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String \(A\) is stretched between two clamps separated by distance \(L\). String \(B\), with the same linear density and under the same tension as string \(A\), is stretched between two clamps separated by distance \(3 L\). Consider the first eight harmonics of string \(B\). For which of these eight harmonics of \(B\) (if any) does the frequency match the frequency of (a) \(A\) 's first harmonic, (b) \(A\) 's second harmonic, and (c) \(A\) 's third harmonic?

Short Answer

Expert verified
B's 3rd harmonic matches A's 1st, B's 6th matches A's 2nd, none match A's 3rd.

Step by step solution

01

Understanding Harmonics

To compare frequencies, we first need to know the frequency formula for harmonics on a string: for a string of length \(L\), the frequency of the \(n\)-th harmonic is \(f_n = \frac{nv}{2L}\), where \(v\) is the wave speed along the string determined by tension and linear density.
02

Frequency of A's Harmonics

For string \(A\), the first harmonic frequency is \(f_1^A = \frac{v}{2L}\), the second harmonic frequency is \(f_2^A = \frac{v}{L}\), and the third harmonic frequency is \(f_3^A = \frac{3v}{2L}\).
03

Frequency of B's Harmonics

String \(B\) has a length of \(3L\), so its \(n\)-th harmonic frequency is \(f_n^B = \frac{nv}{6L}\). We need to find some \(f_n^B\) that matches the frequencies of \(A\).
04

Finding B's Harmonic Matching A's First Harmonic

Set \(f_n^B = f_1^A\): \(\frac{nv}{6L} = \frac{v}{2L}\). Solving gives \(n = 3\), so the third harmonic of string \(B\) matches string \(A\)'s first harmonic.
05

Finding B's Harmonic Matching A's Second Harmonic

Set \(f_n^B = f_2^A\): \(\frac{nv}{6L} = \frac{v}{L}\). Solving gives \(n = 6\), so the sixth harmonic of string \(B\) matches string \(A\)'s second harmonic.
06

Finding B's Harmonic Matching A's Third Harmonic

Set \(f_n^B = f_3^A\): \(\frac{nv}{6L} = \frac{3v}{2L}\). Solving gives \(n = 9\), which exceeds the 8 harmonics considered, so there's no match for the third harmonic within the first 8 harmonics of \(B\).

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Key Concepts

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

Wave Speed
The wave speed on a string is a key factor influencing the frequency of harmonics. Wave speed is determined by both the tension in the string and its linear density. Mathematically, wave speed is given by the equation \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the string and \( \mu \) is the linear density.
This concept is especially important when comparing strings, such as string \( A \) and string \( B \), where both strings might have the same tension and linear density, but different lengths. Even if tension and density are constant, a change in string length can influence the harmonics through the wave speed's role in the frequency formula \( f_n = \frac{nv}{2L} \).
Always remember that wave speed is constant for a string under uniform tension and density, and it directly influences how swiftly the wave can travel along the string.
Linear Density
Linear density, expressed as \( \mu \), is a measure of a string's mass per unit length. It's a critical factor in calculating wave speed and, consequently, harmonic frequencies.
The relationship between linear density and wave speed is inverse; as linear density increases, wave speed decreases, assuming tension is constant. This is pivotal in the formula \( v = \sqrt{\frac{T}{\mu}} \).
For strings \( A \) and \( B \), having equal linear densities shows that their wave speeds depend mostly on their tension. When analyzing problems involving harmonics, consistent linear density across strings suggests that frequency variations depend on other factors, such as the length of the string or differing harmonic modes.
String Tension
String tension, denoted as \( T \), is the force applied along the string's length that helps keep it taut. It crucially affects the wave speed, influencing the harmonic frequencies we calculate.
High tension typically leads to a higher wave speed, which in turn affects the harmonic series of the string. This reliance is clear in the equation \( v = \sqrt{\frac{T}{\mu}} \), where tension directly impacts the speed.
In exercises where two strings (\( A \) and \( B \)) share the same tension, we can expect their harmonic frequencies to be influenced by other variables such as length \( L \). Ensuring correct tension helps understanding scenarios where harmonic frequencies may match despite differing physical setups of the strings.
String Harmonics
String harmonics describe the various vibrational states of a string, characterized by their frequency relationships. These harmonics depend on various conditions such as the string's length, tension, and linear density.
Harmonics are integer multiples of the fundamental frequency, meaning each harmonic corresponds to a specific vibrational pattern. For a string of length \( L \), the frequency of the \( n \)-th harmonic is \( f_n = \frac{nv}{2L} \).
In the context of string \( A \) and string \( B \), identifying harmonics involves aligning these frequencies. The exercise demonstrates how the third harmonic of string \( B \) aligns with the first harmonic of string \( A \) due to equal tension and density, but varying lengths.
  • The first harmonic implies the simplest vibration, occurring when \( n = 1 \).
  • As \( n \) increases, the complexity of the vibration pattern grows, resulting in higher-frequency harmonics.
  • Understanding harmonics is essential for predicting and matching frequencies, especially when analyzing or comparing strings of different lengths and setups.

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Most popular questions from this chapter

A \(100 \mathrm{~g}\) wire is held under a tension of \(220 \mathrm{~N}\) with one end at \(x=0\) and the other at \(x=10.0 \mathrm{~m}\). At time \(t=0\), pulse 1 is sent along the wire from the end at \(x=10.0 \mathrm{~m}\). At time \(t=30.0 \mathrm{~ms}\), pulse 2 is sent along the wire from the end at \(x=0 .\) At what position \(x\) do the pulses begin to meet?

A generator at one end of a very long string creates a wave given by $$ y=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}\left[\left(2.00 \mathrm{~m}^{-1}\right) x+\left(6.00 \mathrm{~s}^{-1}\right) t\right] $$ and a generator at the other end creates the wave $$ y=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}\left[\left(2.00 \mathrm{~m}^{-1}\right) x-\left(6.00 \mathrm{~s}^{-1}\right) t\right] $$ Calculate the (a) frequency, (b) wavelength, and (c) speed of each wave. For \(x \geq 0\), what is the location of the node having the (d) smallest, (e) second smallest, and (f) third smallest value of \(x\) ? For \(x \geq 0\), what is the location of the antinode having the \((\mathrm{g})\) smallest, (h) second smallest, and (i) third smallest value of \(x\) ?

A string along which waves can travel is \(2.70 \mathrm{~m}\) long and has a mass of \(130 \mathrm{~g}\). The tension in the string is \(36.0 \mathrm{~N}\). What must be the frequency of traveling waves of amplitude \(7.70 \mathrm{~mm}\) for the average power to be \(170 \mathrm{~W}\) ?

Use the wave equation to find the speed of a wave given by $$ y(x, t)=(3.00 \mathrm{~mm}) \sin \left[\left(3.00 \mathrm{~m}^{-1}\right) x-\left(8.00 \mathrm{~s}^{-1}\right) t\right] . $$

A sinusoidal wave of frequency \(500 \mathrm{~Hz}\) has a speed of \(320 \mathrm{~m} / \mathrm{s}\). (a) How far apart are two points that differ in phase by \(\pi / 3 \mathrm{rad}\) ? (b) What is the phase difference between two displacements at a certain point at times \(1.00 \mathrm{~ms}\) apart?

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