Question

A traveling wave propagating on a string is described by the following equation: $$ y(x, t)=(5.00 \mathrm{~mm}) \sin \left(\left(157.08 \mathrm{~m}^{-1}\right) x-\left(314.16 \mathrm{~s}^{-1}\right) t+0.7854\right) $$ a) Determine the minimum separation, \(\Delta x_{\min }\), between two points on the string that oscillate in perfect opposition of phases (move in opposite directions at all times). b) Determine the separation, \(\Delta x_{A B}\), between two points \(A\) and \(B\) on the string, if point \(B\) oscillates with a phase difference of 0.7854 rad compared to point \(A\). c) Find the number of crests of the wave that pass through point \(A\) in a time interval \(\Delta t=10.0 \mathrm{~s}\) and the number of troughs that pass through point \(B\) in the same interval. d) At what point along its trajectory should a linear driver connected to one end of the string at \(x=0\) start its oscillation to generate this sinusoidal traveling wave on the string?

Short answer

The minimum separation between two points oscillating in perfect opposition of phases is 2.00 cm. b) What is the separation between two points A and B with phase difference 0.7854 rad? The separation between points A and B with a phase difference of 0.7854 rad is 5.00 mm. c) How many crests pass through point A and how many troughs pass through point B in a time interval of 10.0 s? In a time interval of 10.0 s, 50 crests pass through point A and 50 troughs pass through point B. d) What is the initial position of a linear driver to generate the given wave? The initial position of the linear driver to generate the given wave is 3.54 mm.

Step-by-step solution

Identify the wave equation

The given wave equation is $$y(x, t)=(5.00 \mathrm{~mm}) \sin \left(\left(157.08 \mathrm{~m}^{-1}\right) x-\left(314.16 \mathrm{~s}^{-1}\right) t+0.7854\right) $$

Find the difference in phase for points in opposition

For two points oscillating in perfect opposition of phases, the phase difference between them should be an odd multiple of \(\pi\) (180 degrees). We can represent it as \(\Delta \phi = (2n+1)\pi\), where \(n\) is an integer.

Find the difference in the position of these points, \(\Delta x_{\min}\)

To find the difference in position, we can compare the phase difference with the wavenumber \(k = 157.08 \mathrm{~m}^{-1}\): $$\Delta \phi = k \Delta x_{\min}$$Solving for \(\Delta x_{\min}\), we get: $$\Delta x_{\min} = \frac{(2n+1)\pi}{157.08 \mathrm{~m}^{-1}}$$For the minimum separation, we choose \(n=0\): $$\Delta x_{\min} = \frac{\pi}{157.08 \mathrm{~m}^{-1}} = 2.00 \times 10^{-2} \mathrm{~m} = 2.00 \mathrm{~cm}$$ b) Separation between points A and B with phase difference 0.7854 rad

Find the difference in position, \(\Delta x_{AB}\)

Using the given phase difference \(\Delta \phi = 0.7854\) rad and the wavenumber \(k = 157.08 \mathrm{~m}^{-1}\), find the difference in position: $$\Delta x_{AB} = \frac{0.7854}{157.08 \mathrm{~m}^{-1}} = 5.00 \times 10^{-3} \mathrm{~m} = 5.00 \mathrm{~mm}$$ c) Number of crests passing point A and number of troughs passing point B in 10.0 s

Identify the angular frequency

The wave equation is given as $$y(x, t)=(5.00 \mathrm{~mm}) \sin \left(\left(157.08 \mathrm{~m}^{-1}\right) x-\left(314.16 \mathrm{~s}^{-1}\right) t+0.7854\right) $$From the equation, we can read off the angular frequency \(\omega = 314.16 \mathrm{~s}^{-1}\).

