semantics><mrow><mi>n<\/mi><mo>=<\/mo><mn>2<\/mn><\/mrow><annotation encoding=\"application\/x-tex\">n=2<\/annotation><\/semantics><\/math> dizemos que este \u00e9 o <strong>segundo harm\u00f4nico<\/strong> ou <strong>primeiro sobretom<\/strong>, quando <math data-latex=\"n=3\"><semantics><mrow><mi>n<\/mi><mo>=<\/mo><mn>3<\/mn><\/mrow><annotation encoding=\"application\/x-tex\">n=3<\/annotation><\/semantics><\/math> temos o terceiro harm\u00f4nico ou o segundo sobretom e assim por diante. Note que conforme a figura acima o valor de <math data-latex=\"n\"><semantics><mi>n<\/mi><annotation encoding=\"application\/x-tex\">n<\/annotation><\/semantics><\/math> representa o n\u00famero de antin\u00f3s do padr\u00e3o de vibra\u00e7\u00e3o.<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>n\\frac{\\lambda}{2}=L \\\\.\n\\\\\n\\lambda = \\frac{2L}{n}, \\quad \\text{onde }n=1,2,3 \\dots<\/pre><\/div>\n\n\n\n<p>Sabendo-se que, <\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>f_{n}=\\frac{v}{\\lambda} \\\\.\n\\\\\nf_{n}=n\\frac{v}{2L}=n\\frac{1}{2L}\\sqrt{\\frac{F}{\\mu}}<\/pre><\/div>\n\n\n\n<p>A <strong>frequ\u00eancia fundamental<\/strong> \u00e9 dada por<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>f_{1}=\\frac{v}{2L}=\\frac{1}{2L}\\sqrt{\\frac{F}{\\mu}}<\/pre><\/div>\n\n\n\n<h2 class=\"wp-block-heading\">Princ\u00edpio da superposi\u00e7\u00e3o de Ondas Progressivas Harm\u00f4nicas com Amplitudes e Constantes de Fase Diferentes  <\/h2>\n\n\n\n<p>Tomemos duas ondas progressivas com a mesma frequ\u00eancia, mas com amplitudes e constantes de fase diferentes, a forma das fun\u00e7\u00f5es de onda s\u00e3o<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\begin{cases}\n   y_{1}(x,t)=A_{1}\\sin(kx-\\omega t +\\phi_{1}) \\\\\n   y_{2}(x,t)=A_{2}\\sin(kx-\\omega t +\\phi_{2})\n\\end{cases}<\/pre><\/div>\n\n\n\n<p>considerando que o termo comum <math data-latex=\"kx-\\omega t\"><semantics><mrow><mi>k<\/mi><mi>x<\/mi><mo>\u2212<\/mo><mi>\u03c9<\/mi><mi>t<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">kx-\\omega t<\/annotation><\/semantics><\/math> aparece nas duas ondas, a <strong>diferen\u00e7a de fases<\/strong> <math data-latex=\"\\Delta \\phi\"><semantics><mrow><mrow><mi mathvariant=\"normal\">\u0394<\/mi><\/mrow><mi>\u03d5<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">\\Delta \\phi<\/annotation><\/semantics><\/math> est\u00e1 relacionada \u00e0s constantes de fase com <math data-latex=\"\\Delta \\phi=\\phi_{2}-\\phi_{1}\"><semantics><mrow><mrow><mi mathvariant=\"normal\">\u0394<\/mi><\/mrow><mi>\u03d5<\/mi><mo>=<\/mo><msub><mi>\u03d5<\/mi><mn>2<\/mn><\/msub><mo>\u2212<\/mo><msub><mi>\u03d5<\/mi><mn>1<\/mn><\/msub><\/mrow><annotation encoding=\"application\/x-tex\">\\Delta \\phi=\\phi_{2}-\\phi_{1}<\/annotation><\/semantics><\/math>, assim, utilizando-se o princ\u00edpio da superposi\u00e7\u00e3o temos a onda resultante dada por<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>y(x,t)=y_{1}(x,t)+y_{2}(x,t)<\/pre><\/div>\n\n\n\n<p>No diagrama abaixo, podemos visualizar as rela\u00e7\u00f5es geom\u00e9tricas entre os fasores de cada onda<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full\"><img