Transactions of the Volume 1-5 - 6 Angebote vergleichen

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Von Händler/Antiquariat, Amazon.com.
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1900 edition. Excerpt: ...been solved before. In its main features it is similar to that which Lagrange solved in bis Mecanique Analytique, pt. II, sec. VI, the problem, namely, of the free vibrations of a weightless string, fixed at its two ends, and loaded at equidistant points by equal weights. But it is much more general because frictional resistances are taken into consideration, and also because forced as well as free oscillations are considered. Let Xu and Lx, i?rt and Rx, C„ and Cx be the inductance, ohmic resistance, capacity of A and B, respectively. Let L, It, C be the corresponding quantities of the coils and condensers in the several component circuits. The real part of Ee'1' is the e. m. f. impressed by alternator A. Let a?, xn be the currents in the component circuits;,-PB_, the differences of potential in tbe line condensers, C, C2, Cn_,; P0 and P' the differences of potential in the condensers in A and B respectively; and?i i i £„_i the condenser currents. A. Forced oscillations. Stating the law of equality of action and reaction for each component circuit, we obtain the following n differential equations: (Z, + 22,) + (P, + 2 7?),„-2P„+ P' = 0. When the steady state lias been reached, the currents will be, just like the impressed e. m. f., simple harmonics of the time t, that is, where Al, A2, are complex quantities. Differentiating each member of (2) and substituting from (3) and (1) we obtain: "kxt + %,-0 = IJ, Equation (3) of § I suggests the following solution: (6«) xm = Kx cos 2(n--m) 0 + Ki sin 2(«--m)0. If h + 2 = 2 cos 20, then all the equations except the first and the last will be satisfied for all values of 7T, and 1£2. These two equations, which correspond to the boundary equations of... Paperback, Label: RareBooksClub.com, RareBooksClub.com, Produktgruppe: Book, Publiziert: 2012-08-06, Studio: RareBooksClub.com.

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Von Händler/Antiquariat, super_star_seller.
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1900 edition. Excerpt: ...been solved before. In its main features it is similar to that which Lagrange solved in bis Mecanique Analytique, pt. II, sec. VI, the problem, namely, of the free vibrations of a weightless string, fixed at its two ends, and loaded at equidistant points by equal weights. But it is much more general because frictional resistances are taken into consideration, and also because forced as well as free oscillations are considered. Let Xu and Lx, i?rt and Rx, C„ and Cx be the inductance, ohmic resistance, capacity of A and B, respectively. Let L, It, C be the corresponding quantities of the coils and condensers in the several component circuits. The real part of Ee'1' is the e. m. f. impressed by alternator A. Let a?, xn be the currents in the component circuits;,-PB_, the differences of potential in tbe line condensers, C, C2, Cn_,; P0 and P' the differences of potential in the condensers in A and B respectively; and?i i i £„_i the condenser currents. A. Forced oscillations. Stating the law of equality of action and reaction for each component circuit, we obtain the following n differential equations: (Z, + 22,) + (P, + 2 7?),„-2P„+ P' = 0. When the steady state lias been reached, the currents will be, just like the impressed e. m. f., simple harmonics of the time t, that is, where Al, A2, are complex quantities. Differentiating each member of (2) and substituting from (3) and (1) we obtain: "kxt + %,-0 = IJ, Equation (3) of § I suggests the following solution: (6«) xm = Kx cos 2(n--m) 0 + Ki sin 2(«--m)0. If h + 2 = 2 cos 20, then all the equations except the first and the last will be satisfied for all values of 7T, and 1£2. These two equations, which correspond to the boundary equations of... Paperback, Label: RareBooksClub.com, RareBooksClub.com, Produktgruppe: Book, Publiziert: 2012-08-06, Studio: RareBooksClub.com.

