The extension of the spring on the left is x0−x1, and the extension of the spring on the right is x2−x0: \[\begin{aligned} \sum F_x = -k_1(x_0-x_1) + k_2 (x_2 - x_0) &= 0\\ -k_1x_0+k_1x_...The extension of the spring on the left is x0−x1, and the extension of the spring on the right is x2−x0: ∑Fx=−k1(x0−x1)+k2(x2−x0)=0−k1x0+k1x1+k2x2−k2x0=0−(k1+k2)x0+k1x1+k2x2=0∴k1x1+k2x2=(k1+k2)x0 Note that if the mass is displaced from x0 in any direction, the net force on the mass will be in the direction of the equilibrium position, and will act to “restore” the position of…
