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What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a wire that is \(10.0 \mathrm{~m}\) long, has a mass of \(100 \mathrm{~g}\), and is stretched under a tension of \(250 \mathrm{~N}\) ?

Short Answer

Expert verified
(a) 7.91 Hz, (b) 15.81 Hz, (c) 23.72 Hz.

Step by step solution

01

Calculate Linear Density

First, calculate the linear density (mass per unit length) of the wire. The mass of the wire is 100 g, which is equal to 0.1 kg (since 1 g = 0.001 kg), and the length of the wire is 10.0 m.Linear density, \( \mu \), is given by:\[ \mu = \frac{\text{mass}}{\text{length}} = \frac{0.1 \text{ kg}}{10.0 \text{ m}} = 0.01 \text{ kg/m} \]
02

Calculate Wave Speed

Next, calculate the speed of the wave on the wire using the formula:\[ v = \sqrt{\frac{T}{\mu}} \]where \( T \) is the tension in the wire.Substituting the values:\[ v = \sqrt{\frac{250 \text{ N}}{0.01 \text{ kg/m}}} = \sqrt{25000} = 158.11 \text{ m/s} \]
03

Determine Frequencies for Standing Waves

Standing wave frequencies on a string fixed at both ends are given by:\[ f_n = \frac{n \cdot v}{2L} \]where \( n \) is the mode number, \( v \) is the wave speed, and \( L \) is the length of the wire (10.0 m).
04

Calculate Lowest Frequency

The lowest frequency (fundamental frequency) corresponds to \( n = 1 \):\[ f_1 = \frac{1 \cdot 158.11 \text{ m/s}}{2 \times 10.0 \text{ m}} = 7.91 \text{ Hz} \]
05

Calculate Second Lowest Frequency

The second lowest frequency corresponds to \( n = 2 \):\[ f_2 = \frac{2 \cdot 158.11 \text{ m/s}}{2 \times 10.0 \text{ m}} = 15.81 \text{ Hz} \]
06

Calculate Third Lowest Frequency

The third lowest frequency corresponds to \( n = 3 \):\[ f_3 = \frac{3 \cdot 158.11 \text{ m/s}}{2 \times 10.0 \text{ m}} = 23.72 \text{ Hz} \]

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Key Concepts

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

Wave Speed Calculation
Understanding how to calculate the wave speed is crucial when dealing with waves on a string or wire. The wave speed dictates how quickly the wave travels through the medium. We calculate it using the formula: \[ v = \sqrt{\frac{T}{\mu}} \] where \( v \) is the wave speed, \( T \) is the tension in the wire, and \( \mu \) is the linear density.- **Tension (\( T \))**: This is the force stretching the wire. It is usually measured in newtons (N).- **Linear Density (\( \mu \))**: This is a measure of mass per unit length, usually in kilograms per meter (kg/m). For example, in the given problem, the tension is 250 N, and the linear density is calculated as 0.01 kg/m. Plugging these values into the formula, we obtain the wave speed \( v \approx 158.11 \text{ m/s} \). This value gives us how fast each pulse of the wave travels along the wire.
Linear Density
Linear density, often represented by the Greek letter \( \mu \), is an essential parameter in wave-related calculations. It measures how much mass is distributed along a unit length of the string or wire. To compute linear density, use the formula:\[ \mu = \frac{\text{mass}}{\text{length}} \]Consider the wire in the exercise, which has:- **Mass**: 100 g = 0.1 kg (since converting grams to kilograms involves multiplying by 0.001)- **Length**: 10.0 mThus, the linear density becomes \( \mu = \frac{0.1 \text{ kg}}{10.0 \text{ m}} = 0.01 \text{ kg/m} \).This value is critical because it affects the wave speed. A higher linear density indicates that the wire has more mass per unit length, typically resulting in slower wave speed if the tension remains constant.
Fundamental Frequency
The fundamental frequency is the lowest frequency at which a system can naturally vibrate. It is crucial for solving problems involving standing waves. The formula to determine the fundamental frequency \( f_1 \) on a wire fixed at both ends is:\[ f_1 = \frac{v}{2L} \]Where:- **\( v \)** is the wave speed- **\( L \)** is the length of the wireFor the given problem, substituting the values:- **Wave Speed, \( v \)**: 158.11 m/s- **Length, \( L \)**: 10.0 mThe fundamental frequency \( f_1 \) is calculated as \( f_1 = \frac{158.11 \text{ m/s}}{2 \times 10.0 \text{ m}} \approx 7.91 \text{ Hz} \).This is the lowest or first harmonic frequency at which standing waves can form on the wire. Understanding this concept is essential to determine other harmonics, which are integer multiples of the fundamental frequency.

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Most popular questions from this chapter

In an experiment on standing waves, a string \(90 \mathrm{~cm}\) long is attached to the prong of an electrically driven tuning fork that oscillates perpendicular to the length of the string at a frequency of 60 \(\mathrm{Hz}\). The mass of the string is \(0.044 \mathrm{~kg}\). What tension must the string be under (weights are attached to the other end) if it is to oscillate in four loops?

A sinusoidal transverse wave traveling in the negative direction of an \(x\) axis has an amplitude of \(1.00 \mathrm{~cm}\), a frequency of \(550 \mathrm{~Hz}\), and a speed of \(330 \mathrm{~m} / \mathrm{s}\). If the wave equation is of the form \(y(x, t)=y_{m} \sin (k x \pm \omega t)\), what are (a) \(y_{m},(\mathrm{~b}) \omega,(\mathrm{c}) k\), and (d) the correct choice of sign in front of \(\omega\) ?

Two identical traveling waves, moving in the same direction, are out of phase by \(\pi / 2\) rad. What is the amplitude of the resultant wave in terms of the common amplitude \(y_{m}\) of the two combining waves?

A string under tension \(\tau_{i}\) oscillates in the third harmonic at frequency \(f_{3}\), and the waves on the string have wavelength \(\lambda_{3}\). If the tension is increased to \(\tau_{f}=4 \tau_{i}\) and the string is again made to oscillate in the third harmonic, what then are (a) the frequency of oscillation in terms of \(f_{3}\) and (b) the wavelength of the waves in terms of \(\lambda_{3} ?\)

A wave has a speed of \(240 \mathrm{~m} / \mathrm{s}\) and a wavelength of \(3.2 \mathrm{~m}\). What are the (a) frequency and (b) period of the wave?

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