Simple harmonic motion. To evaluate the numerical answer, the following values are used: gravitational acceleration: g = 9.8 m/s 2, air resistance proportionality constant: K = 110 kg/s. This expression for the position function can be rewritten using the trigonometric identity cos(α – β) = cos α cos β + sin α sin β, as follows: The phase angle, φ, is defined here by the equations cos φ = 3/ 5 and sin φ = 4/ 5, or, more briefly, as the first‐quadrant angle whose tangent is 4/ 3 (it's the larger acute angle in a 3–4–5 right triangle). 0000012015 00000 n
The net force on the block is , so Newton's Second Law becomes, because m = 1. <]>>
The original differential equation (*) for the LRC circuit was nonhomogeneous, so a particular solution must still be obtained. 0000013093 00000 n
We will discuss here some of the techniques used for obtaining the second-order differential equation for an RLC Circuit. We will see that the DE's are identical to those for the mechanical systems studied earlier. 0000017611 00000 n
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The block can be set into motion by pulling or pushing it from its original position and then letting go, or by striking it (that is, by giving the block a nonzero initial velocity). A survey is presented on the applications of differential equations in some important electrical engineering problems. 0000010587 00000 n
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Consider a spring fastened to a wall, with a block attached to its free end at rest on an essentially frictionless horizontal table. » ], In the underdamped case , the roots of the auxiliary polynomial equation can be written as, and consequently, the general solution of the defining differential equation is. 0000015681 00000 n
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��O�Y�a��G͘��NH����/5-[sS��i���E�l�}��C�>FǾ��M��xi��0�᎐���A|����^Z������{�Cf�����Q��SQ,p%�?¹
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e!Z"���YƅEG��찜.T��d$xK�ߺ�:��`�?�:��>A�Aw�!Fk��"K~Yu�W|��E)�ێ�>�{⿴�l�!E>*(�"��!eD�R�� The restoring force here is proportional to the displacement ( F = −kx α x), and it is for this reason that the resulting periodic (regularly repeating) motion is called simple harmonic. (Again, recall the sky diver falling with a parachute. If this spring‐block apparatus is submerged in a viscous fluid medium which exerts a damping force of – 4 v (where v is the instantaneous velocity of the block), sketch the curve that describes the position of the block as a function of time. 0000016208 00000 n
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Courses 192 102
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There's no signup, and no start or end dates. At what minimum altitude must her parachute open so that she slows to within 1% of her new (much lower) terminal velocity ( v 2) by the time she hits the ground? Freely browse and use OCW materials at your own pace. For example, ⋅ (“ s dot”) denotes the first derivative of s with respect to t, and (“ s double dot”) denotes the second derivative of s with respect tot. 0000052181 00000 n
Materials include course notes, Javascript Mathlets, and a problem set with solutions. Made for sharing. But this seems reasonable: Damping reduces the speed of the block, so it takes longer to complete a round trip (hence the increase in the period). 0000013645 00000 n
It is pulled 3/ 10 m from its equilibrium position and released from rest. This function is periodic, which means it repeats itself at regular intervals. (Recall that if, say, x = cosθ, then θ is called the argument of the cosine function.) This will always happen in the case of underdamping, since will always be lower than. 0000064076 00000 n
The angular frequency of this periodic motion is the coefficient of. 0000074439 00000 n
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Since the period specifies the length of time per cycle, the number of cycles per unit time (the frequency) is simply the reciprocal of the period: f = 1/ T. Therefore, for the spring‐block simple harmonic oscillator. Differential Equations Home a�Z���yß�E���I�&[w�x�恻ѩ���L���FL�ݳ15Ac��7�-�&^�Z�@AIŦ}/��H�ޒ���d�[���P����9'�o�˼����/-K3#4Ӑ��FE�3���P�8fٌ~q�*�Jb?54����nC�'� �Ј���Zy���B .�@�=�?���Cv����ɬ��U�����$�}?2*���� �qٽ:P�jn*�I At the relatively low speeds attained with an open parachute, the force due to air resistance was given as Kv, which is proportional to the velocity.). the general solution of (**) must be, by analogy, But the solution does not end here. Now, if an expression for i( t)—the current in the circuit as a function of time—is desired, then the equation to be solved must be written in terms of i. In this case, the frequency (and therefore angular frequency) of the transmission is fixed (an FM station may be broadcasting at a frequency of, say, 95.5 MHz, which actually means that it's broadcasting in a narrow band around 95.5 MHz), and the value of the capacitance C or inductance L can be varied by turning a dial or pushing a button. 192 0 obj <>
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Another important characteristic of an oscillator is the number of cycles that can be completed per unit time; this is called the frequency of the motion [denoted traditionally by v (the Greek letter nu) but less confusingly by the letter f]. Modify, remix, and reuse (just remember to cite OCW as the source. Therefore, set v equal to (1.01) v 2 in equation (***) and solve for t; then substitute the result into (**) to find the desired altitude. �>p�E�g��1̱��:z�)�&/��>���g��ƞUZ���?�[ꃬ�� 0
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Since the roots of the auxiliary polynomial equation, , are, the general solution of the differential equation is. 0000011295 00000 n
bookmarked pages associated with this title. Then Newton's Second Law ( F net = ma) becomes mg – Kv = ma, or, since v = and a =, This situation is therefore described by the IVP, The differential equation is second‐order linear with constant coefficients, and its corresponding homogeneous equation is, where B = K/m. 0000045171 00000 n
As you probably already know, electric circuits can consist of a wide variety of complex components. In particular, assuming that the inductance L, capacitance C, resistance R, and voltage amplitude V are fixed, how should the angular frequency ω of the voltage source be adjusted to maximized the steady‐state current in the circuit? Let y denote the vertical distance measured downward form the point at which her parachute opens (which will be designated time t = 0). 0000076857 00000 n
Note that the period does not depend on where the block started, only on its mass and the stiffness of the spring. 0000051308 00000 n
���. from your Reading List will also remove any We will also use complex techniques to define and understand impedance in these circuits. This resistance would be rather small, however, so you may want to picture the spring‐block apparatus submerged in a large container of clear oil. 0000014695 00000 n
Once the block is set into motion, the only horizontal force that acts on it is the restoring force of the spring. 0000032732 00000 n
The presence of the decaying exponential factor e −2 t in the equation for x( t) means that as time passes (that is, as t increases), the amplitude of the oscillations gradually dies out. 0000080422 00000 n
Once the transient current becomes so small that it may be neglected, under what conditions will the amplitude of the oscillating steady‐state current be maximized? where B = K/m. and solving this second‐order differential equation for s. [You may see the derivative with respect to time represented by a dot. %PDF-1.4
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Find materials for this course in the pages linked along the left. Finally, a resistor opposes the flow of current, creating a voltage drop equal to iR, where the constant R is the resistance. Kirchhoff's Loop Rule states that the algebraic sum of the voltage differences as one goes around any closed loop in a circuit is equal to zero. the inductance L, the capacitance C and the resistor R in a closed form in terms of the three- parameters Mittag-Leffer function. 0000011203 00000 n
A Second-order circuit cannot possibly be solved until we obtain the second-order differential equation that describes the circuit.