/*! 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 27 A transverse wave is traveling o... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
The tension in the string is 2.5 N.

Step by step solution

01

Identify wave properties

The wave equation is \( y = (0.021 \, \text{m}) \sin(25t - 2.0x) \). From this equation, we can identify the wave number \( k = 2.0 \) and the angular frequency \( \omega = 25 \).
02

Calculate wave speed

Wave speed \( v \) can be found using the formula \( v = \frac{\omega}{k} \). Here, \( \omega = 25 \) and \( k = 2.0 \). Thus, the speed \( v = \frac{25}{2.0} = 12.5 \, \text{m/s} \).
03

Use the wave speed to find tension

The wave speed \( v \) on a string is related to the tension \( T \) and linear density \( \mu \) by the formula \( v = \sqrt{\frac{T}{\mu}} \). Rearrange this to find \( T = v^2 \mu \).
04

Substitute values to find tension

Given \( v = 12.5 \, \text{m/s} \) and \( \mu = 1.6 \times 10^{-2} \, \text{kg/m} \), substitute these values into \( T = v^2 \mu \) to get \( T = (12.5)^2 \times (1.6 \times 10^{-2}) \). Calculate \( T = 2.5 \, \text{N} \).

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Key Concepts

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

Transverse Wave
A transverse wave is a type of wave where particles of the medium move perpendicular to the direction of the wave's travel. Imagine a rope lying flat on the ground. When you give one end a quick upward flick, the wave travels along the rope, but the movement of the rope segments is vertical. This vertical movement is why it's called a transverse wave.

Transverse waves are common in everyday life. Light waves and waves on a string are typical examples. The equation for a transverse wave, such as the one given in the problem, can describe the displacement of any point on the wave at any time.
  • The displacement in our wave equation, \(y=(0.021 \, \text{m}) \sin(25t-2.0x)\), is a sine function, reflecting how the wave oscillates over both time and space.
  • The parts of the equation: amplitude, wave number, and angular frequency define the wave's characteristics.
Understanding these components helps in determining properties like wave speed and tension in the medium. Understanding how the wave moves is crucial for applications like transmitting signals through cables.
Wave Speed
Wave speed is how fast the wave travels through a medium. To find it, we can use the relationship between the wave's angular frequency and wave number. Angular frequency (\(\omega\)) and wave number (\(k\)) characterize how quickly the wave oscillates in time and space.

The formula for wave speed (\(v\)) is \(v = \frac{\omega}{k}\). In our example, the values are \(\omega = 25 \, \text{radians/s}\) and \(k = 2.0 \, \text{radians/m}\). So, the wave speed is \(v = \frac{25}{2.0} = 12.5 \, \text{m/s}\).
  • This speed indicates how fast the energy is being transferred along the string.
  • Wave speed is crucial in determining the tension of the string, as faster waves imply different forces acting within the medium.
Knowing wave speed can also help in applications like tuning musical instruments and designing effective communication systems.
Linear Density
Linear density (\(\mu\)), is a measure of how much mass is distributed along a unit length of string or material. It is expressed as mass per unit length (\( \text{kg/m} \)). In the context of waves, linear density plays a key role in determining wave speed and tension.

For our problem, the linear density is given as \(1.6 \times 10^{-2} \, \text{kg/m}\). This tells us how much mass of the string is present over each meter, which influences how the wave travels.
  • A higher linear density means more mass, which affects how the wave moves and the tension needed to maintain the wave.
  • The wave speed \(v\) and tension \(T\) are related by the equation \(v = \sqrt{\frac{T}{\mu}}\).
Understanding linear density helps in applications such as designing cables for specific loads and ensuring efficient signal transmission.
Angular Frequency
Angular frequency (\(\omega\)) measures how quickly a wave oscillates in time, usually expressed in radians per second. It is a cornerstone concept for describing oscillations and waves. The angular frequency relates to how many cycles happen per unit of time.

In the given wave equation, \(\omega = 25\), indicating that the sine wave completes 25 radians worth of phase in one second. This frequency directly impacts the wave speed and how the medium will respond to applied tension.
  • Angular frequency is essential when calculating wave properties alongside wave number.
  • It emphasizes temporal characteristics of the wave, such as how often a particle returns to its starting position.
Understanding angular frequency is crucial in fields like acoustics and electronics, where time-based waves are common. It allows engineers and physicists to predict and control wave behavior effectively.

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Most popular questions from this chapter

An aircraft carrier has a speed of \(13.0 \mathrm{~m} / \mathrm{s}\) relative to the water. A jet is catapulted from the deck and has a speed of \(67.0 \mathrm{~m} / \mathrm{s}\) relative to the water. The engines produce a \(1550-\mathrm{Hz}\) whine, and the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is the frequency of the sound heard by the crew on the ship?

So Concept Questions Example 4 in the text discusses an ultrasonic ruler that 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 you use this ruler underwater, rather than in air. (a) Is the speed of sound in water greater than, less than, or equal to the speed of sound in air? (b) Is the reading on the ruler greater than, less than, or equal to the actual distance? Provide reasons for your answers. Problem The actual distance from the ultrasonic ruler to an object is \(25.0 \mathrm{~m} .\) The adiabatic bulk modulus and density of seawater are \(B_{\mathrm{ad}}=2.37 \times 10^{9} \mathrm{~Pa}\) 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, and determine the distance reading on its display. Verify that your answer is consistent with your answers to the Concept Questions.

You are riding your bicycle directly away from a stationary source of sound and hear a frequency that is \(1.0 \%\) lower than the emitted frequency. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is your speed?

A listener doubles his distance from a source that emits sound uniformly in all directions. By how many decibels does the sound intensity level change?

A rocket, starting from rest, travels straight up with an acceleration of \(58.0 \mathrm{~m} / \mathrm{s}^{2}\). When the rocket is at a height of \(562 \mathrm{~m}\), it produces sound that eventually reaches a ground-based monitoring station directly below. The sound is emitted uniformly in all directions. The monitoring station measures a sound intensity \(I .\) Later, the station measures an intensity \(\frac{1}{3} I\). Assuming that the speed of sound is \(343 \mathrm{~m} / \mathrm{s},\) find the time that has elapsed between the two measurements.

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