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Undamped oscillators that are driven at resonance have unusual (and nonphysical) solutions.(a) To investigate this, find the synchronous solution $A \cos \Omega t+B \sin \Omega t$ to the generic forced oscillatorequation(7)$$m y^{\prime \prime}+b y^{\prime}+k y=\cos \Omega t$$(b) Sketch graphs of the coefficients $A$ and $B,$ as functions of $\Omega,$ for $m=1, b=0,1,$ and $k=25$(c) Now set $b=0$ in your formulas for $A$ and $B$ and resketch the graphs in part (b), with $m=1,$ and $k=25 .$ What happens at $\Omega=5 ?$ Notice that the amplitudes of the synchronous solutions grow without bound as $\Omega$ approaches 5 .(d) Show directly, by substituting the form $A \cos \Omega t+$ $B \sin \Omega t$ into equation $(7),$ that when $b=0$ there are no synchronous solutions if $\Omega=\sqrt{k / m}$(e) Verify that $(2 m \Omega)^{-1} t \sin \Omega t$ solves equation (7) when $b=0$ and $\Omega=\sqrt{k / m}$. Notice that this nonsynchronous solution grows in time, without bound.Clearly one cannot neglect damping in analyzing an oscillator forced at resonance, because otherwise the solutions, as shown in part (e), are nonphysical. This behavior will be studied later in this chapter.

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Calculus 2 / BC

Chapter 4

Linear Second-Order Equations

Section 1

Introduction: The Mass-Spring Oscillator

Differential Equations

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We have mass time to a plus a is b times y prime plus k and equal 12 point for a party of us solving using big y d equal in cosine omega 7. A d y e is equal to the minus, a omemee plus de omega omega t and then, which is equal to men, a man, a square foot minus, because your sine hoe situation below and you're able to get e time, k minus t times, maka square times. Cosine e times minus time, maguscosine mass multiplied by sine or to that plus e times little beat omega times, cosine omeativetmi 8 times no bin omega omega the which is equal to cosine and solve equations. In the equation you produce are 8 times minus m magus square plus b, a e tineton 8 and the answer to produce is going to be a minus m times. Meta, squarefoot, omega square square, omega square and a p will be equal to v. Omens, omegusthis omega divide, minus times, omega square square square; temomega squarefoot, you perpotet graph, this equation these points 0.1 and masseno and 25. So i replace the functions in. Forgive you to a equal to using the values that you place it with above here. Looking 25 minus omega square minus omega square square plus 2.01 is planned, 1 multiplied divided by 25, minus omega square square 0.01 omega square. Then then you got to plot the graph. So here it is first you you have omega here on the x axis and in the other 1 is either a or b. So at 5 we made a equals. I it's going to be undefined. So, as you go from the left side go to the left, it will burn 0 but as you go close to cost 5, you keep going upward infinity. But as you go as you go from 5 all the weight to infinity it's going from negative infinity. All the way to just for these 2 here they repose a but along this axis b, is shown. It'S kept. Constant is supposed to note, as it goes, because from infinite infinity and now for part c for c it will be using b equals 0 m equaling. 1 and in 25, and you replace it with 8 equals 1 divided by 25 minus omega square and b equal to 0 apook, like simply looks like a line of curve going from infinity or 0 to infinity and that's where x, equals as where omega equals. Sorry, omega holsome equals 5, but as it goes further as omega t 0, that this here represents a the other 1. The second line it's going to be 0, so this is this represents e, and this represents a b d b equals 0 and will megan equal square root of a divided by m, then a equals b and b will equal 0 and then brusore equaling 2 times. As meg in very multiplied by t sine omega t use y prime, which is equal to 2 times times, omega, inverse multiplied by sining t plus o a t o line omega t y double prime is colin 2 tenses, and this is the inverse multiplied by 2. Omega 3 minus a square tin to make a wood set between makes your else b is equals here here. So it will be mass times 2 times omega times, inverse to omega close sine t minus omega square times. T sine omega t plus k times 2 and 2 times times omega. This is the inverse t times sine omega t using omega equal in square root of k. Divided by this will be equal to both sides will make the cane a non secretive solution.

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