Physics · Oscillations · NEET
Parallel is always stiffer. In parallel both springs pull together on the same block, so the restoring force adds and k = k1 + k2 (bigger than either spring). In series the springs are one after another, so each one stretches and the total stretch is larger for the same force, giving 1/k = 1/k1 + 1/k2 (smaller than either spring). Rule: parallel k is larger than any single spring, series k is smaller than any single spring.
Think about what is shared. In parallel the two springs stretch by the SAME amount x, and the forces add: F = k1x + k2x = (k1+k2)x, so k = k1 + k2. In series the same force F passes through BOTH springs, and the total stretch is the sum: x = x1 + x2 = F/k1 + F/k2, so 1/k = 1/k1 + 1/k2. Same displacement gives addition; same force gives reciprocal addition.
Spring constant is inversely proportional to length: k is proportional to 1/L. So a shorter piece is stiffer. If a spring of constant k is cut into two equal halves, each half has constant 2k (twice as stiff). If cut into n equal parts, each part has constant nk. This is the most common trap in cut-spring problems, so always convert to segment constants first before combining them.
No. The effective spring constant depends only on the springs (their k values and how they are connected), not on the mass. Mass only enters later, in the time period T = 2 pi times root(m/k_eff). So first find k_eff from the springs, then plug it and m into the time period formula.
Two steps. First combine the springs into one effective constant k_eff (parallel add, series reciprocal add). Then use T = 2 pi times root(m/k_eff). Parallel makes k_eff larger, so T is smaller (faster oscillation). Series makes k_eff smaller, so T is larger (slower oscillation).
A spring of force constant k is cut into lengths in the ratio 1 : 2 : 3. These pieces are first connected in series, giving force constant k', and then in parallel, giving force constant k''. The ratio k' : k'' is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For springs in series the reciprocals add: 1/k_eff = 1/k1 + 1/k2. For two springs this gives k_eff = (k1 times k2) / (k1 + k2). The combined spring is softer than either single spring.
For springs in parallel the constants add directly: k_eff = k1 + k2. The combined spring is stiffer than either single spring because both share the same displacement and their forces add.
Series gives a larger time period. Series lowers k_eff, and since T = 2 pi root(m/k_eff), a smaller k_eff means a longer (slower) period. Parallel raises k_eff and shortens the period.
Each half becomes twice as stiff, so the new constant is 2k. Spring constant is inversely proportional to length, so halving the length doubles the constant.
No. The effective spring constant is set only by the springs and how they are connected. Mass only appears in the time period formula T = 2 pi root(m/k_eff).