decoding=\"async\" width=\"604\" height=\"930\" src=\"https:\/\/fiziko.net\/wp-content\/uploads\/2026\/02\/onda-003.jpg\" alt=\"\" class=\"wp-image-1220\" srcset=\"https:\/\/fiziko.net\/wp-content\/uploads\/2026\/02\/onda-003.jpg 604w, https:\/\/fiziko.net\/wp-content\/uploads\/2026\/02\/onda-003-195x300.jpg 195w\" sizes=\"(max-width: 604px) 100vw, 604px\" \/><\/figure>\n<\/div>\n\n\n<p>Na figura, o fasor verde representa a fun\u00e7\u00e3o de onda <math data-latex=\"y_{1}(x,t)\"><semantics><mrow><msub><mi>y<\/mi><mn>1<\/mn><\/msub><mo form=\"prefix\" stretchy=\"false\">(<\/mo><mi>x<\/mi><mo separator=\"true\">,<\/mo><mi>t<\/mi><mo form=\"postfix\" stretchy=\"false\">)<\/mo><\/mrow><annotation encoding=\"application\/x-tex\">y_{1}(x,t)<\/annotation><\/semantics><\/math> e o fasor azul representa a fun\u00e7\u00e3o de onda <math data-latex=\"y_{2}(x,t)\"><semantics><mrow><msub><mi>y<\/mi><mn>2<\/mn><\/msub><mo form=\"prefix\" stretchy=\"false\">(<\/mo><mi>x<\/mi><mo separator=\"true\">,<\/mo><mi>t<\/mi><mo form=\"postfix\" stretchy=\"false\">)<\/mo><\/mrow><annotation encoding=\"application\/x-tex\">y_{2}(x,t)<\/annotation><\/semantics><\/math>, e o fasor vermelho \u00e9 a onda resultante <math data-latex=\"y(x,t)\"><semantics><mrow><mi>y<\/mi><mo form=\"prefix\" stretchy=\"false\">(<\/mo><mi>x<\/mi><mo separator=\"true\">,<\/mo><mi>t<\/mi><mo form=\"postfix\" stretchy=\"false\">)<\/mo><\/mrow><annotation encoding=\"application\/x-tex\">y(x,t)<\/annotation><\/semantics><\/math> que ter\u00e1 a amplitude <math data-latex=\"A\"><semantics><mi>A<\/mi><annotation encoding=\"application\/x-tex\">A<\/annotation><\/semantics><\/math>. Os \u00e2ngulos <math data-latex=\"\\delta\"><semantics><mi>\u03b4<\/mi><annotation encoding=\"application\/x-tex\">\\delta<\/annotation><\/semantics><\/math> e <math data-latex=\"\\tau\"><semantics><mi>\u03c4<\/mi><annotation encoding=\"application\/x-tex\">\\tau<\/annotation><\/semantics><\/math> s\u00e3o iguais e o \u00e2ngulo <math data-latex=\"\\gamma=\\Delta \\phi=\\phi_{2}-\\phi_{1}\"><semantics><mrow><mi>\u03b3<\/mi><mo>=<\/mo><mrow><mi mathvariant=\"normal\">\u0394<\/mi><\/mrow><mi>\u03d5<\/mi><mo>=<\/mo><msub><mi>\u03d5<\/mi><mn>2<\/mn><\/msub><mo>\u2212<\/mo><msub><mi>\u03d5<\/mi><mn>1<\/mn><\/msub><\/mrow><annotation encoding=\"application\/x-tex\">\\gamma=\\Delta \\phi=\\phi_{2}-\\phi_{1}<\/annotation><\/semantics><\/math>, onde <math data-latex=\"\\delta=\\tau=180\u00b0-\\gamma=180\u00b0-\\Delta\\phi\"><semantics><mrow><mi>\u03b4<\/mi><mo>=<\/mo><mi>\u03c4<\/mi><mo>=<\/mo><mn>180<\/mn><mi>\u00b0<\/mi><mo>\u2212<\/mo><mi>\u03b3<\/mi><mo>=<\/mo><mn>180<\/mn><mi>\u00b0<\/mi><mo>\u2212<\/mo><mrow><mi mathvariant=\"normal\">\u0394<\/mi><\/mrow><mi>\u03d5<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">\\delta=\\tau=180\u00b0-\\gamma=180\u00b0-\\Delta\\phi<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n<p>Utilizando-se a Lei dos Cossenos para determinar a amplitude da onda resultante, temos<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>A^2=A_{1}^2+A_{2}^2-2A_{1}A_{2}\\cos(\\delta)\\\\.