{"id":1778,"date":"2020-10-15T08:22:08","date_gmt":"2020-10-15T08:22:08","guid":{"rendered":"http:\/\/physicscatalyst.com\/article\/?p=1778"},"modified":"2021-06-05T13:45:19","modified_gmt":"2021-06-05T13:45:19","slug":"coupled-pendulums","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/graduation\/coupled-pendulums\/","title":{"rendered":"Coupled Pendulums: two pendulums connected by a spring"},"content":{"rendered":"\n

Question<\/strong>
Obtain the equations of motion of coupled pendulum (two pendulums connected by a spring) using the Lagrangian method.
Solution<\/strong>
Consider a system of coupled pendulums as shown below in the figure
\"\"<\/a>
The displacement of A is $x_{1}$ and B is $x_{2}$ , condition being \\({x_1}\\) < \\({x_2}\\).<\/p>\n\n\n\n

In such state the spring gets stretched. The lengths of the strings of both the pendulums are same (say l).<\/p>\n\n\n\n

The angular displacement of A is \\({\\theta _1}\\) and that of B is \\({\\theta _2}({\\theta _2} > {\\theta _1})\\).
Therefore<\/p>\n\n\n\n

\\(\\begin{array}{l}{x_1} = l{\\theta _1} \\Rightarrow {\\theta _1} = \\frac{{{x_1}}}{l}{\\rm{ (1)}}\\\\{x_2} = l{\\theta _2} \\Rightarrow {\\theta _2} = \\frac{{{x_2}}}{l}{\\rm{ (2)}}\\end{array}\\)<\/p>\n\n\n\n

As the spring gets stretched, it is clear from the figure that restoring force works along the direction of displacement\\({\\theta _1}\\) and opposite to the direction of displacement \\({\\theta _2}\\) .<\/p>\n\n\n\n

Now A and B at zero potential level, the total potential energy of the system is given as<\/p>\n\n\n\n

\\(V = mgl(1 – \\cos {\\theta _1}) + mgl(1 – \\cos {\\theta _2}) + \\frac{1}{2}k{({x_2} – {x_1})^2}\\)<\/p>\n\n\n\n

Where m is the mass of each one of the bob and k is the spring constant.<\/p>\n\n\n\n

Since \\({\\theta _1}\\)and \\({\\theta _2}\\)are small so,<\/p>\n\n\n\n

\\(\\begin{array}{l}\\cos {\\theta _1} = 1 – \\frac{{\\theta _1^2}}{2} + \\frac{{\\theta _1^4}}{4} + ……\\\\\\cos {\\theta _2} = 1 – \\frac{{\\theta _2^2}}{2} + \\frac{{\\theta _2^4}}{4} + ……\\end{array}\\)<\/p>\n\n\n\n

Neglecting the higher powers other than squares of \\({\\theta _1}\\)and \\({\\theta _2}\\)the expression of potential energy can be written as<\/p>\n\n\n\n

\\(\\begin{array}{l}V = mgl\\frac{{\\theta _1^2}}{2} + mgl\\frac{{\\theta _2^2}}{2} + \\frac{1}{2}k{({x_2} – {x_1})^2}\\\\{\\rm{ = }}\\frac{{mgx_1^2}}{{2l}} + \\frac{{mgx_2^2}}{{2l}} + \\frac{1}{2}k{({x_2} – {x_1})^2}\\end{array}\\)<\/p>\n\n\n\n

Also the kinetic energy of whole system is<\/p>\n\n\n\n

\\(T = \\frac{1}{2}m\\dot x_1^2 + \\frac{1}{2}m\\dot x_2^2 = \\frac{1}{2}m(\\dot x_1^2 + \\dot x_2^2)\\)<\/p>\n\n\n\n

Hence Lagrangian L would be<\/p>\n\n\n\n

\\(\\begin{array}{l}L = T – V \\\\L = \\frac{1}{2}m(\\dot x_1^2 + \\dot x_2^2) – \\frac{{mgx_1^2}}{{2l}} – \\frac{{mgx_2^2}}{{2l}} – \\frac{1}{2}k{({x_2} – {x_1})^2}\\end{array}\\)<\/p>\n\n\n\n

