A metallic rod of length 1 m is rigidly clamped at its mid-point. Longitudinal stationary waves are

Q: Why is the center of the rod a node? Shouldn't it be an antinode since we strike the rod?

The clamp physically restricts the movement of the rod's particles at the center. Because the clamp is rigid, the displacement of the particles at $x = 0$ is forced to be zero at all times, which is the definition of a displacement node.

Q: How did we get 5 quarter-wavelengths in the half-length of the rod?

From the center (Node) to the end (Antinode), we are told there are 2 nodes in between. The sequence of nodes and antinodes is: Node (center) -> Antinode -> Node -> Antinode -> Node -> Antinode (end). The distance between any consecutive node and antinode is $\lambda/4$. Counting these intervals gives exactly 5 intervals of $\lambda/4$.

Q: Why is the wave speed formula $v = \sqrt{Y/\rho}$ used instead of $v = \sqrt{T/\mu}$?

The formula $v = \sqrt{T/\mu}$ is for transverse waves on a stretched string under tension $T$. For longitudinal sound waves propagating through a solid bulk medium or rod, the speed depends on the elastic modulus (Young's Modulus $Y$) and the volume density $\rho$.

Q: Why does the standing wave equation use a sine function for space instead of cosine?

Since we chose our origin $x = 0$ at the center clamp (which is a node), the displacement must be zero at $x = 0$. Since $\sin(0) = 0$ and $\cos(0) = 1$, only the sine function satisfies this boundary condition.

Q: What is the physical meaning of the constituent waves?

The constituent waves are the individual travelling waves that propagate back and forth along the rod. They continuously reflect off the free ends and superpose. The standing wave is the net result of this continuous superposition.

Question

A metallic rod of length $1	ext{ m}$ is rigidly clamped at its mid-point. Longitudinal stationary waves are set-up in the rod in such a way that there are two nodes on either side of the mid-point. The amplitude of an antinode is $2 	imes 10^{-6}	ext{ m}$. Write the equation of motion at a point $2	ext{ cm}$ from the mid-point and those of the constituent waves in the rod. (Young's modulus of the material of the rod $= 2 	imes 10^{11}	ext{ Nm}^{-2}$; density $= 8000	ext{ kg m}^{-3}$)

Sigma Percentile · Accepted Answer

### Introduction: The Symphony of Standing Waves

Imagine holding a solid metallic rod and striking it.

The clear, metallic hum you hear is not just random noise; it is a beautifully ordered physical phenomenon.

Inside the rod, sound waves are bouncing back and forth between the ends, interfering with each other to create **standing waves**.

In this problem, we explore the physics of a $1	ext{ m}$ long metallic rod clamped rigidly at its center.

By analyzing the boundary conditions and the material properties, we will unlock the exact mathematical equations that describe this microscopic dance of atoms.

### The Physics of the Clamp: Boundary Conditions

When a rod is clamped rigidly at a point, the clamp exerts massive restoring forces that prevent any displacement of the particles at that location.

Therefore, the clamped point must always be a **displacement node** ($N$).

Conversely, the free ends of the rod are completely unrestricted and can vibrate with maximum amplitude, forming **displacement antinodes** ($A$).

Our rod is clamped at its mid-point, which we will define as our origin, $x = 0$.

We are told that there are two nodes on either side of this mid-point.

This means that as we move from the center to either end, the wave pattern must pass through two nodes before reaching the free end.

Let's trace this spatial pattern from the center ($x = 0$) to the right end ($x = 0.5	ext{ m}$):

$$N ightarrow A ightarrow N ightarrow A ightarrow N ightarrow A$$

Since the distance between a node and the adjacent antinode is always $\frac{\lambda}{4}$, we can count the total number of quarter-wavelengths in this half-length:

$$	ext{Total distance} = 5 	imes \frac{\lambda}{4}$$

Setting this equal to the physical half-length of $0.5	ext{ m}$:

$$\frac{5\lambda}{4} = 0.5	ext{ m} \implies \lambda = 0.4	ext{ m}$$

This is a beautiful geometric realization! The wavelength of our standing wave is exactly $0.4	ext{ m}$.

### Calculating the Speed of Sound in Solids

To find the temporal frequency of these oscillations, we must first determine how fast longitudinal waves travel through our metallic rod.

