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

Expert verified
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

01

Identify the wave equation

The given wave equation is $$y(x, t)=(5.00 \mathrm{~mm}) \sin \left(\left(157.08 \mathrm{~m}^{-1}\right) x-\left(314.16 \mathrm{~s}^{-1}\right) t+0.7854\right) $$
02

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.
03

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
04

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
05

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 \(\omega = 314.16 \mathrm{~s}^{-1}\).
06

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: $$\text{Number of cycles} = \frac{\omega \Delta t}{2 \pi}$$
07

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
08

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 \(\phi_0=0.7854\).
09

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.

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