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A sinusoidal wave travels down a taut, horizontal string with a linear mass density of \(\mu=0.060 \mathrm{kg} / \mathrm{m} .\) The \(\begin{array}{lllllll}\text { maximum } & \text { vertical } & \text { speed } & \text { of } & \text { the } & \text { wave } & \text { is }\end{array}\),\(v_{y \max }=0.30 \mathrm{cm} / \mathrm{s} .\) The wave is modeled with the wave equation \(\quad y(x, t)=A \sin \left(6.00 \mathrm{m}^{-1} x-24.00 \mathrm{s}^{-1} t\right)\).What is the amplitude of the wave? (b) What is the tension in the string?

Short Answer

Expert verified
(a) The amplitude of the wave is \(1\, \mathrm{m}\). (b) The tension in the string is \(0.96\, \mathrm{N}\).

Step by step solution

01

Identify the amplitude in the wave equation

The wave equation is given as: \(y(x, t)=A \sin \left(6.00 \mathrm{m}^{-1} x-24.00 \mathrm{s}^{-1} t\right)\). In this equation, A is the amplitude of the wave. So, our first task is to identify the value of A in the equation.
02

Determine the value of A

In the given wave equation, \(y(x, t)=A \sin \left(6.00 \mathrm{m}^{-1} x-24.00 \mathrm{s}^{-1} t\right)\), the amplitude A is the coefficient of the sine function. We can see that A = 1 (when a value is not shown, it is equal to 1). So the amplitude of the wave is 1 m.
03

Calculate the wave speed

In order to find the tension in the string, we need to calculate the speed of the wave. From the given wave equation, we have the angular wavenumber \(k = 6.00\,\mathrm{m}^{-1}\) and the angular frequency \(\omega = 24.00\,\mathrm{s}^{-1}\). We can find the wave speed (v) using the formula \(v = \frac{\omega}{k}\). Plugging in the given values: \(v = \frac{24.00\,\mathrm{s}^{-1}}{6.00\,\mathrm{m}^{-1}} = 4.00\,\mathrm{m/s}\).
04

Find the tension in the string

Now that we have the wave speed (v) and the mass density \(\mu\), we can calculate the tension (T) in the string using the formula for wave speed on a string: \(v = \sqrt{\frac{T}{\mu}}\). Rearranging this formula to solve for T, we get: \(T = \mu v^2\). Plugging in the given values: \(T = 0.060\,\mathrm{kg/m} \times (4.00\,\mathrm{m/s})^2 = 0.060\,\mathrm{kg/m} \times 16.00\,\mathrm{m^2/s^2} = 0.96\,\mathrm{N}\). In conclusion: (a) The amplitude of the wave is 1 m. (b) The tension in the string is 0.96 N.

