/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 The mass of a string is \(5.0 \t... [FREE SOLUTION] | 91Ó°ÊÓ

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The mass of a string is \(5.0 \times 10^{-3} \mathrm{kg}\), and it is stretched so that the tension in it is \(180 \mathrm{N}\). A transverse wave traveling on this string has a frequency of \(260 \mathrm{Hz}\) and a wavelength of \(0.60 \mathrm{m} .\) What is the length of the string?

Short Answer

Expert verified
The length of the string is about 0.676 m.

Step by step solution

01

Identify the Variables

We are given the mass of the string as \( m = 5.0 \times 10^{-3} \) kg, the tension \( T = 180 \) N, the frequency \( f = 260 \) Hz, and the wavelength \( \lambda = 0.60 \) m. We need to find the length \( L \) of the string.
02

Calculate Wave Speed

Use the wave equation \( v = f \cdot \lambda \) to find the wave speed \( v \). Substitute \( f = 260 \) Hz and \( \lambda = 0.60 \) m: \[ v = 260 \times 0.60 = 156.0 \text{ m/s} \]
03

Determine String Speed from Tension

The speed \( v \) of a wave on a string can also be calculated using the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( \mu \) is the linear mass density of the string.
04

Calculate Linear Mass Density

The linear mass density \( \mu \) is given by \( \mu = \frac{m}{L} \), but since we know the wave speed, we use \( v = \sqrt{\frac{T}{\mu}} \). Rearrange to find \( \mu \): \[ \mu = \frac{T}{v^2} = \frac{180}{156.0^2} \approx 0.007386 \text{ kg/m} \]
05

Find String Length

Using the linear density \( \mu = \frac{m}{L} \), solve for \( L \): \[ L = \frac{m}{\mu} = \frac{5.0 \times 10^{-3}}{0.007386} \approx 0.676 \text{ m} \]

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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 a fundamental concept when dealing with waves on a string, or any medium for that matter. In this context, it tells us how fast a wave travels along a string. To calculate the wave speed, we use the formula that connects frequency and wavelength:
  • Frequency (\( f \)): The number of wave cycles that pass a point per second, measured in Hertz (\( \text{Hz} \)).
  • Wavelength (\( \lambda \)): The distance between two consecutive points at the same phase in meters (m).
The wave speed (\( v \)) is given by the equation:\[ v = f \times \lambda \]Applying this formula to find the wave speed of a wave with a frequency of 260 Hz and a wavelength of 0.60 m, we calculate:\[ v = 260 \times 0.60 = 156.0 \, \text{m/s} \]This speed indicates how quickly the wave front moves across the medium, crucial for understanding wave behavior on strings, like in musical instruments.
Linear Mass Density
Linear mass density is a key property of a string that affects how waves travel through it. It describes the mass of the string per unit length and is crucial in calculating the wave speed when considering the tension applied to the string.We represent linear mass density as (\( \mu \)), where its relation to mass (\( m \)) and length (\( L \)) of the string is given by:\[ \mu = \frac{m}{L} \]However, in some cases, like when given the wave speed and the tension, it is more practical to determine (\( \mu \)) using the modified equation derived from the wave speed formula:\[ \mu = \frac{T}{v^2} \]Where (\( T \)) is tension in the string. Applying this to our scenario with tension of 180 N and wave speed of 156.0 m/s, we have:\[ \mu = \frac{180}{156.0^2} \approx 0.007386 \, \text{kg/m} \]Understanding linear mass density is also essential when designing strings for specific frequencies or tensions, impacting how smoothly and efficiently waves propagate.
Tension in String
Tension in a string is the force applied along the string's length. It plays a significant role in the speed at which waves travel through the string. Generally, the higher the tension, the faster the wave can move because the string is tighter and supports quicker oscillations.The relationship between tension (\( T \)) and wave speed (\( v \)) is expressed through the wave speed formula:\[ v = \sqrt{\frac{T}{\mu}} \]Where (\( \mu \)) is the string's linear mass density. As we see, the wave speed is directly influenced by the tension, following the square root relationship. For instance, in the case of our string with a tension of 180 N and a linear mass density of approximately 0.007386 kg/m, wave speed is calculated as:\[ v = \sqrt{\frac{180}{0.007386}} \approx 156.0 \, \text{m/s} \]This demonstrates how tension affects wave speed, providing insights into the necessary tension needed to achieve desired wave properties on a string.

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Most popular questions from this chapter

A transverse wave is traveling on a string. The displacement \(y\) of a particle from its equilibrium position is given by \(y=(0.021 \mathrm{m}) \sin (25 t-\) \(2.0 x) .\) Note that the phase angle \(25 t-2.0 x\) is in radians, \(t\) is in seconds, and \(x\) is in meters. The linear density of the string is \(1.6 \times 10^{-2} \mathrm{kg} / \mathrm{m}\). What is the tension in the string?

An ultrasonic ruler, such as the one discussed in Example 4 in Section \(16.6,\) displays the distance between the ruler and an object, such as a wall. The ruler sends out a pulse of ultrasonic sound and measures the time it takes for the pulse to reflect from the object and return. The ruler uses this time, along with a preset value for the speed of sound in air, to determine the distance. Suppose that you use this ruler under water, rather than in air. The actual distance from the ultrasonic ruler to an object is \(25.0 \mathrm{m}\). The adiabatic bulk modulus and density of seawater are \(B_{a d}=2.37 \times 10^{\circ} \mathrm{Pa}\) and \(\rho=1025 \mathrm{kg}\) \(\mathrm{m}^{3},\) respectively. Assume that the ruler uses a preset value of \(343 \mathrm{m} / \mathrm{s}\) for the speed of sound in air. Determine the distance reading that the ruler displays.

The tension in a string is \(15 \mathrm{N}\), and its linear density is \(0.85 \mathrm{kg} / \mathrm{m}\). A wave on the string travels toward the \(-x\) direction; it has an amplitude of \(3.6 \mathrm{cm}\) and a frequency of \(12 \mathrm{Hz} .\) What are the (a) speed and (b) wavelength of the wave? (c) Write down a mathematical expression (like Equation 16.3 or 16.4 ) for the wave, substituting numbers for the variables \(A, f,\) and \(\lambda\).

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at a location \(22 \mathrm{m}\) away from the sound source is \(3.0 \times 10^{-4} \mathrm{W} / \mathrm{m}^{2} .\) What is the intensity at a spot that is \(78 \mathrm{m}\) away?

A wave traveling along the \(x\) axis is described mathematically by the equation \(y=0.17 \sin (8.2 \pi t+0.54 \pi x),\) where \(y\) is the displacement (in meters), \(t\) is in seconds, and \(x\) is in meters. What is the speed of the wave?

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