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String A has twice the tension but half the linear mass density as string \(B\), and both strings have the same length. (a) The frequency of the first harmonic on string A is (1) four times, (2) twice, (3) half, (4) \(1 / 4\) times that of string B. Explain. (b) If the lengths of the strings are \(2.5 \mathrm{~m}\) and the wave speed on string \(\mathrm{A}\) is \(500 \mathrm{~m} / \mathrm{s},\) what are the frequencies of the first harmonic on both strings?

Short Answer

Expert verified
(a) Frequency on string A is twice that of string B. (b) Frequencies: A = 100 Hz, B = 50 Hz.

Step by step solution

01

Understanding Tension and Linear Mass Density

String A has twice the tension (A) and half the linear mass density (A) as string B. Thus, if \( T_B \) is the tension in string B and \( \mu_B \) is its linear mass density, then for string A, \( T_A = 2T_B \) and \( \mu_A = \frac{1}{2}\mu_B \).
02

Formula for Wave Speed

The wave speed \( v \) on a string is given by the formula:\[v = \sqrt{\frac{T}{\mu}}\]Apply this to string A:\[v_A = \sqrt{\frac{T_A}{\mu_A}} = \sqrt{\frac{2T_B}{(1/2)\mu_B}} = \sqrt{4\frac{T_B}{\mu_B}} = 2v_B\]Thus, wave speed on string A is twice that on string B.
03

Frequency of the First Harmonic

The frequency \( f \) of the first harmonic is determined by the formula:\[f = \frac{v}{2L}\]Since the length \( L \) of both strings is the same, the frequency on string A is:\[f_A = \frac{2v_B}{2L} = \frac{v_B}{L} = 2f_B\]Thus, option (2) is correct: The frequency of the first harmonic on string A is twice that of string B.
04

Calculate Frequency for String A

We know the wave speed \( v_A = 500 \, \mathrm{m/s} \) for string A and its length \( L_A = 2.5 \, \mathrm{m} \). Using the formula for the frequency of the first harmonic:\[f_A = \frac{v_A}{2L_A} = \frac{500}{2 \times 2.5} = \frac{500}{5} = 100 \, \mathrm{Hz}\]Therefore, the frequency of the first harmonic for string A is \( 100 \, \mathrm{Hz} \).
05

Calculate Frequency for String B Using Ratio

Since we know \( f_A = 2f_B \) from Step 3, and \( f_A = 100 \, \mathrm{Hz} \) from Step 4, we set up the equation:\[100 = 2f_B\]Solving for \( f_B \):\[f_B = \frac{100}{2} = 50 \, \mathrm{Hz}\]Thus, the frequency of the first harmonic for string B is \( 50 \, \mathrm{Hz} \).

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Key Concepts

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

Wave Speed
Wave speed is an essential part of understanding how waves behave on a string. It's determined by the medium through which the wave is traveling, and in a string, the main factors that define wave speed are the tension in the string and its linear mass density. The formula for calculating wave speed on a string is:\[v = \sqrt{\frac{T}{\mu}}\]In this equation, \( v \) stands for wave speed, \( T \) represents the tension in the string, and \( \mu \) is the linear mass density. This formula shows that wave speed increases with higher tension and decreases with higher mass density. When a string is tightened, the wave can travel faster along it. Similarly, a lighter string (lower mass density) will allow the wave to travel quicker compared to a heavier one. Understanding these relationships will help you predict how changing different properties of a string will affect wave speed.
Harmonics in Strings
Harmonics are the different frequencies at which a string can naturally vibrate, primarily determined by the string's length, mass, and tension. The first harmonic, also known as the fundamental frequency, is the lowest frequency at which the string vibrates. When we talk about harmonics, we often refer to a natural sequence of frequencies, each a whole-number multiple of the fundamental frequency.The frequency of the first harmonic is calculated using the formula:\[f = \frac{v}{2L}\]Here, \( f \) is the frequency, \( v \) is the wave speed, and \( L \) is the length of the string. Each additional harmonic can be found as a multiple of this frequency, meaning the second harmonic is twice the first, the third is three times, and so on. Understanding harmonics is crucial for musicians and engineers alike, helping them create instruments with specific resonance characteristics.
Tension in Strings
Tension in a string is a force exerted when the string is stretched tight between two points. This force influences how a wave travels through the string. Greater tension means the wave can travel faster, which in turn affects the pitch of the sound produced. For stringed instruments, adjusting the tension changes the note that's played.To calculate the effect of tension on wave speed, use the formula:\[v = \sqrt{\frac{T}{\mu}}\]When the tension is doubled, as in the case with string A, the wave speed increases. This also shows us that tension directly links to the frequency of waves generated on the string. A higher tension typically results in a higher frequency, leading to a higher pitch. Therefore, tension is a key factor in fine-tuning the output of sound waves in musical contexts.
Mass Density in Physics
Mass density, in the context of wave mechanics, usually refers to linear mass density, which is the mass per unit length of a string. This property significantly influences wave speed. A heavier string, with greater mass density, tends to have slower wave propagation.The formula for wave speed, considering mass density, is:\[v = \sqrt{\frac{T}{\mu}}\]In this case, \( \mu \) represents the linear mass density. If a string's mass density is reduced, waves can move more quickly across it. In the scenario with strings A and B, string A is half as dense as string B, allowing waves to travel faster. Mass density is a crucial factor in designing objects like musical instruments since it helps determine the natural frequencies and therefore what notes and tones are achievable. Understanding how mass density impacts wave behavior is fundamental in both academic physics and practical applications like engineering.

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Most popular questions from this chapter

\(\bullet\bullet\) (a) At what position is the magnitude of the force on a mass in a mass-spring system minimum: \((1) x=0\), (2) \(x=-A,\) or \((3) x=+A ?\) Why? \((b)\) If \(m=0.500 \mathrm{~kg}\) \(k=150 \mathrm{~N} / \mathrm{m},\) and \(A=0.150 \mathrm{~m},\) what are the magnitude of the force on the mass and the acceleration of the mass at \(x=0,0.050 \mathrm{~m},\) and \(0.150 \mathrm{~m} ?\)

If the frequency of the third harmonic of a vibrating string is \(600 \mathrm{~Hz}\), what is the frequency of the first harmonic?

\(\bullet\bullet\) A \(\mathrm{} 0.25-\mathrm{kg}\) object is suspended on a light spring of spring constant \(49 \mathrm{~N} / \mathrm{m}\). The spring is then compressed to a position \(15 \mathrm{~cm}\) above the stretched equilibrium position. How much more energy does the system have at the compressed position than at the stretched equilibrium position?

A 2.0-kg mass resting on a horizontal frictionless surface is connected to a fixed spring. The mass is displaced \(16 \mathrm{~cm}\) from its equilibrium position and released. At \(t=0.50 \mathrm{~s},\) the mass is \(8.0 \mathrm{~cm}\) from its equilibrium position (and has not passed through it yet). (a) What is the period of oscillation of the mass? (b) What are the speed of the mass and the force on the mass at \(t=0.50 \mathrm{~s} ?\)

A simple pendulum is set into small-angle motion, making a maximum angle with the vertical of \(5^{\circ} .\) Its period is 2.21 s. (a) Determine its length. (b) Determine its maximum speed. (c) What is the acceleration of the pendulum bob when it is at the lowest position?

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