Von Händler/Antiquariat, The Book Depository US [58762574], London, United Kingdom.
Language: English Brand New Book ***** Print on Demand *****. This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1900 edition. Excerpt: .been solved before. In its main features it is similar to that which Lagrange solved in bis Mecanique Analytique, pt. II, sec. VI, the problem, namely, of the free vibrations of a weightless string, fixed at its two ends, and loaded at equidistant points by equal weights. But it is much more general because frictional resistances are taken into consideration, and also because forced as well as free oscillations are considered. Let Xu and Lx, i?rt and Rx, C and Cx be the inductance, ohmic resistance, capacity of A and B, respectively. Let L, It, C be the corresponding quantities of the coils and condensers in the several component circuits. The real part of Ee 1 is the e. m. f. impressed by alternator A. Let a?, xn be the currents in the component circuits;, -PB , the differences of potential in tbe line condensers, C, C2, Cn ; P0 and P the differences of potential in the condensers in A and B respectively; and?i i i i the condenser currents. A. Forced oscillations. Stating the law of equality of action and reaction for each component circuit, we obtain the following n differential equations: (Z, + 22, ) + (P, + 2 7?), -2P + P = 0. When the steady state lias been reached, the currents will be, just like the impressed e. m. f., simple harmonics of the time t, that is, where Al, A2, are complex quantities. Differentiating each member of (2) and substituting from (3) and (1) we obtain: kxt + , -0 = IJ, Equation (3) of I suggests the following solution: (6-) xm = Kx cos 2(n--m) 0 + Ki sin 2(---m)0. If h + 2 = 2 cos 20, then all the equations except the first and the last will be satisfied for all values of 7T, and 1 2. These two equations, which correspond to the boundary equations of.

Von Händler/Antiquariat, BuySomeBooks [52360437], Las Vegas, NV, U.S.A.
This item is printed on demand. Paperback. 126 pages. Dimensions: 9.7in. x 7.4in. x 0.3in.This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1900 edition. Excerpt: . . . been solved before. In its main features it is similar to that which Lagrange solved in bis Mecanique Analytique, pt. II, sec. VI, the problem, namely, of the free vibrations of a weightless string, fixed at its two ends, and loaded at equidistant points by equal weights. But it is much more general because frictional resistances are taken into consideration, and also because forced as well as free oscillations are considered. Let Xu and Lx, irt and Rx, C and Cx be the inductance, ohmic resistance, capacity of A and B, respectively. Let L, It, C be the corresponding quantities of the coils and condensers in the several component circuits. The real part of Ee1 is the e. m. f. impressed by alternator A. Let a, xn be the currents in the component circuits;, -PB, the differences of potential in tbe line condensers, C, C2, Cn, ; P0 and P the differences of potential in the condensers in A and B respectively; andi i i i the condenser currents. A. Forced oscillations. Stating the law of equality of action and reaction for each component circuit, we obtain the following n differential equations: (Z, 22, ) (P, 2 7), -2P P 0. When the steady state lias been reached, the currents will be, just like the impressed e. m. f. , simple harmonics of the time t, that is, where Al, A2, are complex quantities. Differentiating each member of (2) and substituting from (3) and (1) we obtain: kxt , -0 IJ, Equation (3) of I suggests the following solution: (6) xm Kx cos 2(n--m) 0 Ki sin 2(--m)0. If h 2 2 cos 20, then all the equations except the first and the last will be satisfied for all values of 7T, and 12. These two equations, which correspond to the boundary equations of. . . This item ships from La Vergne,TN.

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Von Händler/Antiquariat, Through The Stacks LLC, NY, Brockport, [RE:3].
We ship worldwide with delivery confirmation. We answer all e-mails within one business day. Trade paperback.

Lieferung aus: Vereinigte Staaten von Amerika, plus shipping, Shipping area: DOM.
Von Händler/Antiquariat, Through The Stacks LLC, NY, Brockport, [RE:3].
We ship worldwide with delivery confirmation. We answer all e-mails within one business day. Trade paperback.