Spring Constant and the Effect of Cutting a Spring
Physics · Oscillations · NEET
Spring constant k tells you how stiff a spring is: force F = k times x, so a bigger k means a harder-to-stretch spring. When you cut a spring into a shorter piece, that piece becomes stiffer, so its k goes UP. The rule is simple: k times L stays constant, so if you cut a spring to half its length, its spring constant doubles. Memory hook: shorter spring, stronger spring.
A full spring of length L and constant k, when cut in half, gives a piece of length L/2 with a doubled constant 2k. Since k times L stays constant, a shorter spring is always stiffer.
Your doubts, answered
Does cutting a spring increase or decrease the spring constant?
Cutting a spring INCREASES the spring constant. The spring constant k measures how much force is needed per unit stretch. In a shorter piece, the same force produces less stretch, so it feels stiffer, which means k is larger. Short spring means larger k. Long spring means smaller k.
Why does a shorter spring have a higher spring constant?
Think of a spring as many small coils joined one after another, like resistors in series. Each coil stretches a little. A long spring has many coils, so the total stretch is large and k is small. A short piece has fewer coils, so the total stretch is small for the same force, so k is large. Fewer coils means stiffer spring.
If a spring is cut in half, what is the new spring constant?
If a spring of constant k is cut into two equal halves, each half has spring constant 2k. This comes from the rule k times L = constant. Original: k times L. Each half has length L/2, so new k times (L/2) must equal k times L, giving new k = 2k. Cut into n equal parts and each part has constant n times k.
What is the formula for spring constant when a spring is cut into pieces?
Use k times L = constant, so k is inversely proportional to length: k proportional to 1/L. If a full spring of constant k and length L is cut into a piece of length L1, that piece has constant k1 = k times (L/L1). For a spring cut in the ratio a : b : c, the pieces have constants in the ratio (1/a) : (1/b) : (1/c).
Does the spring constant depend on the length of the spring?
Yes. Spring constant is NOT a fixed property of the material alone; it depends on the length (number of coils), the wire thickness, coil diameter and material. For the same spring cut into pieces, only the length changes, so k is inversely proportional to length. This is why cutting matters in NEET problems.
⚠️ The NEET trap ✗ A spring is cut in half, so each half becomes weaker (smaller k), because it is smaller. ✓ Each half becomes STIFFER, not weaker. Smaller length means larger k. A half spring has k that is doubled (2k), so its time period T = 2 pi root(m/k) becomes shorter. 🧠 Small size tricks you. Physics says: shorter spring, stronger spring. Length down, k up.
Real NEET questions
NEET 2017
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
A · A. 1 : 6
B · B. 1 : 9
C · C. 1 : 11 ✓
D · D. 1 : 14
Solution: Step 1: Spring constant is inversely proportional to length, k proportional to 1/L. The spring is cut into 6 equal parts total (1 + 2 + 3 = 6). A full spring of length L has constant k, so one part of length L/6 has constant 6k.
Step 2: Piece lengths 1, 2, 3 parts give constants k1 = 6k, k2 = 3k, k3 = 2k (each = 6k divided by number of parts in that piece).
Step 3: In series, 1/k' = 1/6k + 1/3k + 1/2k = (1 + 2 + 3)/6k = 6/6k = 1/k, so k' = k.
Step 4: In parallel, k'' = 6k + 3k + 2k = 11k.
Step 5: Ratio k' : k'' = k : 11k = 1 : 11. Answer C.
NEET 2021
A spring is stretched by 5 cm under a force of 10 N. The time period of oscillation when a mass of 2 kg is suspended from it is
A · A. 3.14 s
B · B. 0.628 s ✓
C · C. 0.0628 s
D · D. 6.28 s
Solution: Step 1: Find the spring constant from F = k times x. Here F = 10 N and x = 5 cm = 0.05 m, so k = F/x = 10/0.05 = 200 N/m.
Step 2: Time period of a spring-mass system, T = 2 pi root(m/k) = 2 pi root(2/200) = 2 pi root(0.01).
Step 3: root(0.01) = 0.1, so T = 2 pi times 0.1 = 0.628 s. Answer B.
Solved Oscillations NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The spring constant k is the force needed to stretch or compress a spring by one unit of length. From F = k times x, a large k means a stiff, hard-to-stretch spring. Its unit is newton per metre (N/m).
What happens to the time period when a spring is cut?
Cutting a spring shorter raises k. Since T = 2 pi root(m/k), a larger k gives a SMALLER time period. So a shorter spring oscillates faster with the same mass.
A spring of constant k is cut into three equal parts. What is the constant of each part?
Each part has constant 3k. For n equal parts, each part has constant n times k, because k is inversely proportional to length and each part is 1/n of the length.
Is spring constant the same as stiffness?
Yes, in everyday NEET language spring constant and stiffness mean the same thing. A larger k means a stiffer spring.
Does the mass hung on a spring change its spring constant?
No. The spring constant depends on length, coil size, wire thickness and material, not on the load mass hung from it. Adding more mass changes the time period, not k.