Calculate the number of wave cycles that pass through points A and B in 10.0 s

The number of wave cycles that pass through points A and B in \(\Delta t = 10.0 \mathrm{~s}\) can be calculated using the angular frequency and the time interval: $$\text{Number of cycles} = \frac{\omega \Delta t}{2 \pi}$$

Calculate the number of crests and troughs

For point A: $$\text{Number of crests} = \frac{(314.16 \mathrm{~s}^{-1})(10.0 \mathrm{~s})}{2\pi} = 50$$For point B: $$\text{Number of troughs} = \text{Number of crests} = 50$$ d) Initial position of the linear driver to generate the wave

Find the initial phase at x=0

The wave equation is given as $$y(x, t)=(5.00 \mathrm{~mm}) \sin \left(\left(157.08 \mathrm{~m}^{-1}\right) x-\left(314.16 \mathrm{~s}^{-1}\right) t+0.7854\right) $$When \(x=0\), the initial phase for the driver is \(\phi_0=0.7854\).

Calculate the initial position of the driver

To generate a sinusoidal wave, the driver should start its trajectory at the same position as when the initial phase is \(0\). But, taking into account the phase \(\phi_0=0.7854\), we start the oscillation at $$y(0,0) = (5.00 \mathrm{~mm}) \sin (0.7854) = 3.54 \mathrm{~mm}$$This is the initial position of the linear driver required to generate this sinusoidal traveling wave on the string.

Key concepts

Wave Equation

Traveling waves play a crucial role in understanding wave phenomena on a string. The wave equation provides a mathematical formula to describe the position of such waves over time. In this case, the equation is given as:\[ y(x, t) = (5.00 \mathrm{~mm}) \sin \left(157.08 \mathrm{~m}^{-1} x - 314.16 \mathrm{~s}^{-1} t + 0.7854\right) \]This equation represents a sinusoidal wave traveling along a string, where:

• \( y(x, t) \) describes the wave's displacement at position \( x \) and time \( t \).
• The amplitude is 5.00 mm, indicating the maximum displacement from the rest position.

The wave equation helps visualize how wave properties like amplitude, frequency, and phase interact to produce wave motion. Breaking it down into components such as the wavenumber and angular frequency makes it easier to analyze the wave's behavior.

Phase Difference

In wave physics, phase difference is vital in determining how waves interact or move relative to each other. It is the angle that describes how much one wave is ahead or behind another in radians or degrees.For two points on our string wave to be in perfect opposition, they need a phase difference of an odd multiple of \(\pi\) (180 degrees). Mathematically, this phase difference is expressed as \(\Delta \phi = (2n+1)\pi\), where \(n\) is an integer.Using this principle, we find that the minimum distance (\(\Delta x_{\min}\)) between two points in perfect opposition on the wave is calculated. By knowing the phase difference and wavenumber, we derive this distance which helps predict the wave's oscillation behavior at different points.

Wavenumber

The wavenumber \(k\) is a key parameter in the analysis of waves, including those traveling on a string. It represents the spatial frequency of the wave, essentially telling us how many wave crests (or troughs) there are per unit distance.For the given wave equation, the wavenumber is \(k = 157.08 \, \text{m}^{-1}\). The wavenumber is usually defined as:\[ k = \frac{2\pi}{\lambda} \]where \(\lambda\) is the wavelength. The wavenumber helps us understand how tightly the wave is packed along the string. With a higher wavenumber, the wave crests are closer together, indicating a shorter wavelength. It directly relates to the determination of phase differences, as \( \Delta \phi = k \Delta x \). This relationship is essential for calculating the spatial separation of points with a specific phase relationship.

Angular Frequency

Angular frequency, denoted as \(\omega\), is fundamental in wave behavior since it describes the rate of oscillation in radians per second. In our wave equation, \(\omega = 314.16 \, \text{s}^{-1}\), which tells us how fast the wave oscillates over time.The angular frequency is linked to the regular frequency \(f\) by the relation:\[ \omega = 2\pi f \]It signifies the number of cycles a wave completes in a given time frame and is crucial for determining how many wave crests pass a point during a specified duration, such as 10 seconds in this exercise.Understanding \(\omega\) helps analyze full wave characteristics and predict its motion, contributing to comprehensive knowledge of wave propagation on strings.