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The equation of a transverse wave traveling along a very long string is \(y=6.0 \sin (0.020 \pi x+4.0 \pi t)\), where \(x\) and \(y\) are expressed in centimeters and \(t\) is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave, and (f) the maximum transverse speed of a particle in the string. \((\mathrm{g})\) What is the transverse displacement at \(x=3.5 \mathrm{~cm}\) when \(t=0.26 \mathrm{~s}\) ?

Short Answer

Expert verified
(a) 6.0 cm, (b) 100 cm, (c) 2 Hz, (d) 200 cm/s, (e) Negative x-direction, (f) 24\(\pi\) cm/s, (g) 1.854 cm.

Step by step solution

01

Identify the Amplitude

The amplitude of a wave is given directly by the coefficient of the sine function in the wave equation. Here, the wave equation is given by \( y = 6.0 \sin(0.020 \pi x + 4.0 \pi t) \). Thus, the amplitude is \( 6.0 \) cm.
02

Calculate the Wavelength

The wave number \( k \) is related to the wavelength \( \lambda \) by the formula \( k = \frac{2\pi}{\lambda} \). Comparing with the given wave equation \( y = 6.0 \sin(0.020 \pi x + 4.0 \pi t) \), we find \( k = 0.020 \pi \). Therefore, \( 0.020 \pi = \frac{2\pi}{\lambda} \), which gives \( \lambda = \frac{2\pi}{0.020 \pi} = 100 \) cm.
03

Determine the Frequency

The angular frequency \( \omega \) is related to the frequency \( f \) by the formula \( \omega = 2\pi f \). From the given wave equation, we have \( \omega = 4.0 \pi \). Thus, \( 4.0 \pi = 2\pi f \), which simplifies to \( f = 2 \) Hz.
04

Find the Speed of the Wave

The speed of the wave \( v \) is given by the product of its frequency \( f \) and its wavelength \( \lambda \). Using \( \lambda = 100 \) cm and \( f = 2 \) Hz, we find \( v = f \times \lambda = 2 \times 100 = 200 \) cm/s.
05

Determine the Direction of Propagation

The wave equation includes the term \( 0.020 \pi x + 4.0 \pi t \). Since both \( k \) and \( \omega \) terms are positive, the wave moves in the negative x-direction because the phase velocity is given by \( -\frac{\omega}{k} \).
06

Find the Maximum Transverse Speed of a Particle

The maximum transverse speed \( v_{max} \) is the maximum value of the derivative of \( y \) with respect to time \( t \). Differentiate the equation to get \( v_y = \frac{d}{dt}(6.0 \sin(0.020 \pi x + 4.0 \pi t)) = 6.0 \times 4.0\pi \cos(0.020 \pi x + 4.0 \pi t) \). The maximum value of \( \cos \) is 1, so \( v_{max} = 6.0 \times 4.0\pi = 24 \pi \) cm/s.
07

Calculate the Transverse Displacement at Given Point

Substitute \( x = 3.5 \) cm and \( t = 0.26 \) s into the wave equation \( y = 6.0 \sin(0.020 \pi \times 3.5 + 4.0 \pi \times 0.26) \). This simplifies to \( y = 6.0 \sin(0.070 \pi + 1.04 \pi) = 6.0 \sin(1.11 \pi) \). Calculate \( \sin(1.11 \pi) \approx 0.309 \), giving a displacement \( y \approx 1.854 \) cm.