\n\\\\\nA^2=A_{1}^2+A_{2}^2-2A_{1}A_{2}\\cos(180\u00b0-\\Delta\\phi)<\/pre><\/div>\n\n\n\n<p>sabendo que <math data-latex=\"\\cos(a-b)=\\cos a \\cos b + \\sin a \\sin b\"><semantics><mrow><mrow><mi>cos<\/mi><mo>\u2061<\/mo><\/mrow><mo form=\"prefix\" stretchy=\"false\">(<\/mo><mi>a<\/mi><mo>\u2212<\/mo><mi>b<\/mi><mo form=\"postfix\" stretchy=\"false\">)<\/mo><mo>=<\/mo><mrow><mi>cos<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mi>a<\/mi><mrow><mspace width=\"0.1667em\"><\/mspace><mi>cos<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mi>b<\/mi><mo>+<\/mo><mrow><mi>sin<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mi>a<\/mi><mrow><mspace width=\"0.1667em\"><\/mspace><mi>sin<\/mi><mo>\u2061<\/mo><mspace width=\"0.1667em\"><\/mspace><\/mrow><mi>b<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">\\cos(a-b)=\\cos a \\cos b + \\sin a \\sin b<\/annotation><\/semantics><\/math>, logo<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>A^2=A_{1}^2+A_{2}^2+2A_{1}A_{2}\\cos(\\Delta\\phi)<\/pre><\/div>\n\n\n\n<p>A fun\u00e7\u00e3o de onda da onda resultante ser\u00e1<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>y(x,t)=A\\sin(kx-\\omega t +\\phi_{1}+\\beta)<\/pre><\/div>\n\n\n\n<p>para determinarmos o \u00e2ngulo <math data-latex=\"\\beta\"><semantics><mi>\u03b2<\/mi><annotation encoding=\"application\/x-tex\">\\beta<\/annotation><\/semantics><\/math> basta utilizarmos a Lei dos Senos, obteremos<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\frac{\\sin \\beta}{A_{2}}=\\frac{\\sin \\tau}{A} \\\\ \n\\\\\n\\sin \\beta=\\frac{A_{2}}{A} \\sin \\Delta\\phi<\/pre><\/div>\n\n\n\n<p>Podemos obter a equa\u00e7\u00e3o da intensidade da onda resultante, que \u00e9 dada por<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>I=I_{1}+I_{2}+2\\sqrt{I_{1}I_{2}}\\cos(\\Delta \\phi)<\/pre><\/div>\n\n\n\n<p>No caso de uma <strong>interfer\u00eancia construtiva<\/strong>, teremos<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>I_{max}=(\\sqrt{I_{1}}+\\sqrt{I_{2}})^2, \\quad \\Delta \\phi=2n\\pi; \\quad n=0, \\pm1, \\pm2,...<\/pre><\/div>\n\n\n\n<p>E para o caso de uma <strong>interfer\u00eancia destrutiva<\/strong>, teremos<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>I_{min}=(\\sqrt{I_{1}}-\\sqrt{I_{2}})^2, \\quad \\Delta \\phi=(2n+1)\\pi; \\quad n=0, \\pm1, \\pm2,...<\/pre><\/div>\n\n\n\n<h2 class=\"wp-block-heading\">Reflex\u00e3o de Ondas Transversais em 1D<\/h2>\n\n\n\n<p>Consideramos um pulso transversal propagando-se em um corda vibrante de dimens\u00e3o finita na dire\u00e7\u00e3o x negativa e \u00e9 representada pela fun\u00e7\u00e3o<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>y_{1}(x,t)=\\frac{b^3}{b^2+(x+vt-4)^2}<\/pre><\/div>\n\n\n\n<p>A segunda onda se propagar\u00e1 no sentido x positivo e \u00e9 representada por<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>y_{2}(x,t)=\\frac{b^3}{b^2+(x-vt+4)^2}<\/pre><\/div>\n\n\n\n<p>A onda resultante representar\u00e1 a reflex\u00e3o em uma haste, como veremos nos v\u00eddeos a seguir.