A spring whose unstretched length is l has a force constant k. The spring is cut into two pieces of

Q: Why is the force constant inversely proportional to length?

Think of a spring as a series of identical smaller springs connected end-to-end. When you pull the whole spring, each small segment stretches a little bit. The total stretch is the sum of all these small stretches. A longer spring has more segments, so for the same applied force, it stretches more overall. Since $k = \frac{F}{x}$, a larger stretch $x$ means a smaller $k$.

Q: Does cutting the spring change its material properties?

No, cutting the spring only changes its macroscopic geometric length. The material properties (like Young's modulus or shear modulus) and the microscopic geometry (wire thickness, coil radius) remain exactly the same.

Q: What happens to the total energy stored if we stretch both cut pieces by the same amount $x$?

The energy stored in a spring is $U = \frac{1}{2}kx^2$. Since the shorter piece has a larger $k$, it will store more energy than the longer piece for the exact same extension $x$.

Q: Can we use $k_1 + k_2 = k$?

No! That is a common misconception. When springs are connected in series (which is what a single uncut spring is), their equivalent spring constant is given by $\frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2}$.

Q: What if the spring was cut into three pieces?

The same principle applies. $k_1 l_1 = k_2 l_2 = k_3 l_3 = k l$. You can find the force constant of any piece by dividing the constant product $k l$ by the length of that specific piece.

Question

A spring whose unstretched length is $l$ has a force constant $k$. The spring is cut into two pieces of unstretched lengths $l_1$ and $l_2$ where, $l_1 = n l_2$ and $n$ is an integer. The ratio $\frac{k_1}{k_2}$ of the corresponding force constants $k_1$ and $k_2$ will be

Options:

(A) $n$

(B) $\frac{1}{n^2}$

(C) $\frac{1}{n}$

(D) $n^2$

Sigma Percentile · Accepted Answer

### The Anatomy of a Spring

Imagine holding a long, uniform spring in your hands. It has a natural, unstretched length denoted by $l$, and a characteristic stiffness known as the force constant, $k$. This force constant $k$ is a measure of how much force is required to stretch or compress the spring by a unit distance.

But what happens when we alter the physical dimensions of this spring? The stiffness of a spring is not just a property of the material it's made of; it is intimately tied to its macroscopic geometry, specifically its length.

### The Inverse Relationship

Here is the golden rule of uniform springs: **the force constant of a spring is inversely proportional to its natural length**.

Mathematically, we express this as:
$$k \propto \frac{1}{l}$$

Why does this happen? Think of a spring as a series of identical, smaller microscopic springs connected end-to-end. When you apply a force to the ends of the entire spring, that force is transmitted equally through every single microscopic segment. Each segment stretches by a tiny amount. The total extension of the spring is the sum of all these tiny stretches.

A longer spring has more of these segments. Therefore, for the exact same applied force, a longer spring will stretch more overall than a shorter spring. Since $k = \frac{F}{x}$, a larger total stretch $x$ means a smaller force constant $k$. Thus, a shorter spring is stiffer, and a longer spring is looser. This implies that the product of the force constant and the length is a constant for any piece cut from the same uniform spring:
$$k \cdot l = 	ext{constant}$$

### The Mathematical Execution

In our problem, the original spring is cut into two unequal pieces with lengths $l_1$ and $l_2$, and corresponding force constants $k_1$ and $k_2$.

Because the product of stiffness and length is constant, we can write:
$$k_1 \cdot l_1 = k_2 \cdot l_2$$

Rearranging this gives us the ratio of their force constants:
$$\frac{k_1}{k_2} = \frac{l_2}{l_1}$$

Notice how the indices flip! The ratio of the force constants is the inverse of the ratio of their lengths.

The problem provides a specific constraint: the length of the first piece is $n$ times the length of the second piece, meaning $l_1 = n l_2$. Let's substitute this constraint into our ratio equation:
$$\frac{k_1}{k_2} = \frac{l_2}{n l_2}$$

The $l_2$ terms in the numerator and denominator cancel out beautifully, leaving us with our final, elegant result:
$$\frac{k_1}{k_2} = \frac{1}{n}$$

This tells us that since the first piece is $n$ times longer than the second piece, it must be $n$ times less stiff!

Select Answer:

Visualized Solution

$k and l of Original Spring$

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

$Cutting the Spring$

$Ratio of Force Constants$

$Substituting l_{1} = n l_{2}$

$The Way Forward$

The Sigma Insight: Simple Harmonic Motion

The Anatomy of a Spring

The Inverse Relationship

The Mathematical Execution

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