The speed of sound in a solid medium is determined by a battle between its elasticity (which pulls particles back to equilibrium) and its inertia (which resists acceleration).

This is mathematically expressed by the classic formula:

$$v = \sqrt{\frac{Y}{ho}}$$

where $Y$ is the Young's modulus of the material and $ho$ is its mass density.

Substituting our given values:

$$v = \sqrt{\frac{2 	imes 10^{11}	ext{ Nm}^{-2}}{8000	ext{ kg m}^{-3}}} = \sqrt{25 	imes 10^6} = 5000	ext{ m/s}$$

This incredibly high speed—five kilometers per second—is characteristic of sound propagation in stiff metals like steel.

### Formulating the Standing Wave Equation

Now we have all the ingredients to construct our wave parameters:

1. **Wave number ($k$):** This represents the spatial frequency of the wave.

$$k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.4} = 5\pi	ext{ rad/m}$$

2. **Angular frequency ($\omega$):** This represents the temporal frequency of the oscillations.

$$\omega = 2\pi f = 2\pi \left(\frac{v}{\lambda}ight) = 2\pi \left(\frac{5000}{0.4}ight) = 25000\pi	ext{ rad/s}$$

Since the origin $x = 0$ is a node, the displacement must be zero at $x = 0$ for all times.

This requires our spatial wave function to be a sine function, $\sin(kx)$.

Thus, the general equation of motion for our standing wave is:

$$y(x, t) = 2A \sin(kx) \sin(\omega t)$$

where $2A$ is the maximum amplitude at any antinode, given as $2 	imes 10^{-6}	ext{ m}$.

Substituting our parameters, we get the complete standing wave equation:

$$y(x, t) = (2 	imes 10^{-6}) \sin(5\pi x) \sin(25000\pi t)$$

### Finding the Motion at $2	ext{ cm}$ from the Mid-point

To find the equation of motion at a specific point located $2	ext{ cm}$ from the clamp, we simply substitute $x = 2	ext{ cm} = 0.02	ext{ m}$ into our standing wave equation:

$$y(0.02, t) = (2 	imes 10^{-6}) \sin(5\pi 	imes 0.02) \sin(25000\pi t)$$

Simplifying the spatial term:

$$5\pi 	imes 0.02 = 0.1\pi$$

$$y(0.02, t) = 2 	imes 10^{-6} \sin(0.1\pi) \sin(25000\pi t)$$

This elegant equation tells us that the particles at this location oscillate harmonically with an amplitude of $2 	imes 10^{-6} \sin(0.1\pi)	ext{ m}$ and an angular frequency of $25000\pi	ext{ rad/s}$.

### Decomposing into Constituent Waves

A standing wave is actually an optical and physical illusion created by the superposition of two identical travelling waves moving in opposite directions.

Using the trigonometric identity:

$$2 \sin(A) \sin(B) = \cos(A - B) - \cos(A + B)$$

We can write our standing wave as the sum of two travelling waves:

$$y_1(x, t) = 10^{-6} \sin(25000\pi t - 5\pi x)$$

$$y_2(x, t) = 10^{-6} \sin(25000\pi t + 5\pi x)$$

Here, $y_1$ represents a wave propagating in the positive $x$-direction, and $y_2$ represents a wave propagating in the negative $x$-direction.

When they meet, they interfere constructively and destructively to form the stationary pattern we observed.

Visualized Solution

Visualizing the Physical Setup

Calculating the Speed of Longitudinal Waves

Determining the Wavelength $λ$

Calculating Wave Number $k$ and Angular Frequency $ω$

Formulating the Standing Wave Equation

Equation of Motion at $2 cm$ from Mid-point

Decomposing into Constituent Travelling Waves

Final Summary of Equations

The Way Forward

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction

The Physics of the Clamp

Calculating the Speed of Sound in Solids

Formulating the Standing Wave Equation

Finding the Motion at $2 cm$ from the Mid-point

Decomposing into Constituent Waves

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Visualized Solution

Visualizing the Physical Setup

Calculating the Speed of Longitudinal Waves

Determining the Wavelength .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

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Equation of Motion at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2 cm from Mid-point

Decomposing into Constituent Travelling Waves

Final Summary of Equations

The Way Forward

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction

The Physics of the Clamp

Calculating the Speed of Sound in Solids

Formulating the Standing Wave Equation

Finding the Motion at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2 cm from the Mid-point

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