Two springs of spring constants K1 and K2 are joined in series. The effective spring constant of the combination is given by:

Explanation

When two springs are joined in series, they are connected end to end. In this configuration, the total displacement of the combined system is the same for both springs. Now, let's consider the individual spring constants: K1 and K2. According to Hooke's Law, the force exerted by a spring is directly proportional to its displacement. Mathematically, we can express this as: F = -K1 * x1 (for the first spring) F = -K2 * x2 (for the second spring) Here, x1 and x2 represent the displacements of the first and second springs, respectively. Since the total displacement of the combined system is the same, we can equate x1 and x2: x1 = x2 Now, let's apply a force F to the combined system. The total force exerted by the springs is the sum of the forces exerted by each spring: F = -K1 * x1 - K2 * x2 Since x1 = x2, we can rewrite the equation as: F = -(K1 + K2) * x1 Comparing this equation with Hooke's Law (F = -Keffective * x1), we can see that the effective spring constant (Keffective) is given by: Keffective = K1 + K2 Therefore, the effective spring constant of two springs joined in series is given by the formula: Keffective = (K1 * K2) / (K1 + K2)