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Key Concepts

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

Amplitude
In wave motion, amplitude is a crucial concept that represents the maximum displacement of a point on the wave from its rest position. It tells us how "tall" the wave is, indicating its maximum vertical reach. In the context of the wave equation, amplitude is the coefficient of the sine (or cosine) function. The equation we are focusing on is \[ y(x, t) = A \sin(6.00 \mathrm{m}^{-1} x - 24.00 \mathrm{s}^{-1} t) \]Here, the amplitude \(A\) is not explicitly given within the equation, so we infer it as \(A = 1\,\mathrm{m}\) since no other value is specified. This indicates that the wave extends 1 meter above and below its equilibrium position. Understanding amplitude helps in analyzing wave energy, as it is directly proportional to the square of amplitude. Thus, a wave with a greater amplitude carries more energy.
Wave Equation
The wave equation describes how a wave travels through space and time. It uses mathematical functions to represent the wave's behavior. In this instance, the wave equation is given by:\[ y(x, t) = A \sin(6.00 \mathrm{m}^{-1} x - 24.00 \mathrm{s}^{-1} t) \]The arguments of the sine function, \(6.00 \mathrm{m}^{-1}\) and \(24.00 \mathrm{s}^{-1}\), are crucial. The number \(6.00 \mathrm{m}^{-1}\) is the wave number \(k\), which describes the wave's spatial frequency, or how many wave cycles fit into a unit of distance. The number \(24.00 \mathrm{s}^{-1}\) is the angular frequency \(\omega\), which defines how many cycles occur in a unit of time. By plugging into the equation, we get detailed information about how the wave propagates. Each component, \(x\) and \(t\), help determine the position and evolution of the wave over time.
Wave Speed
Wave speed is a fundamental aspect in the study of wave motion. It tells us at what speed the wave propagates through a medium. In transverse waves on a string, such as our scenario, the wave speed \(v\) is linked to both the wave's frequency and wavelength. Here is a useful formula:\[ v = \frac{\omega}{k} \]Where \(\omega = 24.00 \mathrm{s}^{-1}\) is the angular frequency, and \(k = 6.00 \mathrm{m}^{-1}\) is the wave number. By plugging these values into the formula, you find that this wave travels at \(4.00\,\mathrm{m/s}\). Understanding wave speed is essential for determining how quickly energy is transferred through a medium, which is a common topic in physics and engineering.
Tension in String
Tension in a string is an important concept when examining wave motion through strings. It is influenced by the speed of the wave and the linear density of the string \(\mu\). For a wave on a taut string, we can relate wave speed \(v\), mass density \(\mu\), and tension \(T\) using the equation:\[ v = \sqrt{\frac{T}{\mu}} \]To find tension, rearrange this equation to solve for \(T\):\[ T = \mu v^2 \]With a linear mass density of \(\mu = 0.060\,\mathrm{kg/m}\) and a wave speed of \(4.00\,\mathrm{m/s}\), we substitute these into our equation, resulting in a tension \(T\) of \(0.96\,\mathrm{N}\). Knowing the tension is crucial for applications, such as musical instruments and engineering systems, where this knowledge helps in maximizing efficiency and safety.

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Most popular questions from this chapter

Two strings are attached between two poles separated by a distance of \(2.00 \mathrm{m}\) as shown below, both under the same tension of \(600.00 \mathrm{N}\). String 1 has a linear density of \(\mu_{1}=0.0025 \mathrm{kg} / \mathrm{m}\) and string 2 has a linear mass density of \(\quad \mu_{2}=0.0035 \mathrm{kg} / \mathrm{m} . \quad\) Transverse wave pulses are generated simultaneously at opposite ends of the strings. How much time passes before the pulses pass one another?

A string, fixed on both ends, is \(5.00 \mathrm{m}\) long and has a mass of \(0.15 \mathrm{kg}\). The tension if the string is \(90 \mathrm{N}\). The string is vibrating to produce a standing wave at the fundamental frequency of the string. (a) What is the speed of the waves on the string? (b) What is the wavelength of the standing wave produced? (c) What is the period of the standing wave?

A quality assurance engineer at a frying pan company is asked to qualify a new line of nonstick-coated frying pans. The coating needs to be \(1.00 \mathrm{mm}\) thick. One method to test the thickness is for the engineer to pick a percentage of the pans manufactured, strip off the coating, and measure the thickness using a micrometer. This method is a destructive testing method. Instead, the engineer decides that every frying pan will be tested using a nondestructive method. An ultrasonic transducer is used that produces sound waves with a frequency of \(f=25 \mathrm{kHz}\). The sound waves are sent through the coating and are reflected by the interface between the coating and the metal pan, and the time is recorded. The wavelength of the ultrasonic waves in the coating is \(0.076 \mathrm{m}\). What should be the time recorded if the coating is the correct thickness \((1.00 \mathrm{mm}) ?\)

Two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the resultant wave is $$y_{R}(x, t)=0.35 \mathrm{cm} \sin \left(6.28 \mathrm{m}^{-1} x-1.57 \mathrm{s}^{-1} t+\frac{\pi}{4}\right)$$.What are the period, wavelength, amplitude, and phase shift of the individual waves?

A seismograph records the S- and P-waves from an earthquake \(20.00 \mathrm{s}\) apart. If they traveled the same path at constant wave speeds of \(v_{S}=4.00 \mathrm{km} / \mathrm{s} \quad\) and \(v_{P}=7.50 \mathrm{km} / \mathrm{s}, \quad\) how far away is the epicenter of the earthquake?

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