![]() ![]() So, how would I apply this equation to this case of a pendulum? Well, I wouldn't use X. 'cause usually you can get away with not using that one. ![]() Times cosine or sine, I'm just gonna write cosine, of two pi divided by the period, times the time and you can if you wantĪdd a phase constant. Was described by an equation that looked like this, X, some variable X is a function of time was equal to some amplitude There's a restoring force proportional to the displacement and we mean that its motion can be described by the simple So, what do we mean that the pendulum is a simple harmonic oscillator? Well, we mean that We can learn a lot about the motion just by looking at this case. We've got enough things to study by just studying simple pendulums. But really complicated toĭescribe mathematically. If you've never seen it, look up double pendulum, Physicists call chaotic, which is kind of cool. Let's say you connect another string, with another mass down here. You could have more complicated examples. And technically speaking, I should say that this is actually a simple pendulum because this is simply a Simple harmonic oscillator and so that's why we study it when we study simple harmonic oscillators. So, this is gonna swingįorward and then backward, and then forward and backward. And a pendulum is just a mass, m, connected to a string of some length, L, that you can then pullīack a certain amount and then you let it swing back and forth. So, that's what I wanna talk to you about in this video. The most common example, but the next most commonĮxample is the pendulum. Simple harmonic oscillators go, masses on springs are However, here is a link to the derivation: ) (I'm sorry, as typing out differential equations would be tedious. However, there are methods of approximating unsolvable differential equations (Euler's method, for example), that can get much closer to the exact answer than would the traditional period formula. Therefore, as of right now, there is no absolute solution to your question. This bottleneck is the very reason why the period formula works best when θs are smallest if you were to look at the graphs of y=sin(x) and y=x, they are closest to each other the smaller x is (thus more accurate), and farther with bigger x values. To alter this differential equation into a solvable one, you can write sin(θ) ≈ θ via the small-angle approximation (while sacrificing a bit of accuracy). You can check the cases when displacement is negative.If you were to try and derive the period of the pendulum (which involves setting up differential equations), you eventually get this term, sin(θ), which makes the whole differential equation unsolvable. and if displacement is positive and decreasing, the velocity vector would be That negative sign comes in because when displacement is positive and increasing, the velocity vector would be positive and decreasing. Recall that in SHM acceleration is proportional to the You shouldīe able to see that the velocity vectors diminish more rapidly when the particles are farther from the center, indicating greater acceleration when farther from the center of the circle. Do notice how the velocity vectors of the cyan and orange particles vary. But the x and y component motions then would not be simple harmonic. Please note that even if θ varies in a more complicated manner this would be true. You should be able to see that eliminating time from x and y equations gives you the equation of the circle. Where A is the amplitude of SHM which is equal to the radius of the circle and ωt+δ is called the phase. X = A Cos(ωt+δ) and y = A Sin(ωt+δ) respectivley ![]() The cyan and orange particles motion can be represented as Phase would increase continuously, going up by 2π radians for every revolution. In the applet the phase is reset to zero afterĮvery revolution. θ is shown by the small white arc at the center. Where ω is the angular speed and δ is the initial angular position of the red particle. But since the motion of the particle round the circle is uniform θ = ωt+δ, If the angular position of the red particle is denotedīy θ then its x and y co-ordinates are R cos θ and R sin θ (With x to the right and y upward and origin at the center of the circle). The red particle is going round the circular path in anti-clock wise sense (considered positive) with a uniform speed and its x-projection and y - projection are shown in cyan and orange respectively. ![]()
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