Animated Solution for Physics - Waves: A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to

Imagine a string stretched tightly between two fixed walls. When we pluck it or use an external vibrator, waves travel back and forth along the string. Because the ends are fixed, the waves reflect and interfere with each other, creating a beautiful pattern known as a standing wave.

In this problem, we are given a string of length $L=1\text{ m}$ and a total mass $m=5\text{ g}$. It is held under a tension $T=8.0\text{ N}$ and is forced to vibrate at a frequency $f=100\text{ Hz}$. Our goal is to find the distance between two consecutive points that do not move at all—these points are called nodes.

Before we can understand the standing wave pattern, we need to know how fast the individual waves are traveling along the string. The speed of a transverse wave on a stretched string depends on two factors: how tightly the string is pulled (tension) and how heavy it is (linear mass density).

First, let's calculate the linear mass density, denoted by $\mu$. This is simply the mass per unit length. We must ensure our units are in standard SI format, so we convert $5\text{ g}$ to $5\times 10^{-3}\text{ kg}$.

Now, we use the wave speed formula:

Substituting our values into the equation:

The waves are zipping along the string at $40\text{ m/s}$!

Now that we have the wave speed $v$ and the frequency $f$, we can easily find the wavelength $\lambda$. The fundamental wave equation connects these three quantities:

Rearranging for wavelength:

Converting this to centimeters, we get $\lambda =40\text{ cm}$.

A full wavelength consists of one complete "up" loop and one complete "down" loop. However, the question asks for the separation between successive nodes. A node is a point of zero displacement, and the distance between two adjacent nodes is exactly half of a wavelength.

We found our answer: the nodes are $20.0\text{ cm}$ apart. But let's take a step back and look at the whole string. The total length of the string is $100\text{ cm}$. If each loop (the distance between nodes) is $20\text{ cm}$ long, how many loops fit on the string?

This tells us that the string is vibrating in its 5th harmonic! The external vibrator has perfectly tuned into this specific resonant frequency, creating a stable pattern of exactly five vibrating segments.