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If \(n_{1}, n_{2}\) and \(n_{3}\) are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency \(n\) of the string is given by (A) \(n=n_{1}+n_{2}+n_{3}\) (B) \(\frac{1}{n}=\frac{1}{n_{1}}+\frac{1}{n_{2}}+\frac{1}{n_{3}}\) (C) \(\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n_{1}}}+\frac{1}{\sqrt{n_{2}}}+\frac{1}{\sqrt{n_{3}}}\) (D) \(\sqrt{n}=\sqrt{n_{1}}+\sqrt{n_{2}}+\sqrt{n_{3}}\)

Short Answer

Expert verified
The original fundamental frequency \(n\) of the string is given by \(\frac{1}{n}=\frac{1}{n_{1}}+\frac{1}{n_{2}}+\frac{1}{n_{3}}\). So, the correct option is B.

Step by step solution

01

Analyzing the Options

The options provided are presumably the formulas that express relationships between the fundamental frequency of a string and the frequencies of its sections. The correct option should hold true in all cases.
02

The Nature of Fundamental Frequency

It's necessary to understand that the fundamental frequency of a string depends on its length, tension, and linear density. Namely, when a string's length is divided into segments, each segment has its own fundamental frequency. Let's consider each option.
03

Exploring Option A

Option A suggests that the original frequency is the sum of the frequencies of three segments. It's important to note, however, that the fundamental frequency is inversely proportional to the length. Therefore, adding the frequencies of different segments would not yield the original frequency of the whole string. Hence, option A is incorrect.
04

Exploring Option B

Option B suggests that the inverses of the frequencies sum up to give the inverse of the original frequency. This is in line with our understanding that the fundamental frequency is inversely proportional to the length of the string assuming the tension and the linear density are kept constant. Hence, if the lengths of the three segments sum up to the length of the whole string, then the sum of the inverses of their fundamental frequencies should be equal to the inverse of the fundamental frequency of the whole string. Therefore, option B is correct.
05

Eliminating the Remaining Options

Since we have found our correct option, options C and D are automatically incorrect as they do not maintain the correct relationship between the lengths and the fundamental frequencies.

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Key Concepts

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

String Vibration
When a string vibrates, it creates waves along its length, which are responsible for producing sound. These vibrations are fundamental to musical instruments like violins and guitars. The nature of string vibration depends on several factors, including:
  • The length of the string
  • The tension applied to the string
  • The linear density (mass per unit length) of the string
Understanding string vibration is critical in physics because it helps explain phenomena like the production of musical notes. As these variables change, the string vibrates differently, leading to changes in pitch. A shorter string, for instance, produces higher pitches, while a longer string produces lower pitches.
Mathematically, the vibration of a string is expressed using the wave equation, where factors like length and tension influence the fundamental frequency, which is pivotal in generating a specific pitch for sound.
Fundamental Frequency
The concept of the fundamental frequency is central to understanding how strings produce sound. The fundamental frequency, often denoted as the first harmonic, is the lowest frequency at which a string vibrates naturally. It is influenced by:
  • The tension of the string: More tension results in a higher fundamental frequency.
  • The length of the string: Decreasing the length increases the frequency.
  • The mass of the string: A heavier string vibrates at a lower frequency.
This frequency is critical because it determines the pitch of the note produced by the string. For example, guitarists adjust the tension or change the length of strings to tune their instruments to the desired pitch.
The fundamental frequency is mathematically related to these variables by the formula: \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( L \) is the length, \( T \) is the tension, and \( \mu \) is the linear density of the string.
Physics Problem Solving
Solving physics problems like the one presented in the original exercise involves analyzing different variables and understanding their relationships. In the case of string vibration and fundamental frequency, the key is to grasp how changes in string characteristics alter the total frequency of segment divisions.Effective problem-solving in physics follows a systematic process:
  • Identify the given data and the unknowns you need to find.
  • Understand the relevant formulas that connect these variables. For string vibration, this includes the relationship between fundamental frequency, length, tension, and density.
  • Eliminate options and solve step-by-step, using logical reasoning. For example, the exercise illustrates that the correct relationship for segments of a string is expressed by option (B): \( \frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3} \).
By methodically applying these steps, you can demystify complex physics problems and gain a deeper understanding of the laws governing wave phenomena.

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Most popular questions from this chapter

A stationary source of sound is emitting waves of frequency \(30 \mathrm{~Hz}\) towards a stationary wall. There is an observer standing between the source and the wall. If the wind blows from the source to the wall with a speed \(30 \mathrm{~m} / \mathrm{s}\) then the number of beats heard by the observer is (velocity of sound with respect to wind is \(330 \mathrm{~m} / \mathrm{s})\) (A) 10 (B) 3 (C) 6 (D) Zero

A tuning fork of frequency \(340 \mathrm{~Hz}\) is vibrated just above a cylindrical tube of length \(120 \mathrm{~cm}\). Water is slowly poured in the tube. If the speed of sound is 340 \(\mathrm{m} / \mathrm{s}\), then the minimum height of water required for resonance is (A) \(25 \mathrm{~cm}\) (B) \(45 \mathrm{~cm}\) (C) \(75 \mathrm{~cm}\) (D) \(95 \mathrm{~cm}\)

An open glass tube is immersed in mercury in such a way that a length of \(8 \mathrm{~cm}\) extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional \(46 \mathrm{~cm}\). What will be length of the air column above mercury in the tube now? (Atmospheric pressure \(=76 \mathrm{~cm}\) of \(\mathrm{Hg}\) ) (A) \(16 \mathrm{~cm}\) (B) \(22 \mathrm{~cm}\) (C) \(38 \mathrm{~cm}\) (D) \(6 \mathrm{~cm}\)

Let a disturbance \(y\) be propagated as a plane wave along the \(x\)-axis. The wave profiles at the instants \(t=t_{1}\) and \(t=t_{2}\) are represented, respectively, as \(y_{1}=f\left(x_{1}-v t_{1}\right)\) and \(y_{2}=f\left(x_{2}-v t_{2}\right)\). The wave is propagating without change of shape. (A) The velocity of the wave is \(2 v\). (B) The velocity of the wave is \(v=\frac{x_{2}-x_{1}}{t_{2}}\). (C) The particle velocity is \(v_{p}=v\). (D) None of these.

The total energy of a particle, executing simple harmonic motion is (A) independent of \(x\) (B) \(\propto x^{2}\) (C) \(\propto x\) \((\mathrm{D}) \propto x^{1 / 2}\) where \(x\) is the displacement from the mean position.

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