
{"id":314,"date":"2016-09-23T03:42:38","date_gmt":"2016-09-23T03:42:38","guid":{"rendered":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/?p=314"},"modified":"2016-09-27T23:09:05","modified_gmt":"2016-09-27T23:09:05","slug":"week-4","status":"publish","type":"post","link":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/2016\/09\/23\/week-4\/","title":{"rendered":"Week 4"},"content":{"rendered":"<p><strong>Forced Damped Simple Harmonic Motion<\/strong><\/p>\n<p>This week we began by reviewing the simple harmonic oscillator including dissipative damping via a viscous force. \u00a0I encourage you to read last week&#8217;s blog posts to see the material associated with this, as well as this weeks mathematica notebook:<\/p>\n<p>https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-content\/uploads\/sites\/5\/2016\/09\/Week-4.nb<\/p>\n<p>The new material for this week is the forced damped simple harmonic oscillator. \u00a0For this case, we are now continuously injecting energy into and out of the system via some sort of periodic driving mechanism.<\/p>\n<p>The <strong>demonstration\u00a0<\/strong>associated with this is the mass on spring in water, where the spring is hung from a motor and piston that drives the entire mechanism up and down sinusoidally. \u00a0Experimentally, we verified that at high frequencies, there is not a great deal of steady-state response (by steady-state, we mean the motion that persists a long time after the driving force has turned on). \u00a0 Similar behavior was noted for low frequencies. \u00a0If, however, we drove the system at a frequency that is close to the natural frequency (the frequency of oscillation if there were no damping or driving) that the response is significant (the mass oscillates with a large steady state amplitude).<\/p>\n<p>The phenomenon of large response at a given characteristic frequency is called\u00a0<strong>resonance. \u00a0<\/strong>You are all likely familiar with resonance from using swings as children, and driving your legs at the right frequency leads to a more exciting (larger amplitude) ride.<\/p>\n<p>How does the mathematics work out for this scenario? \u00a0There is now one additional force, the force that is an <strong>external<\/strong> input into the system. \u00a0This force is what it is, independent of the displacement of the oscillator from equilibrium:<\/p>\n<p><img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cvec%7BF%7D_%5Ctext%7Bdrive%7D+%3D+F_0+%5Ccos+%5Comega+t&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\vec{F}_\\text{drive} = F_0 \\cos \\omega t' title='\\vec{F}_\\text{drive} = F_0 \\cos \\omega t' class='latex' \/>.<\/p>\n<p>Note that here <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 not a characteristic of the oscillator itself, but rather of the external input. \u00a0This force is\u00a0<strong>not dependent on x(t).<\/strong><\/p>\n<p>The total force then (let&#8217;s say the system is a mass on a spring with spring constant k, and there is a viscous damping coefficient b) is given by:<\/p>\n<img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cvec%7BF%7D+%3D+F_0+%5Ccos+%5Comega+t+-+k+x%28t%29+-+b+%5Cdot%7Bx%7D%28t%29&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\vec{F} = F_0 \\cos \\omega t - k x(t) - b \\dot{x}(t)' title='\\vec{F} = F_0 \\cos \\omega t - k x(t) - b \\dot{x}(t)' class='latex' \/>\n<p>Applying Newton&#8217;s 2nd law, we have an equation of motion given by<\/p>\n<img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cvec%7BF%7D+%3D+F_0+%5Ccos+%5Comega+t+-+k+x%28t%29+-+b+%5Cdot%7Bx%7D%28t%29+%3D+m+%5Cddot%7Bx%7D%28t%29&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\vec{F} = F_0 \\cos \\omega t - k x(t) - b \\dot{x}(t) = m \\ddot{x}(t)' title='\\vec{F} = F_0 \\cos \\omega t - k x(t) - b \\dot{x}(t) = m \\ddot{x}(t)' class='latex' \/>\n<p>We rearrange the equation of motion, dividing by <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' \/>, and we have:<\/p>\n<img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cddot%7Bx%7D%28t%29+%2B+%5Cgamma+%5Cdot%7Bx%7D%28t%29+%2B+%5Comega_0%5E2+x%28t%29+%3D+%5Cfrac%7BF_0%7D%7Bm%7D+%5Ccos+%5Comega+t&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\ddot{x}(t) + \\gamma \\dot{x}(t) + \\omega_0^2 x(t) = \\frac{F_0}{m} \\cos \\omega t' title='\\ddot{x}(t) + \\gamma \\dot{x}(t) + \\omega_0^2 x(t) = \\frac{F_0}{m} \\cos \\omega t' class='latex' \/>\n<p>where <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cgamma+%3D+b%2Fm&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\gamma = b\/m' title='\\gamma = b\/m' class='latex' \/> and <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Comega_0%5E2+%3D+k%2Fm&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\omega_0^2 = k\/m' title='\\omega_0^2 = k\/m' class='latex' \/>.<\/p>\n<p>Note the important difference from the equations we have looked at before &#8211; there is now a term in the equation that does not depend on x(t) at all. \u00a0To solve this equation, let us presume that the long-time behavior of the system (a long time after the driving force is turned on) is given by oscillation with the same frequency as the driving force. \u00a0Let us express this in the complex plane, however, when the equation of motion is expressed as<\/p>\n<p><img src='https:\/\/s0.wp.com\/latex.php?latex=%5Cddot%7Bz%7D%28t%29+%2B+%5Cgamma+%5Cdot%7Bz%7D%28t%29+%2B+%5Comega_0%5E2+z%28t%29+%3D+%5Cfrac%7BF_0%7D%7Bm%7D+e%5E%7Bi+%5Comega+t%7D&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\ddot{z}(t) + \\gamma \\dot{z}(t) + \\omega_0^2 z(t) = \\frac{F_0}{m} e^{i \\omega t}' title='\\ddot{z}(t) + \\gamma \\dot{z}(t) + \\omega_0^2 z(t) = \\frac{F_0}{m} e^{i \\omega t}' class='latex' \/>.<\/p>\n<p>Then our guess that the <strong>response\u00a0<\/strong>is characterized by the same frequency is equivalent to the mathematical statement<\/p>\n<p><img src='https:\/\/s0.wp.com\/latex.php?latex=z%28t%29+%3D+Z_0+e%5E%7Bi+%5Comega+t%7D&#038;bg=T&#038;fg=000000&#038;s=0' alt='z(t) = Z_0 e^{i \\omega t}' title='z(t) = Z_0 e^{i \\omega t}' class='latex' \/>.<\/p>\n<p>Here <img src='https:\/\/s0.wp.com\/latex.php?latex=Z_0&#038;bg=T&#038;fg=000000&#038;s=0' alt='Z_0' title='Z_0' class='latex' \/> is an as yet undetermined constant. \u00a0If we plug this guess into the equation of motion, we obtain an equation for <img src='https:\/\/s0.wp.com\/latex.php?latex=Z_0&#038;bg=T&#038;fg=000000&#038;s=0' alt='Z_0' title='Z_0' class='latex' \/>:<\/p>\n<img src='https:\/\/s0.wp.com\/latex.php?latex=Z_0+%3D+%5Cfrac%7B+F_0%2Fm%7D%7B%28%5Comega_0%5E2-%5Comega%5E2%29+%2B+i+%5Comega+%5Cgamma%7D&#038;bg=T&#038;fg=000000&#038;s=0' alt='Z_0 = \\frac{ F_0\/m}{(\\omega_0^2-\\omega^2) + i \\omega \\gamma}' title='Z_0 = \\frac{ F_0\/m}{(\\omega_0^2-\\omega^2) + i \\omega \\gamma}' class='latex' \/>\n<p>Now this looks like a relatively complicated formula, with both real and complex parts, but let us unpack it a bit. \u00a0First, let us recall that any complex number can be expressed as an amplitude and a phase:<\/p>\n<img src='https:\/\/s0.wp.com\/latex.php?latex=Z_0+%3D+A+e%5E%7Bi+%5Calpha%7D&#038;bg=T&#038;fg=000000&#038;s=0' alt='Z_0 = A e^{i \\alpha}' title='Z_0 = A e^{i \\alpha}' class='latex' \/>\n<p>We can find <img src='https:\/\/s0.wp.com\/latex.php?latex=A&#038;bg=T&#038;fg=000000&#038;s=0' alt='A' title='A' class='latex' \/> first:<\/p>\n<img src='https:\/\/s0.wp.com\/latex.php?latex=A+%3D+%5Cfrac%7BF_0%2Fm%7D%7B%5Csqrt%7B%5Comega_0%5E2-%5Comega%5E2%29%5E2%2B%5Comega%5E2+%5Cgamma%5E2%7D%7D&#038;bg=T&#038;fg=000000&#038;s=0' alt='A = \\frac{F_0\/m}{\\sqrt{\\omega_0^2-\\omega^2)^2+\\omega^2 \\gamma^2}}' title='A = \\frac{F_0\/m}{\\sqrt{\\omega_0^2-\\omega^2)^2+\\omega^2 \\gamma^2}}' class='latex' \/>\n<p>You can do some geometry in the complex plane to convince yourself that<\/p>\n<img src='https:\/\/s0.wp.com\/latex.php?latex=%5Calpha+%3D+%5Carctan+%5Cleft%5B+%5Cfrac%7B+-+%5Cgamma+%5Comega%7D%7B%5Comega_0%5E2-%5Comega%5E2%7D+%5Cright%5D&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\alpha = \\arctan \\left[ \\frac{ - \\gamma \\omega}{\\omega_0^2-\\omega^2} \\right]' title='\\alpha = \\arctan \\left[ \\frac{ - \\gamma \\omega}{\\omega_0^2-\\omega^2} \\right]' class='latex' \/>\n<p>Be careful that the signs in the numerator and denominator have meaning, as the arctan function is double valued. \u00a0You must figure out which quadrant the point is in based on the signs of the numerator and denominator separately.<\/p>\n<p>So what does all this mean in terms of the\u00a0<strong>physics<\/strong>??<\/p>\n<p>We are characterizing a <strong>response\u00a0<\/strong>(the motion of the mass) due to a <strong>stimulus<\/strong> (the applied driving force). \u00a0As we have found a solution, we see that for a given stimulus (an <img src='https:\/\/s0.wp.com\/latex.php?latex=F_0&#038;bg=T&#038;fg=000000&#038;s=0' alt='F_0' title='F_0' class='latex' \/> and an <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' \/>, there is a response with amplitude <img src='https:\/\/s0.wp.com\/latex.php?latex=A&#038;bg=T&#038;fg=000000&#038;s=0' alt='A' title='A' class='latex' \/> that leads the stimulus\u00a0in phase <img src='https:\/\/s0.wp.com\/latex.php?latex=%5Calpha&#038;bg=T&#038;fg=000000&#038;s=0' alt='\\alpha' title='\\alpha' class='latex' \/>. \u00a0To get a sense of what this means, please take a look at this weeks mathematica file, in which you can vary the stimulus frequency and watch the stimulus and response change with respect to each other in the complex plane.<\/p>\n<p>Note that the complete solution can be written as<\/p>\n<p><img src='https:\/\/s0.wp.com\/latex.php?latex=z+%3D+C_%2B+e%5E%7B%5Crho_%2B+t%7D+%2B+C_-+e%5E%7B%5Crho_-+t%7D+%2B+Z_0+e%5E%7Bi+%5Comega+t%7D&#038;bg=T&#038;fg=000000&#038;s=0' alt='z = C_+ e^{\\rho_+ t} + C_- e^{\\rho_- t} + Z_0 e^{i \\omega t}' title='z = C_+ e^{\\rho_+ t} + C_- e^{\\rho_- t} + Z_0 e^{i \\omega t}' class='latex' \/>,<\/p>\n<p>where the first two parts of the solution are damped, and thus not relevant for the\u00a0behavior of the system at times that are large in comparison with the damping times. \u00a0These are frequently referred to as\u00a0<strong>transients\u00a0<\/strong>of the system.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Forced Damped Simple Harmonic Motion This week we began by reviewing the simple harmonic oscillator including dissipative damping via a viscous force. \u00a0I encourage you to read last week&#8217;s blog posts to see the material associated with this, as well &hellip; <a href=\"https:\/\/jhubisz.expressions.syr.edu\/phy360\/2016\/09\/23\/week-4\/\">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\/314"}],"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=314"}],"version-history":[{"count":15,"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/posts\/314\/revisions"}],"predecessor-version":[{"id":336,"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/posts\/314\/revisions\/336"}],"wp:attachment":[{"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/media?parent=314"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/categories?post=314"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/jhubisz.expressions.syr.edu\/phy360\/wp-json\/wp\/v2\/tags?post=314"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}