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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

Extract Wave Information

The wave can be described by the equation \( y = 0.021 \sin(25t - 2.0x) \). Here, the angular frequency \( \omega \) is 25 rad/s, and the wave number \( k \) is 2.0 rad/m. We will use these values to find the wave speed and tension in the string.
02

Determine Wave Speed

The wave speed \( v \) is given by the relation \( v = \frac{\omega}{k} \). Substitute \( \omega = 25 \) rad/s and \( k = 2.0 \) rad/m into this equation to find \( v \): \[ v = \frac{25}{2.0} = 12.5 \text{ m/s} \].
03

Use Wave Speed Formula

The wave speed on a string can also be expressed in terms of the tension \( T \) and linear density \( \mu \) as \( v = \sqrt{\frac{T}{\mu}} \). Rearrange this formula to solve for \( T \): \( T = \mu v^2 \).
04

Substitute Known Values

Substitute \( \mu = 1.6 \times 10^{-2} \) kg/m and \( v = 12.5 \) m/s into the rearranged formula: \[ T = (1.6 \times 10^{-2}) \times (12.5)^2 \].
05

Calculate Tension

Calculate the tension using the values substituted: \[ T = 1.6 \times 10^{-2} \times 156.25 = 2.5 \text{ N} \].

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Key Concepts

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

Transverse Waves
Transverse waves are a fascinating phenomenon where particles in the medium move perpendicular to the direction of the wave's travel. This is what happens on a string, like the one described in our exercise. When you pluck a string or a rope, it creates a wave that travels along the length of the string while the particles of the string move up and down.

Picture a ripple on a pond, where the water moves up and down, and the ripple spreads out in a circle. That's similar to a transverse wave. In mathematics, we describe these waves using a sinusoidal function, like the one given in the problem, \(y = 0.021 \sin(25t - 2.0x)\).

Understanding transverse waves helps us grasp how energy and information travel. Many everyday phenomena, like light and electromagnetic waves, are transverse waves.
Wave Speed
Wave speed is an essential concept in wave dynamics, describing how fast the wave travels through the medium. The formula for the speed of a transverse wave is particularly insightful. It's given by the relation \(v = \frac{\omega}{k}\), where \(\omega\) is the angular frequency, and \(k\) is the wave number.

In our exercise, we derived the wave speed using known values for \(\omega = 25\) rad/s and \(k = 2.0\) rad/m. Calculating this results in a wave speed of \(12.5\) m/s. This information helps us understand how quickly the wave pulses move through the string.

Knowing the wave speed can help solve various physics problems, such as determining the time it takes for a wave to travel a certain distance.
Tension in Strings
The tension in a string plays a crucial role in defining how waves travel along it. In this context, tension refers to the force stretched through the string. The tension determines both the wave speed and the dynamic behavior of the transverse wave.

For our calculation, we used the formula \(T = \mu v^2\), where \(\mu\) is the linear density of the string and \(v\) is the wave speed.

Using the given values \(\mu = 1.6 \times 10^{-2}\) kg/m and \(v = 12.5\) m/s, we found the tension to be \(2.5\) N.
  • Higher tension typically results in faster wave speeds.
  • The tension can be adjusted by altering how much the string is stretched.
Such an understanding is practically applied in musical instruments, where string tension is adjusted to alter the pitch.
Angular Frequency
Angular frequency is a key element when describing how waves oscillate. It tells us how quickly a wave completes its cycles of motion in radians per second. In our exercise, angular frequency, \(\omega\), is given as \(25\) rad/s. This means the wave oscillates with a rapid pace in time.

Angular frequency is closely linked to the period and frequency of a wave. For instance:
  • Frequency is the number of oscillations per second, measured in hertz (Hz).
  • The period is the time it takes for one complete oscillation.
The connection between angular frequency and regular frequency \(f\) is given by \(\omega = 2\pi f\). Understanding this allows us to convert between angular and regular frequencies with ease.
Linear Density
Linear density is a measure of mass distribution along the length of an object, specifically in kilograms per meter (kg/m). It describes how much mass is contained in a unit length of the string. This property significantly impacts wave dynamics.

In our problem, the linear density \(\mu\) is \(1.6 \times 10^{-2}\) kg/m. Linear density affects the wave speed and tension on the string. By the relationship \(v = \sqrt{\frac{T}{\mu}}\), we see that lower linear density allows for a higher wave speed for the same tension.

In practical applications:
  • Knowing the linear density is vital for designing structures that involve cable or string tension.
  • It also assists in the engineering of stringed musical instruments, ensuring consistent sound quality.
Understanding linear density helps in calculating other related properties, making it a pivotal aspect of wave dynamics.

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Most popular questions from this chapter

(a) A uniform rope of mass \(m\) and length \(L\) is hanging straight down from the ceiling. A small-amplitude transverse wave is sent up the rope from the bottom end. Derive an expression that gives the speed \(v\) of the wave on the rope in terms of the distance \(y\) above the bottom end of the rope and the magnitude \(g\) of the acceleration due to gravity. (b) Use the expression that you have derived to calculate the speeds at distances of \(0.50 \mathrm{m}\) and \(2.0 \mathrm{m}\) above the bottom end of the rope.

A loudspeaker in a parked car is producing sound whose frequency is \(20510 \mathrm{Hz} .\) A healthy young person with normal hearing is standing nearby on the sidewalk but cannot hear the sound because the frequency is too high. When the car is moving, however, this person can hear the sound. (a) Is the car moving toward or away from the person? Why? (b) If the speed of sound is \(343 \mathrm{m} / \mathrm{s},\) what is the minimum speed of the moving car?

An observer stands \(25 \mathrm{m}\) behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of the bullets is \(840 \mathrm{m} / \mathrm{s},\) and the air temperature is \(20^{\circ} \mathrm{C} .\) How far does each bullet travel before the observer hears the report of the rifle? Assume that the bullets encounter no obstacles during this interval, and ignore both air resistance and the vertical component of the bullets' motion.

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 \(\frac{1}{3} I .\) There are no reflections. Assuming that the speed of sound is \(343 \mathrm{m} / \mathrm{s}\) find the time that has elapsed between the two measurements.

Two wires are parallel, and one is directly above the other. Each has a length of \(50.0 \mathrm{m}\) and a mass per unit length of \(0.020 \mathrm{kg} / \mathrm{m} .\) However, the tension in wire \(\mathrm{A}\) is \(6.00 \times 10^{2} \mathrm{N},\) and the tension in wire \(\mathrm{B}\) is \(3.00 \times 10^{2} \mathrm{N} .\) Transverse wave pulses are generated simultaneously, one at the left end of wire \(\mathrm{A}\) and one at the right end of wire \(\mathrm{B}\). The pulses travel toward each other. How much time does it take until the pulses pass each other?

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