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串联为简单起见,以两根弹簧为例。设甲弹簧的倔强系数为 K_1,乙弹簧的倔强系数为 K_2,串联后弹簧组的总倔强系数为 K,在外力 F 作用下,甲、乙的伸长分别为 x_1和 x_2,弹簧组总的伸长为 x。1.在不考虑弹簧本身重量的情况下,图1所示之弹簧组有如下规律。①串联的甲、乙两弹簧受力相同,都等于 F。②由胡克定律知:F=K_1x_1,F=K_2x_2,F=Kx_0图1 即 x_1/x_2=K_2/K_1在串联情况下,每一弹簧的伸长与其本身的倔强系数成反比。③总伸长 x=x_1+x_2由②x_1=F/K_1,x_2=F/K_2,x=F/K。代入上式得:F/K=(F/K_1)+(F/K_2)即1/K=(1/K_1)+(1/K_2)。(或 K=(K_1K_2)/(K_1+K_2))串联弹簧组总倔强系数的倒数等于串联的
For the sake of simplicity, two springs are taken as examples. The stubborn coefficient of the set A spring is K_1, the stub coefficient of the B spring is K_2, and the total coefficient of the spring group after the series is K. Under the action of the external force F, the elongations of the A and B are x_1 and x_2, respectively, and the spring group is total. Extend to x. 1. Regardless of the weight of the spring itself, the spring set shown in Figure 1 has the following rules. 1 The two springs A and B in the series have the same force and are equal to F. 2 by Hooke’s Law: F = K_1x_1, F = K_2x_2, F = Kx_0 Figure 1 That is, x_1/x_2 = K_2/K_1 In the series case, the elongation of each spring is inversely proportional to its own marginal coefficient. 3 Total elongation x=x_1+x_2 From 2x_1=F/K_1, x_2=F/K_2, x=F/K. Substituting into the above formula: F/K=(F/K_1)+(F/K_2) ie 1/K=(1/K_1)+(1/K_2). (or K=(K_1K_2)/(K_1+K_2)) The reciprocal of the total coefficient of series springs is equal to the serial number.