A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is

Q: Why do we need to convert grams to kilograms and centimeters to meters?

Standard formulas like $v = \sqrt{T/\mu}$ require SI units (Newtons, kilograms, meters) to yield the correct wave speed in meters per second. Mixing units like grams and centimeters will lead to incorrect orders of magnitude.

Q: What is the physical meaning of a node?

A node is a point along a standing wave where the wave has minimum (or zero) amplitude. At these points, the destructive interference between the incident and reflected waves is complete, meaning the string particles do not vibrate at all.

Q: Why is the distance between successive nodes $\lambda/2$ and not $\lambda$?

A full wavelength $\lambda$ consists of one complete upward loop and one complete downward loop (a total of two loops). Since each loop is bounded by nodes, there are nodes at the start, middle, and end of a full wavelength. Thus, the distance between two consecutive nodes is exactly half of a wavelength.

Q: Does the frequency of the standing wave change if we change the tension?

The frequency of the standing wave is determined by the external vibrator, which is $100\text{ Hz}$ in this case. If you change the tension, the wave speed and wavelength will change, causing the string to vibrate in a different harmonic mode, but the frequency remains fixed by the source.

Q: What would happen if the string was not fixed at both ends?

If one end were free, that end would form a displacement antinode (maximum vibration) instead of a node. This would change the boundary conditions, and the allowed wavelengths would follow a different harmonic series.

Question

A $20	ext{ cm}$ long string, having a mass of $1.0	ext{ g}$, is fixed at both the ends. The tension in the string is $0.5	ext{ N}$. The string is set into vibration using an external vibrator of frequency $100	ext{ Hz}$. Find the separation (in cm) between the successive nodes on the string.

Sigma Percentile · Accepted Answer

### Introduction to Standing Waves

Imagine a guitar string clamped tightly at both ends.

When you pluck it, waves travel outward, hit the rigid boundaries, and reflect back.

These incoming and reflected waves overlap, creating a beautiful pattern of constructive and destructive interference known as a **standing wave**.

In this problem, we explore the physics of a $20	ext{ cm}$ long string set into vibration by an external source.

Our goal is to find the physical distance between successive points of zero motion, known as **nodes**.

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### Step-by-Step Mathematical Walkthrough

To find the distance between nodes, we must first determine the wavelength of the wave traveling along the string.

Let's break this down into simple, logical steps.

#### 1. Calculating the Linear Mass Density ($\mu$)

The speed of a wave on a string depends heavily on how heavy the string is per unit length.

This is called the **linear mass density** ($\mu$):

$$\mu = \frac{m}{L}$$

Given:
* Mass of the string, $m = 1.0	ext{ g} = 1.0 	imes 10^{-3}	ext{ kg}$
* Length of the string, $L = 20	ext{ cm} = 0.2	ext{ m}$

Substituting these values:

$$\mu = \frac{1.0 	imes 10^{-3}	ext{ kg}}{0.2	ext{ m}} = 5.0 	imes 10^{-3}	ext{ kg/m}$$

#### 2. Finding the Wave Speed ($v$)

The speed of a transverse wave on a stretched string is determined by the tension ($T$) and the linear mass density ($\mu$):

$$v = \sqrt{\frac{T}{\mu}}$$

Given the tension $T = 0.5	ext{ N}$:

$$v = \sqrt{\frac{0.5}{5.0 	imes 10^{-3}}} = \sqrt{100} = 10	ext{ m/s}$$

This means any disturbance on this string travels at a speed of exactly $10	ext{ m/s}$.

#### 3. Determining the Wavelength ($\lambda$)

Using the fundamental wave relation connecting speed ($v$), frequency ($f$), and wavelength ($\lambda$):

$$\lambda = \frac{v}{f}$$

Given the vibrator frequency $f = 100	ext{ Hz}$:

$$\lambda = \frac{10	ext{ m/s}}{100	ext{ Hz}} = 0.1	ext{ m} = 10	ext{ cm}$$

#### 4. Finding the Node Separation ($d$)

In any standing wave, a single complete loop represents half of a wavelength.

Since nodes exist at the boundaries of each loop, the distance $d$ between two successive nodes is exactly half of the wavelength:

$$d = \frac{\lambda}{2}$$

Substituting our calculated wavelength:

$$d = \frac{10	ext{ cm}}{2} = 5	ext{ cm}$$

Thus, the separation between successive nodes is **$5	ext{ cm}$**.

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### Deep Physical Insights

Let's look at the bigger picture.

Since the total length of the string is $20	ext{ cm}$ and the distance between successive nodes is $5	ext{ cm}$, the string accommodates exactly:

$$\frac{20	ext{ cm}}{5	ext{ cm}} = 4	ext{ loops}$$

This tells us that the string is vibrating in its **4th harmonic** (or 3rd overtone).

This perfectly satisfies the boundary conditions where both clamped ends ($x = 0$ and $x = 20	ext{ cm}$) are forced to be nodes.

Enter Numerical Value:

Visualized Solution

Visualizing the Standing Wave

Linear Mass Density $μ$

Substituting Mass and Length

Calculating $μ$

Wave Speed on a Stretched String

Substituting Tension and Density

Calculating Wave Speed

Wavelength of the Wave

Substituting and Solving for $λ$

Separation Between Successive Nodes

Calculating the Node Separation

Analyzing the Harmonic Mode

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction to Standing Waves

Step-by-Step Mathematical Walkthrough

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Deep Physical Insights

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