A spring of force constant k is cut into two pieces such that one piece is double the length of the

Q: Why does cutting a spring make it stiffer? It's made of the same material!

Even though the material is the same, a shorter spring has fewer active coils to share the applied deformation. For a given force, each individual coil stretches by the same amount, but with fewer coils, the total extension of the spring is smaller. Since stiffness is force divided by total extension ($k = F/x$), a smaller extension for the same force means a higher spring constant.

Q: Can I just assume that if a spring is cut into $n$ equal parts, each part has a spring constant of $n \cdot k$?

Yes, absolutely! If a spring is cut into $n$ equal parts, the length of each part becomes $L/n$. Since $k \propto 1/L$, the spring constant of each piece increases by a factor of $n$, becoming $n \cdot k$.

Q: What if the spring is cut into three pieces instead of two?

The same rule applies! No matter how many pieces you cut it into, the relation $k_i \cdot L_i = k \cdot L$ holds true for every single piece $i$, where $L_i$ is the length of that piece and $k_i$ is its new spring constant.

Q: Does the spring constant change if we stretch the spring instead of cutting it?

No. The spring constant $k$ is defined for the *natural (unstretched) length* of the spring. Stretching or compressing a spring within its elastic limit does not change its natural length, so its spring constant remains constant.

Q: Is the mass of the spring neglected in these calculations?

Yes, in standard JEE problems (unless specified otherwise), we assume the spring is massless. This ensures that the tension is uniform throughout the length of the spring, which makes the inverse relationship $k \propto 1/L$ perfectly uniform.

Question

A spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then, the long piece will have a force constant of

Options:

(A) $\frac{2}{3}k$

(B) $\frac{3}{2}k$

(C) $3k$

(D) $6k$

Sigma Percentile · Accepted Answer

### Introduction to Spring Stiffness

Imagine holding a long, flexible metal spring in your hands.

If you pull it, it stretches relatively easily.

But what happens if you cut that spring in half and try to stretch just one of the halves?

You will immediately notice that it feels significantly stiffer!

This is one of the most beautiful and counter-intuitive properties of elastic bodies: **the stiffness of a spring is intimately tied to its physical length**.

In this article, we will explore the physics behind this phenomenon and solve a classic JEE Advanced problem where a spring is cut into two unequal pieces.

---

### The Microscopic Origin of Spring Constant

To understand why a shorter spring is stiffer, we must look at the spring at a microscopic level.

Think of a spring of length $L$ as a series combination of $N$ tiny, identical spring segments, each of length $\Delta x$ and stiffness $k_0$.

When you apply a pulling force $F$ to the entire spring, that same tension $F$ is transmitted through every single segment in series.

Each segment stretches by a tiny amount $\Delta y = \frac{F}{k_0}$.

The total extension of the spring, $Y$, is the sum of the extensions of all these segments:

$$Y = N \cdot \Delta y = N \cdot \frac{F}{k_0}$$

We can rewrite this to find the effective spring constant $k$ of the entire spring:

$$k = \frac{F}{Y} = \frac{k_0}{N}$$

Since the number of segments $N$ is directly proportional to the natural length $L$ of the spring ($L = N \cdot \Delta x$), we arrive at a fundamental law of elasticity:

$$k \propto \frac{1}{L}$$

Or, written as a constant product:

$$k \cdot L = 	ext{constant}$$

This simple relation is our master tool for solving any spring-cutting problem.

---

### Analyzing the Setup

Let's apply this principle to our specific problem.

We start with an uncut spring of length $L$ and spring constant $k$.

This spring is cut into two pieces of lengths $L_1$ and $L_2$.

We are given that **one piece is double the length of the other**.

Let $L_1$ be the longer piece and $L_2$ be the shorter piece.

This gives us our first mathematical constraint:

$$L_1 = 2L_2$$

Since the two pieces together make up the original spring, their lengths must sum to the original length $L$:

$$L_1 + L_2 = L$$

---

### Calculating the Lengths

Now, let's substitute the first constraint into our sum equation to find the exact length of each piece:

$$2L_2 + L_2 = L \implies 3L_2 = L$$

Solving for $L_2$:

$$L_2 = \frac{1}{3}L$$

Now, we can easily find the length of the longer piece, $L_1$:

$$L_1 = 2L_2 = \frac{2}{3}L$$

So, the longer piece is exactly **two-thirds** of the original length, and the shorter piece is **one-third** of the original length.

---

### Finding the New Spring Constant

Let's denote the spring constant of the longer piece as $k_1$.

Using our inverse proportionality relation, the product of the spring constant and length for the longer piece must equal that of the original spring:

$$k_1 \cdot L_1 = k \cdot L$$

Now, we substitute the value of $L_1$ we found earlier:

$$k_1 \cdot \left(\frac{2}{3}Light) = k \cdot L$$

We can cancel the original length $L$ from both sides of the equation:

$$k_1 \cdot \frac{2}{3} = k$$

To isolate $k_1$, we multiply both sides by $\frac{3}{2}$:

$$k_1 = \frac{3}{2}k$$

Thus, the spring constant of the longer piece is **$\frac{3}{2}k$** (or $1.5k$).

This perfectly matches **Option (b)**.

---

### Verification via Series Combination

To be absolutely certain of our result, let's perform a quick sanity check.

If we calculate the spring constant of the shorter piece, $k_2$, using the same method:

$$k_2 \cdot L_2 = k \cdot L \implies k_2 \cdot \left(\frac{1}{3}Light) = k \cdot L \implies k_2 = 3k$$

If we connect these two pieces back together in series, their equivalent spring constant $k_{	ext{eq}}$ must equal the original spring constant $k$.

Let's verify this using the series combination formula:

$$\frac{1}{k_{	ext{eq}}} = \frac{1}{k_1} + \frac{1}{k_2}$$

Substitute the values of $k_1$ and $k_2$:

$$\frac{1}{k_{	ext{eq}}} = \frac{1}{\frac{3}{2}k} + \frac{1}{3k} = \frac{2}{3k} + \frac{1}{3k} = \frac{3}{3k} = \frac{1}{k}$$

Taking the reciprocal, we get:

$$k_{	ext{eq}} = k$$

This beautiful, consistent result confirms that our physical reasoning and calculations are absolutely flawless!

Animated Solution for Physics - Oscillations: A spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then, the long piece will have a force constant of

Select Answer:

Visualized Solution

Visualizing the Original Spring

The Inverse Proportionality Law

Defining the Cut Pieces

Setting up the Length Equations

Calculating the Length of the Long Piece

Applying the Inverse Proportionality Relation

Solving for the New Spring Constant

The Way Forward & Verification

The Sigma Insight: Simple Harmonic Motion (SHM)

Introduction to Spring Stiffness

The Microscopic Origin of Spring Constant

Analyzing the Setup

Calculating the Lengths

Finding the New Spring Constant

Verification via Series Combination

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