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  • Welcome to Electra Online.

  • In this playlist we're going to take a closer look at simple harmonic motion and in particular how the equation is derived what it means physically relative to what's happening when there's simple harmonic motion and then we're also going to look at it in terms of the damping because a lot of simple harmonic motion involves damping and we want to see how the equation itself is derived and how to utilize that equation as well and there's several solutions that we need to look at depending upon what kind of damping we're dealing with.

  • But before we look at the details of damping let's take a look again at the general undamped simple harmonic motion equation.

  • So here we graphically show when we have an object hanging from a spring the spring has a certain amount of spring constant K and the object has a certain amount of mass M.

  • Once we allow it to balance so let's say we hang it on the at the equilibrium point then we can give it a push upward or we can pull downward and then it'll begin to oscillate up and down.

  • The gravity portion is is then negated by the fact that the spring has been extended somewhat before we start the oscillatory motion.

  • Here at this position this position in this position that the mass is what we call at the equilibrium point.

  • That is where there is no net force acting on the object and therefore no acceleration and the position is therefore at zero.

  • That's the equilibrium point and in this particular example let's assume that the motion is upward so velocity in this point is upward.

  • It will then reach its maximum distance away from the equilibrium point when X equals the amplitude of the motion capital A stands for amplitude.

  • At that moment V will be zero and the acceleration will be in the negative direction.

  • As long as the object is above the equilibrium point acceleration will be negative.

  • When it's below the equilibrium point like like it is over here then the acceleration will be in the positive direction.

  • Notice it'll be coming back down moving through the equilibrium point but in this case it's moving on downward negative velocity.

  • At this moment in time the position is zero the acceleration is zero then it reaches its maximum elongation below the equilibrium point when X equals negative the amplitude.

  • So we have positive amplitude negative amplitude at this moment the velocity is zero the acceleration is upward and then it reaches back to the equilibrium point just like it was over here with an upward velocity X is equal to zero and there's no acceleration that moment because there's no net force acting on the mass.

  • And so that's how it is continuous up and down and notice that there's some relationship to a sine or a And then we moved a paper past that pencil at a constant speed as this is oscillating up and down you would actually the pen would actually make that sine wave or that cosine wave as the paper is moving and as the object is oscillating up and down.

  • So there's some relationship between simplomatic motion and the sine or the cosine function.

  • Now we'll see mathematically why that is the case.

  • Starting with Newton's second law where f equals ma we can turn the ma equals f and then we know that the force exerted on the mass by the spring is equal to minus kx. k is the spring constant and X is a distance away from the equilibrium point.

  • Notice the negative sign because if the mass is in a positive position the spring is then pushing in the negative direction that's why the negative is there.

  • Then we realize that the acceleration is essentially the second derivative of position with respect to time so we can replace a by that and then notice that if we move the minus kx to the left we end up with an equation here equal to zero and also notice if we divide everything by m then we have d square x dt square with other words the second derivative of x with respect to time plus k over m times x equals zero.

  • Again this is an undamped case and if we then replace this with x double dot. x double dot simply means the second derivative of x with respect to time and then if we allow Omega to be equal to the square root of k over m which essentially that's the definition of Omega we can now write this as x double dot plus Omega squared x equals zero which is the second order differential equation of an undamped system.

  • Now the solution to that can either be the sine or the cosine.

  • Why is there a potential for a sine or cosine?

  • Well it depends on the initial condition, the initial time.

  • If time equals zero here notice that the amplitude is zero when time equals zero.

  • Well that would happen when we have a sine function.

  • When time equals zero the sine of zero is zero and x will be at zero so that's what we have over here.

  • However if time equals zero when the object is up here then we have a cosine function because notice now when time equals zero the cosine of zero is one and x equals a which is what we have over here.

  • So whether or not we use the sine or the cosine to describe the oscillatory motion or the simple harmonic motion simply depends on the initial condition.

  • Where's the object when time equals zero?

  • Now of course it could be over here on the way down it could be over here at time equals zero and then we have to modify the equation by putting a negative sign in front or by having a phase angle and we'll show you how to do that in the later video.

  • Now we want to show you also that these two solutions are indeed solutions of this second order differential equation.

  • How do we know that?

  • Well first of all we can take the first derivative with respect to time and the derivative of the sine is the cosine and the derivative of the angle omega t is omega so end up with a omega cosine of omega t.

  • If we then take the second derivative the derivative of cosine is the negative sign and again we have to multiply times the derivative of the omega square sine of omega t.

  • Then realizing that a sine omega t, a sine omega t is equal to x we can replace that so we're left with minus omega squared times x.

  • And then if we move to the left side then we end up with x double dot plus omega squared x equals zero which is exactly what we had over there.

  • So therefore we understand that this is indeed a solution to this second order differential equation.

  • We can do the same with the cosine function.

  • Again take the first derivative with respect to time.

  • The derivative of the cosine is the negative sine.

  • The derivative of omega t is omega.

  • Do it a second time.

  • X double dot as we call it is equal to again the derivative of sine is the negative cosine.

  • That's where the negative came from.

  • Oh no I'm sorry.

  • Take that back.

  • The derivative of the sine is the positive cosine.

  • We already had a negative sine.

  • And then we'll multiply the times the derivative of omega t which is omega.

  • That's where omega square comes from.

  • And cosine of omega t is equal to x.

  • We make that substitution.

  • We have x double dot equals minus omega squared x.

  • Move to the left side.

  • Now you can see again we get the exact same second order differential equation showing that x equal a cosine of omega t is also a solution of this differential equation.

  • So that's the basic differential equation describing the oscillator motion of an object.

  • Remember that k is the spring constant. m is the mass.

  • And the square root of k over m. m is indeed the angular speed or the angular frequency of the motion.

  • Again we'll show you more about that.

  • But that's the basic construct of symplemonic motion as we use the equation f equals ma which is then converted to the second order differential equation in terms of omega which is the angular frequency or angular speed of the oscillator motion.

  • And that is how it's done.

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