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Chapter 4: Problem 8

A string has both ends fixed. What is the ratio of the frequencies of the first to the second harmonic?

Short Answer

Expert verified
The ratio of the frequencies of the first to the second harmonic is \( \frac{1}{2} \).

Step by step solution

01

Understanding Harmonics

Harmonics are the natural frequencies at which a string fixed at both ends vibrates. The first harmonic is also called the fundamental frequency.
02

Formula for Frequencies of Harmonics

The frequency of the nth harmonic for a string fixed at both ends is given by: \[ f_n = n \cdot f_1 \] where \( f_n \) is the frequency of the nth harmonic and \( f_1 \) is the fundamental frequency.
03

Calculate the Frequencies

For the first harmonic (fundamental frequency): \[ f_1 = f_1 \] For the second harmonic: \[ f_2 = 2 \cdot f_1 \]
04

Ratio of the Frequencies

To find the ratio of the first harmonic to the second harmonic: \[ \frac{f_1}{f_2} = \frac{f_1}{2 \cdot f_1} = \frac{1}{2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Harmonic
When we talk about the first harmonic, we're referring to the simplest form of vibration a string can have when it's fixed at both ends. This is often called the fundamental frequency. Imagine a guitar string. When you pluck it, it vibrates in a way that creates a single loop, stretching from one end to the other. This is the first harmonic or fundamental frequency. The formula for this fundamental frequency is usually represented as: The key takeaway here is that the first harmonic is the natural, primary vibration of the string.
Second Harmonic
The second harmonic is the next step up in terms of vibrational complexity. When a string vibrates at its second harmonic, it's divided into two equal sections. So, instead of one complete loop, you get two halves. Each of these halves vibrates at the same frequency, but combined, they produce a frequency that's twice the fundamental frequency. The formula for the second harmonic would be: Plainly put, if the fundamental frequency (first harmonic) is represented by f1, then the second harmonic will be 2 * f1. Now you can visualize that double the loops means double the frequency.
Fundamental Frequency
The fundamental frequency (often denoted as f1) is essentially the frequency at which the entire string vibrates as a single segment. This is the starting point for all other harmonics. It serves as the basis from which higher-order harmonics are calculated. The formula for this fundamental frequency can be expressed as: It's important to understand that this is the most basic and lowest frequency at which the string can naturally vibrate. All other harmonics are just multiples of this fundamental frequency. This is why we call it the 'fundamental' frequency鈥攊t lays the foundation for all the higher harmonics.
Frequency Ratio
Understanding the frequency ratio between harmonics is crucial for grasping how complex sound waves are formed. In our specific problem, the frequency ratio between the first harmonic (f1) and the second harmonic (2 * f1) is calculated as follows: This ratio tells us how much higher the second harmonic's frequency is compared to the first harmonic. In this case, the ratio of 1 to 2 means the second harmonic's frequency is twice that of the first harmonic. This simple but pivotal concept helps explain the relationship between different harmonic frequencies.

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Most popular questions from this chapter

Unpolarized light of intensity \(I_{0}\) is incident on a polarizer. A number of other polarizers will be placed in line with the first so that the final transmitted intensity is \(\frac{I_{0}}{100} .\) If each polarizer has its transmission axis rotated by \(10^{\circ}\) with respect to the previous one, how many additional polarizers are required?

The vertical displacement of a point on the string a distance \(x\) from the left end is given by \(y=6.0 \cos (1040 \pi t) \sin (\pi x)\) where \(y\) is in millimetres, \(x\) is in metres and \(t\) is in seconds. Use this expression to: (a) deduce that all points on the string execute SHM with a common frequency and common phase, and determine the common frequency; (b) deduce that different points on the string have different amplitudes; (c) determine the maximum amplitude of oscillation; (d) calculate the length \(L\) of the string: (e) calculate the amplitude of oscillation of the point on the string where \(x=\frac{3}{4} L\).

(a) Write down an equation for the displacement of a particle undergoing SHM with an amplitude equal to \(8.0 \mathrm{cm}\) and a frequency of \(14 \mathrm{Hz}\), assuming that at \(t=0\) the displacement is \(8.0 \mathrm{cm}\) and the particle is at rest. (b) Find the displacement, velocity and acceleration of this particle at a time of \(0.025 \mathrm{s}\).

A particle undergoes SHM with angular frequency \(\omega .\) The initial displacement is \(x_{0}\) and the initial velocity is \(v_{0} .\) Deduce that an expression for the amplitude of this motion is $$A=\sqrt{x_{0}^{2}+\frac{v_{0}^{2}}{\omega^{2}}}$$.

Explain why only transverse waves can be polarized.
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