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Key Concepts

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

Wave Equation
A transverse wave moves in a direction perpendicular to the oscillation of the particles within the medium, often seen in strings or on the surface of water. The wave equation is a mathematical representation of this movement, typically expressed as a sinusoidal function. The general form of a wave equation is \[ y(x, t) = A \, \sin(kx + \omega t + \phi) \]where:- \( A \) is the amplitude, - \( k \) is the wave number,- \( \omega \) is the angular frequency,- and \( \phi \) is the phase constant.
In the given wave equation \( y = 6.0 \sin(0.020 \pi x + 4.0 \pi t) \), it's evident that the sine function oscillates to describe the wave's motion over time \( t \) and position \( x \). Each part of the function corresponds to a physical property of the wave, such as its speed, wavelength, and frequency. Understanding these components helps in identifying the behavior and characteristics of the wave.
Amplitude
Amplitude is a crucial concept when discussing waves. It refers to the maximum extent of a vibration or displacement from the rest position. In simple terms, it refers to the wave's height above or below its central axis.
In the provided wave equation:\[ y = 6.0 \sin(0.020 \pi x + 4.0 \pi t) \]
the amplitude is the coefficient of the sine function. Therefore, the amplitude is \( 6.0 \) cm. This indicates that particles in the medium will move 6.0 cm away from their equilibrium position in the wave's propagation. Amplitude is essential in determining the wave's energy; higher amplitudes equate to waves carrying more energy.
Wavelength
Wavelength is the distance between consecutive crests or troughs in a wave. It is a complete cycle of the wave, and for wave equations like the one given, it contributes to defining the spatial scale of the wave.
Wavelength \( \lambda \) is related to the wave number \( k \) through the equation:\[ k = \frac{2\pi}{\lambda} \].
From the wave equation:\( y = 6.0 \sin(0.020 \pi x + 4.0 \pi t) \)
we identify that \( k = 0.020 \pi \). Solving for \( \lambda \) gives:\[ \lambda = \frac{2\pi}{0.020 \pi} = 100 \text{ cm} \].
This measurement helps understand how the wave travels through the medium and interacts with other waves, especially in interference and diffraction scenarios.
Frequency
Frequency of a wave is the number of complete cycles that occur in a given time interval, generally measured in hertz (Hz). It's a vital parameter in understanding how often particles in the medium vibrate relative to the position of the wave.
The formula relating angular frequency \( \omega \) to frequency \( f \) is:\[ \omega = 2\pi f \].
From the wave equation:\( y = 6.0 \sin(0.020 \pi x + 4.0 \pi t) \)
the angular frequency \( \omega \) is \( 4.0 \pi \). Thus, solving for \( f \), we have:\[ 4.0 \pi = 2\pi f \Rightarrow f = 2 \text{ Hz} \].
This means the wave completes two full oscillations every second. Frequency is pivotal in determining the pitch in sound waves and colors in light waves, each with unique applications in communication and technology.

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Most popular questions from this chapter

A string along which waves can travel is \(2.70 \mathrm{~m}\) long and has a mass of \(260 \mathrm{~g} .\) The tension in the string is \(36.0 \mathrm{~N}\). What must be the frequency of traveling waves of amplitude \(7.70 \mathrm{~mm}\) for the average power to be \(85.0 \mathrm{~W} ?\)

A sand scorpion can detect the motion of a nearby beetle (its prey) by the waves the motion sends along the sand surface (Fig. \(16-29\) ). The waves are of two types: transverse waves traveling at \(v_{t}=50 \mathrm{~m} / \mathrm{s}\) and longitudinal waves traveling at \(v_{l}=150 \mathrm{~m} / \mathrm{s} .\) If a sudden motion sends out such waves, a scorpion can tell the distance of the beetle from the difference \(\Delta t\) in the arrival times of the waves at its leg nearest the beetle. If \(\Delta t=4.0 \mathrm{~ms}\) what is the beetle's distance?

A sinusoidal transverse wave traveling in the positive direction of an \(x\) axis has an amplitude of \(2.0 \mathrm{~cm}\), a wavelength of \(10 \mathrm{~cm}\), and a frequency of \(400 \mathrm{~Hz}\). If the wave equation is of the form \(y(x, t)=y_{m} \sin (k x \pm \omega t)\), what are (a) \(y_{m}\), (b) \(k\), (c) \(\omega\), and (d) the correct choice of sign in front of \(\omega ?\) What are (e) the maximum transverse speed of a point on the cord and (f) the speed of the wave?

(a) What is the fastest transverse wave that can be sent along a steel wire? For safety reasons, the maximum tensile stress to which steel wires should be subjected is \(7.00 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\). The density of steel is \(7800 \mathrm{~kg} / \mathrm{m}^{3}\). (b) Does your answer depend on the diameter of the wire?

Use the wave equation to find the speed of a wave given by $$ y(x, t)=(3.00 \mathrm{~mm}) \sin \left[\left(4.00 \mathrm{~m}^{-1}\right) x-\left(7.00 \mathrm{~s}^{-1}\right) t\right] . $$

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