
{"id":369,"date":"2016-10-14T01:40:29","date_gmt":"2016-10-14T01:40:29","guid":{"rendered":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/?p=369"},"modified":"2016-10-14T02:35:43","modified_gmt":"2016-10-14T02:35:43","slug":"week-7","status":"publish","type":"post","link":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/2016\/10\/14\/week-7\/","title":{"rendered":"Week 7"},"content":{"rendered":"<p><strong>Forced oscillations of systems with 2 dof:<\/strong><\/p>\n<p>Recall the demo! \u00a0We had the usual two pendula with 1kg masses on strings with a weak spring connecting the two masses. \u00a0The additional ingredient in the demo on Tuesday was that we attached the piston motor so that it would push sinusoidally on one of the masses. \u00a0As you showed in your homework, this is equivalent to the addition of a sinusoidal driving force acting on one of the two masses. \u00a0We then slowly scanned through the various different frequencies on the frequency generator that was controlling the motor. \u00a0We found that at around 0.67 Hz, the symmetric normal mode was excited with large steady state amplitude. \u00a0We discovered a resonance of the two-body system, and it coincides with a normal mode excitation! \u00a0We then explored higher frequencies, and we found at at around 0.77 Hz, the second asymmetric normal mode was excited!<\/p>\n<p>We can see this behavior as arising from the equations of motion, as always. \u00a0The equations of motion are now:<\/p>\n<img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cvec%7BF%7D_A+%3D+-+m+%5Comega_0%5E2+x_A+-+k+%28x_A+-+x_B%29+%2B+F_0+%5Ccos+%5Comega+t+%3D+m+%5Cddot%7Bx%7D_A&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\vec{F}_A = - m \\omega_0^2 x_A - k (x_A - x_B) + F_0 \\cos \\omega t = m \\ddot{x}_A' title='\\vec{F}_A = - m \\omega_0^2 x_A - k (x_A - x_B) + F_0 \\cos \\omega t = m \\ddot{x}_A' class='latex' \/>\n<img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cvec%7BF%7D_B+%3D+-+m+%5Comega_0%5E2+x_B+-+k+%28x_B-x_A%29+%3D+m+%5Cddot%7Bx%7D_B&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\vec{F}_B = - m \\omega_0^2 x_B - k (x_B-x_A) = m \\ddot{x}_B' title='\\vec{F}_B = - m \\omega_0^2 x_B - k (x_B-x_A) = m \\ddot{x}_B' class='latex' \/>\n<p>Note that the motor was only attached to the first pendulum, so there is no driving force acting on mass B.<\/p>\n<p>Now we use the same trick &#8211; adding or subtracting the two equations from each other in order to decouple them:<\/p>\n<img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cddot%7B%28x_A+%2B+x_B%29%7D+%2B+%5Comega_0%5E2+%28x_A+%2Bx_B%29+%3D+F_0%2Fm+%5Ccos+%5Comega+t&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\ddot{(x_A + x_B)} + \\omega_0^2 (x_A +x_B) = F_0\/m \\cos \\omega t' title='\\ddot{(x_A + x_B)} + \\omega_0^2 (x_A +x_B) = F_0\/m \\cos \\omega t' class='latex' \/>\n<img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cddot%7B%28x_A+-+x_B%29%7D+%2B+%28%5Comega_0%5E2+%2B+2+k%2Fm%29+%28x_A-x_B%29+%3D+F_0%2Fm+%5Ccos+%5Comega+t&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\ddot{(x_A - x_B)} + (\\omega_0^2 + 2 k\/m) (x_A-x_B) = F_0\/m \\cos \\omega t' title='\\ddot{(x_A - x_B)} + (\\omega_0^2 + 2 k\/m) (x_A-x_B) = F_0\/m \\cos \\omega t' class='latex' \/>\n<p>So the mathematics leads us to two simple harmonic oscillators, both driven by the same driving force. \u00a0When <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Comega&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\omega' title='\\omega' class='latex' \/> coincides with $\\omega_0$, you expect to find resonance for the first (symmetric) mode of oscillation. \u00a0Similarly, when <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Comega&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\omega' title='\\omega' class='latex' \/> is close to the frequency <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Comega_-+%3D+%5Csqrt%7B+%5Comega_0%5E2+%2B+2+k%2Fm+%7D&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\omega_- = \\sqrt{ \\omega_0^2 + 2 k\/m }' title='\\omega_- = \\sqrt{ \\omega_0^2 + 2 k\/m }' class='latex' \/>, you expect to be exciting the resonance for the anti-symmetric normal mode. \u00a0This is exactly what we saw in the demonstration! \u00a0We also used the frequencies of the normal modes to determine <img src='https:\/\/s0.wp.com\/latex.php?latex=k&#038;bg=T&#038;fg=000000&#038;s=0' alt='k' title='k' class='latex' \/>.