If a spring has time period T and is cut into n equal parts, then the time period of each part will

Q: Why does a shorter spring have a larger spring constant?

Think of a spring as a series of tiny springs connected end-to-end. When you pull a long spring, each tiny segment stretches a little bit, adding up to a large total stretch for a given force. If you cut the spring, you have fewer segments. To get the same total stretch, you have to force each remaining segment to stretch much more, which requires a larger force. Hence, a shorter spring is stiffer (larger $k$).

Q: Does the mass change when we cut the spring?

No. The problem implies we are testing the new spring piece with the same mass $m$ that was used on the original spring to compare their time periods.

Q: What if the spring was cut into unequal parts, say in a ratio of 1:2?

The principle remains exactly the same! The spring constant of any part is inversely proportional to its specific length. The part with length $L/3$ would have a spring constant of $3k$, and the part with length $2L/3$ would have a spring constant of $1.5k$.

Q: Is the formula $k \propto 1/L$ always true for any spring?

Yes, as long as you are cutting the same uniform spring. It assumes the material, wire thickness, and coil radius remain identical, and only the number of active coils (which determines the length) is changing.

Q: Why does the time period decrease when the spring is cut?

Cutting the spring makes it stiffer (larger $k$). A stiffer spring exerts a stronger restoring force for the same displacement, pulling the mass back to equilibrium faster. This higher acceleration leads to a shorter time period.

Question

If a spring has time period $T$ and is cut into $n$ equal parts, then the time period of each part will be

Options:

(A) $T\sqrt{n}$

(B) $T/\sqrt{n}$

(C) $nT$

(D) $T$

Sigma Percentile · Accepted Answer

### The Setup
Imagine you have a classic spring-mass system oscillating back and forth with a time period $T$. Now, suppose you take that exact same spring and cut it into $n$ equal pieces. If you attach the same mass to just one of these smaller pieces, how does the time period of oscillation change? This is a classic problem that tests your understanding of what makes a spring "stiff."

### The Stiffness Secret
The secret lies in understanding the spring constant, $k$. The spring constant is a measure of a spring's stiffness. A fundamental property of springs is that their stiffness is inversely proportional to their natural length $L$.

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

Think about it physically: if you have a very long spring, it's relatively easy to stretch it by a few centimeters because that stretch is distributed over many coils. But if you have a very short spring with only a few coils, stretching it by the same amount requires much more force.

### The Mathematical Transformation
Since we cut the original spring into $n$ equal parts, the new length $L'$ of each piece is simply the original length divided by $n$:

$$L' = \frac{L}{n}$$

Because the length is divided by $n$, and stiffness is inversely proportional to length, the new spring constant $k'$ must be multiplied by $n$.

$$k' = n \cdot k$$

This means each smaller piece is $n$ times stiffer than the original, uncut spring.

### The Time Period Formula
Now, let's bring in the master equation for the time period of a spring pendulum. The time period $T$ is given by:

$$T = 2\pi \sqrt{\frac{m}{k}}$$

For our new, shorter spring, the mass $m$ remains unchanged, but the spring constant is now $k' = nk$. Let's substitute this into our formula to find the new time period $T'$:

$$T' = 2\pi \sqrt{\frac{m}{nk}}$$

### The Final Reveal
We can elegantly separate the $n$ term from the rest of the expression:

$$T' = \frac{1}{\sqrt{n}} \left( 2\pi \sqrt{\frac{m}{k}} \right)$$

**Notice the magic here!** The expression inside the parentheses is exactly our original time period $T$. Substituting $T$ back in, we arrive at our final, beautiful result:

$$T' = \frac{T}{\sqrt{n}}$$

By cutting the spring into $n$ pieces, we made it $n$ times stiffer, which in turn reduced the time period by a factor of $\sqrt{n}$.

Animated Solution for Physics - Oscillations: If a spring has time period $T$ and is cut into $n$ equal parts, then the time period of each part will be

Select Answer:

Visualized Solution

Visualizing the Setup

Spring Constant and Length Relation

New Spring Constant

Time Period Formula

Substituting New Values

Final Answer

The Way Forward

The Sigma Insight: Simple Harmonic Motion (SHM)

The Setup Imagine you have a classic spring-mass system oscillating back and forth with a time period $T$

The Stiffness Secret The secret lies in understanding the spring constant, $k$

The Mathematical Transformation

The Time Period Formula Now, let's bring in the master equation for the time period of a spring pendulum

The Final Reveal

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Animated Solution for Physics - Oscillations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a spring has time period T and is cut into n equal parts, then the time period of each part will be