One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massl

Q: Why do we treat the wire and the spring as a series combination instead of parallel?

In a series combination, the components are connected end-to-end, meaning the same tension force acts through both of them, but their elongations add up. In a parallel combination, they would share the load and have the same elongation. Here, the wire and spring are connected sequentially, so they are strictly in series.

Q: How does Young's Modulus relate to a spring constant?

Young's Modulus ($Y$) measures a material's intrinsic stiffness. By relating force to elongation ($F = \frac{YA}{L} \Delta L$), we see that the term $\frac{YA}{L}$ behaves exactly like the spring constant $k$ in Hooke's Law ($F = k \Delta L$).

Q: What happens if the wire is completely rigid?

If the wire is perfectly rigid, its Young's Modulus $Y \to \infty$. Taking the limit of our final time period formula as $Y \to \infty$, the term $\frac{YA + kL}{YAk}$ simplifies to $\frac{1}{k}$, giving $T = 2\pi \sqrt{\frac{m}{k}}$, which is the time period of the spring alone.

Q: Does the mass of the wire or spring affect the time period here?

In this problem, both the wire and the spring are assumed to be massless. If they had mass, we would have to account for their kinetic energy, which would increase the effective mass of the system and alter the time period.

Q: Why does a longer wire have a smaller equivalent spring constant?

Since $k_1 = \frac{YA}{L}$, the spring constant is inversely proportional to the length $L$. A longer wire has more atomic bonds in series, making it more flexible overall and easier to stretch for a given force.

Question

One end of a long metallic wire of length $L$ is tied to the ceiling. The other end is tied to a massless spring of spring constant $k$. A mass $m$ hangs freely from the free end of the spring. The area of cross-section and the Young's modulus of the wire are $A$ and $Y$ respectively. If the mass is slightly pulled down and released, it will oscillate with a time period $T$ equal to

Options:

(A) $2\pi (m/k)^{1/2}$

(B) $2\pi \sqrt{\frac{m(YA + kL)}{YAk}}$

(C) $2\pi [(mYA/kL)]^{1/2}$

(D) $2\pi (mL/YA)^{1/2}$

Sigma Percentile · Accepted Answer

### The Duality of Elasticity: Wire as a Spring

When we think of a metallic wire, we often picture a rigid, unyielding rod.

However, at the microscopic level, the metallic bonds holding the atoms together act like tiny, incredibly stiff springs.

When you pull on a wire, you are stretching these atomic springs.

This means that any elastic material, including a long metallic wire, can be modeled as a spring with its own unique spring constant.

To find this equivalent spring constant, we turn to the definition of **Young's Modulus** ($Y$):

$$Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta L/L}$$

By rearranging this formula to solve for the restoring force $F$, we get:

$$F = \left(\frac{YA}{L}ight) \Delta L$$

Comparing this directly with Hooke's Law ($F = k_1 \Delta L$), we can define the equivalent spring constant of the wire as:

$$k_1 = \frac{YA}{L}$$

This is a beautiful realization: a thicker wire (larger $A$) or a wire made of a stiffer material (larger $Y$) has a higher spring constant, while a longer wire (larger $L$) is easier to stretch and thus has a lower spring constant.

### Analyzing the Series Combination

In our problem, the wire and the spring are connected end-to-end.

This is a classic **series combination**.

Why is it series?

Because if you pull down on the mass, the tension force transmitted through the spring is exactly the same as the tension force transmitted through the wire.

However, both the wire and the spring will stretch by different amounts depending on their stiffness.

The total displacement $\Delta x_{	ext{total}}$ is the sum of the individual displacements:

$$\Delta x_{	ext{total}} = \Delta x_{	ext{wire}} + \Delta x_{	ext{spring}}$$

Using Hooke's Law, this relation leads directly to the equivalent spring constant formula for a series combination:

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

Substituting our derived value for the wire ($k_1 = \frac{YA}{L}$) and the spring constant ($k_2 = k$):

$$\frac{1}{k_{eq}} = \frac{L}{YA} + \frac{1}{k}$$

Finding a common denominator and inverting gives us the total equivalent stiffness of the system:

$$k_{eq} = \frac{YAk}{YA + kL}$$

### Finding the Time Period of Oscillation

Now that we have simplified our complex system into a single equivalent spring-mass system, finding the time period is straightforward.

The time period $T$ of a simple harmonic oscillator is given by:

$$T = 2\pi \sqrt{\frac{m}{k_{eq}}}$$

Substituting our expression for $k_{eq}$:

$$T = 2\pi \sqrt{\frac{m}{\frac{YAk}{YA + kL}}}$$

Simplifying the compound fraction yields the final elegant result:

$$T = 2\pi \sqrt{\frac{m(YA + kL)}{YAk}}$$

This matches option **(b)** perfectly!

Select Answer:

Visualized Solution

Understanding the Physical Setup

Modeling the Wire as a Spring

Deriving the Spring Constant of the Wire

Identifying the Series Connection

Substituting the Spring Constants

Simplifying the Equivalent Spring Constant

The Time Period of SHM

Calculating the Final Time Period

Analyzing Limiting Cases

The Sigma Insight: Force and Energy Method in SHM

The Duality of Elasticity

Analyzing the Series Combination

Finding the Time Period of Oscillation

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