<\/p>\n<p><strong>Normal Modes with 2 dof:<\/strong><\/p>\n<p>We continued our discussion about normal modes of oscillation on Thursday, wrapping up the discussion using a Mathematica demonstration of computing the normal modes of oscillation in a more general case without obvious symmetry. \u00a0In this case, there is a procedure. \u00a0The mathematica file that we used is linked here:<\/p>\n<p><a href=\"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-content\/uploads\/sites\/5\/2016\/10\/Week-7.nb\">https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-content\/uploads\/sites\/5\/2016\/10\/Week-7.nb<\/a><\/p>\n<p>The procedure is to first identify the potential energy function. \u00a0This demonstration was for the &#8220;double pendulum&#8221; where one pendulum is attached to the bottom of the first, and both have equal length <img src='https:\/\/s0.wp.com\/latex.php?latex=l&#038;bg=T&#038;fg=000000&#038;s=0' alt='l' title='l' class='latex' \/> and equal mass <img src='https:\/\/s0.wp.com\/latex.php?latex=m&#038;bg=T&#038;fg=000000&#038;s=0' alt='m' title='m' class='latex' \/>. \u00a0The potential energy function we found was given by (in the small angle approximation):<\/p>\n<img src='https:\/\/s0.wp.com\/latex.php?latex=U%28x_1+%2Cx_2%29+%3D+%5Cfrac%7Bmg%7D%7B2l%7D+%5Cleft%5B+2+x_1%5E2+%2B+%28x_2-x_1%29%5E2+%5Cright%5D&#038;bg=T&#038;fg=000000&#038;s=0' alt='U(x_1 ,x_2) = \\frac{mg}{2l} \\left[ 2 x_1^2 + (x_2-x_1)^2 \\right]' title='U(x_1 ,x_2) = \\frac{mg}{2l} \\left[ 2 x_1^2 + (x_2-x_1)^2 \\right]' class='latex' \/>\n<p>From this one function, we can get everything about the behavior of the system when it is perturbed from equilibrium. \u00a0First we get the forces on each mass by taking the derivative with respect to the two coordinates, then we apply Newton&#8217;s 2nd law. \u00a0The next step is to assume we have identified a normal mode of oscillation of the system, in which case we have<\/p>\n<img src='https:\/\/s0.wp.com\/latex.php?latex=x_1+%3D+A+r+%5Ccos+%5Comega+t&#038;bg=T&#038;fg=000000&#038;s=0' alt='x_1 = A r \\cos \\omega t' title='x_1 = A r \\cos \\omega t' class='latex' \/>\n<img src='https:\/\/s0.wp.com\/latex.php?latex=x_2+%3D+A+%5Ccos+%5Comega+t&#038;bg=T&#038;fg=000000&#038;s=0' alt='x_2 = A \\cos \\omega t' title='x_2 = A \\cos \\omega t' class='latex' \/>\n<p>where <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Comega&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\omega' title='\\omega' class='latex' \/> is the frequency of the normal mode, and <img src='https:\/\/s0.wp.com\/latex.php?latex=r&#038;bg=T&#038;fg=000000&#038;s=0' alt='r' title='r' class='latex' \/> is the ratio of amplitudes when the normal mode is excited. \u00a0Recall that for the symmetric pendulum, <img src='https:\/\/s0.wp.com\/latex.php?latex=r+%3D+1&#038;bg=T&#038;fg=000000&#038;s=0' alt='r = 1' title='r = 1' class='latex' \/> for the symmetric normal mode, and <img src='https:\/\/s0.wp.com\/latex.php?latex=r+%3D+-1&#038;bg=T&#038;fg=000000&#038;s=0' alt='r = -1' title='r = -1' class='latex' \/> for the antisymmetric normal mode.<\/p>\n<p>Plugging this into the equations of motion, we can re-express the equations as algebraic equations that allow us to determine pairs of <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Comega&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\omega' title='\\omega' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=r&#038;bg=T&#038;fg=000000&#038;s=0' alt='r' title='r' class='latex' \/>.<\/p>\n<p>Finally we can plot what the normal modes actually look like, as I did in the mathematica file using the Animate command.<\/p>\n<p>We can also then take arbitrary superpositions of the two normal modes, and adjust the amount of amplitude of each of them.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Forced oscillations of systems with 2 dof: Recall the demo! \u00a0We had the usual two pendula with 1kg masses on strings with a weak spring connecting the two masses. \u00a0The additional ingredient in the demo on Tuesday was that we &hellip; <a href=\"https:\/\/jhubisz.expressions.syr.edu\/phy360\/2016\/10\/14\/week-7\/\">Continue reading <span class=\"meta-nav\">&rarr;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"_links":{"self":[{"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/posts\/369"}],"collection":[{"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/comments?post=369"}],"version-history":[{"count":9,"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/posts\/369\/revisions"}],"predecessor-version":[{"id":379,"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/posts\/369\/revisions\/379"}],"wp:attachment":[{"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/media?parent=369"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/categories?post=369"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/tags?post=369"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}