<\/p>\n\n\n\n<p>Assumimos que a <strong>extremidade presa<\/strong> est\u00e1 localizada no ponto x = 0. A fun\u00e7\u00e3o resultante ser\u00e1<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>y(x,t)=y_{1}(x,t)+y_{2}(x,t)<\/pre><\/div>\n\n\n\n<p>Esta condi\u00e7\u00e3o de contorno, em qualquer instante, \u00e9 expressa por<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>y(0,t)=0<\/pre><\/div>\n\n\n\n<p>Podemos concluir, observando o comportamento do pulso resultante, que a reflex\u00e3o em uma extremidade fixa provoca uma defasagem de <strong>180\u00b0 graus<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-video aligncenter\"><video height=\"1130\" style=\"aspect-ratio: 1622 \/ 1130;\" width=\"1622\" controls src=\"https:\/\/fiziko.net\/wp-content\/uploads\/2026\/02\/Ondas-02.mp4\"><\/video><\/figure>\n\n\n\n<p>Quando a <strong>extremidade est\u00e1 livre<\/strong>, a tens\u00e3o neste ponto n\u00e3o pode ter nenhuma componente <math data-latex=\"F_{y}\"><semantics><msub><mi>F<\/mi><mi>y<\/mi><\/msub><annotation encoding=\"application\/x-tex\">F_{y}<\/annotation><\/semantics><\/math> na dire\u00e7\u00e3o y perpendicular \u00e0 dire\u00e7\u00e3o x da corda em repouso, o que traduzimos pela nova condi\u00e7\u00e3o de contorno<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>-F\\frac{\\partial y (0,t)}{\\partial x}=0<\/pre><\/div>\n\n\n\n<p>Quando ocorre uma reflex\u00e3o numa extremidade livre, um pulso se reflete <strong>sem sofrer invers\u00e3o<\/strong>, isto \u00e9, <strong>sem mudan\u00e7a de fase<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-video\"><video height=\"1082\" style=\"aspect-ratio: 1626 \/ 1082;\" width=\"1626\" controls src=\"https:\/\/fiziko.net\/wp-content\/uploads\/2026\/02\/Ondas-03.mp4\"><\/video><\/figure>\n\n\n\n<p>Esse comportamento ter\u00e1 uma influ\u00eancia profunda na forma\u00e7\u00e3o dos <strong>modos normais de vibra\u00e7\u00e3o<\/strong> ou como dizemos, nas <strong>frequ\u00eancia de resson\u00e2ncia<\/strong> e nas <strong>formas dos harm\u00f4nicos<\/strong>.<\/p>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Uma onda mec\u00e2nica \u00e9 uma perturba\u00e7\u00e3o energ\u00edtica que se desloca atrav\u00e9s de um meio material, no qual a onda se propaga. Ondas n\u00e3o transportam materia, a energia se propaga de mol\u00e9cula a mol\u00e9cula ao longo do meio material por onde a perturba\u00e7\u00e3o energ\u00e9tica \u00e9 transmitida. A forma mais simples de uma onda \u00e9 a onda&hellip; <br \/> <a class=\"read-more\" href=\"https:\/\/fiziko.net\/?page_id=1183\">Leia mais<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-1183","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/fiziko.net\/index.php?rest_route=\/wp\/v2\/pages\/1183","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fiziko.net\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fiziko.net\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fiziko.net\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fiziko.net\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1183"}],"version-history":[{"count":15,"href":"https:\/\/fiziko.net\/index.php?rest_route=\/wp\/v2\/pages\/1183\/revisions"}],"predecessor-version":[{"id":1227,"href":"https:\/\/fiziko.net\/index.php?rest_route=\/wp\/v2\/pages\/1183\/revisions\/1227"}],"wp:attachment":[{"href":"https:\/\/fiziko.net\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1183"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}