A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cro

Q: Why can we treat a solid wire as a spring?

At a microscopic level, the atoms in a solid are held together by interatomic forces that behave like tiny springs. When we apply a macroscopic force, the cumulative stretching of these bonds results in a linear restoring force, perfectly obeying Hooke's Law for small displacements.

Q: What is the physical significance of the equivalent spring constant $k = YA/L$?

It tells us that a thicker wire (larger $A$) or a shorter wire (smaller $L$) is stiffer (higher $k$), while a material with a higher Young's Modulus $Y$ inherently resists stretching more.

Q: Does the mass of the wire affect the oscillation frequency?

In this problem, the wire is specified to have 'negligible mass'. If the wire had a non-negligible mass $m_w$, we would have to include a fraction of its mass (typically $m_w/3$) in the effective mass of the system.

Q: Why did we use angular frequency $\omega$ instead of ordinary frequency $f$?

The problem explicitly states that the 'angular frequency' is $140\text{ rad s}^{-1}$. Angular frequency $\omega$ is measured in radians per second, whereas ordinary frequency $f$ is in Hertz. Always check the units to be sure!

Q: What would happen if the displacement $x$ was very large?

If the displacement is too large, the wire might exceed its elastic limit. Beyond this limit, the deformation becomes plastic (permanent), Hooke's Law no longer holds, and the motion is no longer simple harmonic.

Question

A $0.1	ext{ kg}$ mass is suspended from a wire of negligible mass. The length of the wire is $1	ext{ m}$ and its cross-sectional area is $4.9 	imes 10^{-7}	ext{ m}^2$. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency $140	ext{ rad s}^{-1}$. If the Young's modulus of the material of the wire is $n 	imes 10^9	ext{ Nm}^{-2}$, the value of $n$ is.

Sigma Percentile · Accepted Answer

### Introduction to Elastic Oscillations

Imagine suspending a heavy metal block from a thin metallic wire.

If you pull the block downward slightly and let it go, what happens?

It doesn't just hang there; it bounces up and down in a beautiful, rhythmic dance.

This is **Simple Harmonic Motion (SHM)** driven by the elastic properties of the wire.

In this article, we will explore how a solid wire can act exactly like a mechanical spring, and how we can use the principles of elasticity to determine the fundamental properties of the material.

---

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

At a microscopic level, the atoms in a solid metal wire are held together by interatomic forces.

These forces act like tiny, invisible springs connecting the atoms.

When we apply a macroscopic force to stretch the wire, we are slightly pulling these atomic bonds apart.

For small deformations, this stretching is perfectly elastic and obeys Hooke's Law.

To quantify this, we use **Young's Modulus ($Y$)**, which is defined as the ratio of tensile stress to tensile strain:

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

Where:
- $F$ is the restoring force exerted by the wire.
- $A$ is the cross-sectional area of the wire.
- $x$ is the elongation (displacement from equilibrium).
- $L$ is the original unstretched length of the wire.

By rearranging this equation, we can express the restoring force ($F$) as:

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

Notice the striking similarity between this equation and Hooke's Law for a spring ($F = kx$).

This allows us to define an **equivalent spring constant ($k$)** for the wire:

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

This is a profound realization: any elastic wire can be modeled as a spring with a stiffness that depends directly on its material property ($Y$) and its geometry ($A$ and $L$).

---

### The Dynamics of Simple Harmonic Motion

Once we have established the equivalent spring constant, the dynamics of the system become straightforward.

The angular frequency ($\omega$) of a mass $m$ executing SHM on a spring of constant $k$ is given by the classic formula:

$$\omega = \sqrt{\frac{k}{m}}$$

Substituting our equivalent spring constant $k = \frac{YA}{L}$ into this formula, we get:

$$\omega = \sqrt{\frac{YA}{Lm}}$$

This is our master equation, beautifully linking the dynamic frequency of oscillation with the elastic modulus of the wire.

---

### Step-by-Step Calculation

Let's substitute the values given in the problem into our master equation:
- Mass, $m = 0.1 	ext{ kg}$
- Length, $L = 1 	ext{ m}$
- Area, $A = 4.9 	imes 10^{-7} 	ext{ m}^2$
- Angular frequency, $\omega = 140 	ext{ rad s}^{-1}$
- Young's Modulus, $Y = n 	imes 10^9 	ext{ Nm}^{-2}$

Squaring both sides of the master equation to remove the square root:

$$\omega^2 = \frac{YA}{Lm}$$

Now, substitute the numerical values:

$$(140)^2 = \frac{(n 	imes 10^9)(4.9 	imes 10^{-7})}{(1)(0.1)}$$

Let's simplify the left side:

$$19600 = \frac{n 	imes 4.9 	imes 10^2}{0.1}$$

$$19600 = \frac{490n}{0.1}$$

$$19600 = 4900n$$

Solving for $n$:

$$n = \frac{19600}{4900} = 4$$

Thus, the value of $n$ is **$4$**, meaning the Young's Modulus of the wire is $4 	imes 10^9 	ext{ Nm}^{-2}$.

Enter Numerical Value:

Visualized Solution

Understanding the Physical Setup

Connecting Elasticity to Spring Constant

The Equivalent Spring Constant $k$

Angular Frequency of a Spring-Mass System

Substituting the Given Values

Simplifying the Algebraic Expression

Calculating the Value of $n$

Conclusion

The Sigma Insight: Force and Energy Method in SHM

Introduction to Elastic Oscillations

The Physics of Elasticity

The Dynamics of Simple Harmonic Motion

Step-by-Step Calculation

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