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

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

Think of a spring as a chain of smaller, identical springs linked together. When you pull the whole chain, each small spring stretches a little bit, adding up to a large total extension. A longer chain has more links, so it stretches more for the same applied force, making it 'softer' (lower $k$). Conversely, a shorter piece has fewer links, stretches less, and is therefore 'stiffer' (higher $k$).

Q: Does cutting the spring change its material properties?

No, cutting the spring only changes its macroscopic geometric property (length). The microscopic material properties, like the shear modulus of the wire, and the cross-sectional area remain exactly the same. That's why the product $k \cdot l$ is a constant.

Q: What would be the spring constant of the shorter piece?

Using the same logic, the shorter piece has a length $l_2 = \frac{l}{3}$. Applying $k_2 l_2 = k l$, we get $k_2 \left(\frac{l}{3}\right) = k l$, which gives $k_2 = 3k$. The shorter piece is three times as stiff as the original spring!

Q: Can we use the formula for springs in series here?

Absolutely! The original spring can be thought of as the two cut pieces connected in series. The equivalent spring constant $k_{eq}$ for springs in series is given by $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$. If you plug in $k_1 = \frac{3}{2}k$ and $k_2 = 3k$, you will find that $\frac{1}{k_{eq}} = \frac{2}{3k} + \frac{1}{3k} = \frac{3}{3k} = \frac{1}{k}$, which perfectly matches the original spring constant $k$.

Q: Wait, if I cut a spring in half, does the spring constant double?

Yes, exactly! If you cut a spring exactly in half, the new length is $\frac{l}{2}$. Since $k \propto \frac{1}{l}$, halving the length doubles the stiffness. Each half will have a spring constant of $2k$.

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) $3 k$

(D) $6 k$

Sigma Percentile · Accepted Answer

Have you ever wondered what happens to the stiffness of a spring when you cut it? It might seem intuitive that a smaller piece of a spring would be weaker, but physics tells us the exact opposite! Let's dive into the fascinating relationship between a spring's length and its force constant.

### The Geometry of the Cut

Imagine a standard, uniform spring hanging from a ceiling. It has a natural length $l$ and a specific force constant $k$. This constant $k$ tells us how much force is required to stretch or compress the spring by a unit distance.

Now, we take a pair of wire cutters and snip this spring into two unequal pieces. The problem states that one piece is exactly double the length of the other. Let's translate this physical action into mathematics. If we call the length of the shorter piece $l_2$ and the longer piece $l_1$, we can write:

$$l_1 = 2l_2$$

Since these two pieces originally made up the entire spring, their lengths must add up to the original length $l$:

$$l_1 + l_2 = l$$

By substituting the first equation into the second, we get:

$$2l_2 + l_2 = l$$

$$3l_2 = l \implies l_2 = \frac{l}{3}$$

This tells us the shorter piece is one-third of the original length. Consequently, the longer piece, which is our main focus, has a length of:

$$l_1 = \frac{2l}{3}$$

### The Core Principle: Stiffness vs. Length

Here is where the beautiful physics comes into play. For any uniform spring, the force constant $k$ is **inversely proportional** to its natural length $l$.

Why is this the case? Think of a spring as a series of tiny, identical coils linked together. When you apply a force to the ends of the spring, every single coil stretches by a tiny amount. The total extension of the spring is the sum of the extensions of all these individual coils. A longer spring has more coils, so it stretches more for the same applied force, making it "softer" (a lower $k$). A shorter spring has fewer coils, stretches less, and is therefore "stiffer" (a higher $k$).

Mathematically, this inverse relationship is expressed as:

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

This implies that the product of the force constant and the length is a constant for a given spring material and cross-section:

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

### The Final Calculation

Since cutting the spring doesn't change the material it's made of or the thickness of the wire, this product $k \cdot l$ must remain exactly the same for both of the new pieces.

Let's apply this conservation principle to the longer piece. If its new force constant is $k_1$ and its length is $l_1$, we can write:

$$k_1 \cdot l_1 = k \cdot l$$

Now, we simply substitute the length $l_1$ that we calculated earlier:

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

Notice how the original length $l$ appears on both sides of the equation. It beautifully cancels out, leaving us with:

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

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

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

And there we have it! The longer piece of the spring has a force constant of $\frac{3}{2} k$. By cutting the spring and taking the piece that is two-thirds the original length, we have created a new spring that is $1.5$ times stiffer than the original.

Animated Solution for Physics - Laws of Motion: 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

The Original Spring Setup

The Cutting Process

Setting up the Length Ratio

Calculating the Lengths $l_{1}$ and $l_{2}$

The Spring Constant-Length Relation

Applying the Inverse Relation

Substituting the Length $l_{1}$

Solving for the New Constant $k_{1}$

The Sigma Insight: Newton's Laws of Motion

The Geometry of the Cut

The Core Principle

The Final Calculation

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