Now<\/p>\n\n\n\n

\\(\\begin{array}{l} \\frac{\\partial L}{\\partial {{x}_{1}}}=-\\frac{mg{{x}_{1}}}{l}+k({{x}_{2}}-{{x}_{1}}) \\\\\\end{array}\\)<\/p>\n\n\n\n

\\(\\begin{array}{l}\\frac{\\partial L}{\\partial {{{\\dot{x}}}_{1}}}=m{{{\\dot{x}}}_{1}} \\\\\\end{array}\\)<\/p>\n\n\n\n

\\(\\begin{array}{l}\\therefore \\frac{d}{dt}\\left( \\frac{\\partial L}{\\partial {{{\\dot{x}}}_{1}}} \\right)=\\frac{d}{dt}(m{{{\\dot{x}}}_{1}})=m{{{\\ddot{x}}}_{1}} \\\\\\end{array}\\)<\/p>\n\n\n\n

Hence Lagrangian equation in terms of \\({x_1}\\)is<\/p>\n\n\n\n

\\(\\begin{array}{l}\\frac{d}{{dt}}\\left( {\\frac{{\\partial L}}{{\\partial {{\\dot x}_1}}}} \\right) – \\frac{{\\partial L}}{{\\partial {x_1}}} = 0\\\\or,\\\\m{{\\ddot x}_1} + \\frac{{mg{x_1}}}{l} – k({x_2} – {x_1}) = 0\\\\or,\\\\m{{\\ddot x}_1} = – \\frac{{mg{x_1}}}{l} + k({x_2} – {x_1})\\end{array}\\)<\/p>\n\n\n\n

Also,<\/p>\n\n\n\n

\\(\\begin{array}{l}\\frac{{\\partial L}}{{\\partial {x_2}}} = – \\frac{{mg{x_2}}}{l} – k({x_2} – {x_1})\\\\\\frac{{\\partial L}}{{\\partial {{\\dot x}_2}}} = m{{\\dot x}_2}\\\\and\\\\\\frac{d}{{dt}}\\left( {\\frac{{\\partial L}}{{\\partial {{\\dot x}_2}}}} \\right) = m{{\\ddot x}_2}\\end{array}\\)<\/p>\n\n\n\n

Hence Lagrangian equation in terms of \\({x_2}\\)is<\/p>\n\n\n\n

\\(\\begin{array}{l}\\frac{d}{{dt}}\\left( {\\frac{{\\partial L}}{{\\partial {{\\dot x}_2}}}} \\right) – \\frac{{\\partial L}}{{\\partial {x_2}}} = 0\\\\or,\\\\m{{\\ddot x}_2} = – \\frac{{mg{x_2}}}{l} – k({x_2} – {x_1})\\end{array}\\)<\/p>\n\n\n\n

The equation of motion for given system are<\/p>\n\n\n\n

\\(\\begin{array}{l}m{{\\ddot x}_1} = – \\frac{{mg{x_1}}}{l} + k({x_2} – {x_1})\\\\m{{\\ddot x}_2} = – \\frac{{mg{x_2}}}{l} – k({x_2} – {x_1})\\end{array}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Question Obtain the equations of motion of coupled pendulum using the lagrangian method. Solution Consider a system of coupled pendulums as shown below in the figure<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[18],"tags":[],"class_list":["post-1778","post","type-post","status-publish","format-standard","hentry","category-classical-mechanics"],"yoast_head":"\nCoupled Pendulums: two pendulums connected by a spring<\/title>\n<meta name=\"description\" content=\"Learn to find the equations of motion of coupled pendulums connected by a spring using the lagrangian method.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/physicscatalyst.com\/graduation\/coupled-pendulums\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Coupled Pendulums: two pendulums connected by a spring\" \/>\n<meta property=\"og:description\" content=\"Learn to find the equations of motion of coupled pendulums connected by a spring using the lagrangian method.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/physicscatalyst.com\/graduation\/coupled-pendulums\/\" \/>\n<meta property=\"og:site_name\" content=\"Learn about education and B.Sc. 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