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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. (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) 75.4 cm/s, (g) -5.2 cm.

Step by step solution

01

Identify Amplitude

The amplitude of the wave is the coefficient of the sine function. From the wave equation \( y = 6.0 \sin(0.020\pi x + 4.0\pi t) \), the amplitude is \( 6.0 \) cm.
02

Determine Wavelength

The wave number \( k \) in the equation \( y = A \sin(kx + \omega t) \) corresponds to \( 0.020\pi \). The formula for wavelength \( \lambda \) is \( \lambda = \frac{2\pi}{k} \). Therefore, \( \lambda = \frac{2\pi}{0.020\pi} = 100 \) cm.
03

Calculate Frequency

The angular frequency \( \omega \) in the equation is \( 4.0\pi \). The formula for frequency \( f \) is \( f = \frac{\omega}{2\pi} \). So, \( f = \frac{4.0\pi}{2\pi} = 2 \) Hz.
04

Find Wave Speed

The wave speed \( v \) can be found using the formula \( v = f \cdot \lambda \). Substituting the values, \( v = 2 \times 100 = 200 \) cm/s.
05

Determine Direction of Propagation

The wave function \( y = 6.0 \sin(0.020\pi x + 4.0\pi t) \) has a positive sign in front of \( 4.0\pi t \), indicating the wave propagates in the negative x-direction.
06

Find Maximum Transverse Speed

The maximum transverse speed \( V_{max} \) of a particle is given by \( V_{max} = A \cdot \omega \). Thus, \( V_{max} = 6.0 \times 4.0\pi = 24\pi \approx 75.4 \) cm/s.
07

Calculate Transverse Displacement

To find the transverse displacement at \( x = 3.5 \) cm and \( t = 0.26 \) s, substitute these values into the wave equation: \( y = 6.0 \sin(0.020\pi \times 3.5 + 4.0\pi \times 0.26) \). After calculations, you find \( y \approx -5.2 \) cm.

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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 the motion of the medium is perpendicular to the direction of the wave's travel. Imagine waves on a string, where the string moves up and down while the wave moves along the string. This is different from longitudinal waves, which move the medium parallel to the direction of the wave.

Transverse waves have several key characteristics:
  • **Crests and troughs**: The high points of the wave are called crests, and the low points are called troughs.
  • **Amplitude**: This is the maximum displacement of the medium from its rest position. For the given wave equation, the amplitude is 6.0 cm.
  • **Direction of motion**: In the context of the given wave, it propagates in the negative x-direction.
Understanding transverse waves is essential for analyzing and predicting wave behavior in various mediums, whether on strings, water, or light waves.
Wave Propagation
Wave propagation refers to the movement of waves through a medium. For transverse waves on a string, the wave can move in positive or negative directions along the string. In our wave equation, \(y=6.0\sin (0.020\pi x+4.0\pi t)\), the positive sign with the time component \(4.0\pi t\) indicates that the wave propagates in the negative x-direction.

This means each point on the wave moves to a later x-location as time progresses. The speed of propagation is crucial because it tells us how fast the wave is moving. In our case, it's calculated using the formula \(v = f \cdot \lambda\), where \(f\) is the frequency and \(\lambda\) is the wavelength. With values found to be 2 Hz for frequency and 100 cm for wavelength, the wave speed is 200 cm/s. Knowing the direction and speed helps in practical scenarios like engineering and design of systems where wave behavior is a critical factor.
Amplitude and Frequency
Amplitude and frequency are fundamental properties of waves. The amplitude of a wave is the maximum extent of a vibration or displacement of the medium it travels through from its rest position. It is visually represented by the height of the wave crest above the rest level. For the wave equation provided, amplitude is straightforward to identify — 6.0 cm.

Frequency, on the other hand, tells us how many oscillations or cycles occur in one second. It is measured in Hertz (Hz). In the equation \(y=6.0\sin (0.020\pi x+4.0\pi t)\), the angular frequency \(\omega\) is given as 4.0\pi. From this, the frequency is found using \(f = \frac{\omega}{2\pi}\), resulting in 2 Hz. High frequency means more cycles per second, translating to higher energy. Understanding amplitude and frequency aids in applications ranging from acoustics to telecommunications.
Wavelength Calculation
Wavelength is the distance over which the wave's shape repeats. In other words, it's the length of one wave cycle. Wavelength is crucial in defining how a wave interacts with objects and other waves.

To calculate wavelength, use the wave number, denoted by \(k\). The wave equation gives \(k\) as 0.020\pi. The formula \(\lambda = \frac{2\pi}{k}\) helps calculate wavelength, leading to \(\lambda = \frac{2\pi}{0.020\pi} = 100\) cm for the given wave.

This measure of wavelength helps in understanding how waves propagate through various mediums. Knowing the wavelength allows us to predict behavior in physical mediums, anticipate resonance, and piece together the wave properties in physics designs and calculations.

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Most popular questions from this chapter

If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind. The air pressure variations in the vortexes tend to cause the line to oscillate (gallop), especially if the frequency of the variations matches a resonant frequency of the line. In long lines, the resonant frequencies are so close that almost any wind speed can set up a resonant mode vigorous enough to pull down support towers or cause the line to short out with an adjacent line. If a transmission line has a length of \(347 \mathrm{~m}\), a linear density of \(3.35 \mathrm{~kg} / \mathrm{m}\), and a tension of \(65.2 \mathrm{MN}\), what are (a) the frequency of the fundamental mode and (b) the frequency difference between successive modes?

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A wave on a string is described by $$ y(x, t)=15.0 \sin (\pi x / 8-4 \pi t), $$ where \(x\) and \(y\) are in centimeters and \(t\) is in seconds. (a) What is the transverse speed for a point on the string at \(x=6.00 \mathrm{~cm}\) when \(t=0.250 \mathrm{~s}\) ? (b) What is the maximum transverse speed of any point on the string? (c) What is the magnitude of the transverse acceleration for a point on the string at \(x=6.00 \mathrm{~cm}\) when \(t=0.250 \mathrm{~s}\) ? (d) What is the magnitude of the maximum transverse acceleration for any point on the string?

A continuous traveling wave with amplitude \(A\) is incident on a boundary. The continuous reflection, with a smaller amplitude \(B\), travels back through the incoming wave. The resulting interference pattern is displayed in Fig. 16-51. The standing wave ratio is defined to be \(\mathrm{SWR}=\frac{A+B}{A-B}\) The reflection coefficient \(R\) is the ratio of the power of the reflected wave to the power of the incoming wave and is thus proportional to the ratio \((B / A)^{2} .\) What is the SWR for (a) total reflection and (b) no reflection? (c) For SWR \(=1.50\), what is \(R\) expressed as a percentage?

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) \(y_{m},(\) b) \(\omega,(\mathrm{c}) k\), and \((\mathrm{d})\) the correct choice of sign in front of \(\omega ?\)

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