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Chapter 17: Problem 5

Two waves are traveling in opposite directions on the same string. The displacements caused by the individual waves are given by \(y_{1}=\) \((24.0\) \(\mathrm{mm}) \sin (9.00 \pi t-1.25 \pi x)\) and \(y_{2}=(35.0\) \(\mathrm{mm}) \sin (2.88 \pi t+0.400 \pi x)\) Note that the phase angles \((9.00 \pi t-1.25 \pi x)\) and \((2.88 \pi t+0.400 \pi x)\) are in radians, \(t\) is in seconds, and \(x\) is in meters. At \(t=4.00\) s, what is the net displacement (in mm) of the string at (a) \(x=2.16 \mathrm{m}\) and (b) \(x=2.56 \mathrm{m} ?\) Be sure to include the algebraic sign ( \(+\) or \(-\) ) with your answers.

Short Answer

Expert verified
At both positions, the net displacement is 0 mm.

Step by step solution

01

Evaluate Displacement of First Wave at t = 4.00 s, x = 2.16 m

The displacement for the first wave at time \(t = 4.00\) s and position \(x = 2.16\) m is given by \(y_1 = (24.0 \ \text{mm}) \sin (9.00 \pi (4.00) - 1.25 \pi (2.16))\). Calculate the phase angle: \(9.00 \pi \times 4.00 - 1.25 \pi \times 2.16 = 36.00 \pi - 2.70 \pi = 33.30 \pi\). Now calculate \(y_1 = 24.0 \ \text{mm} \times \sin(33.30 \pi)\). Since \(\sin(33.30 \pi) = 0\), \(y_1 = 0\).
02

Evaluate Displacement of Second Wave at t = 4.00 s, x = 2.16 m

The displacement for the second wave at time \(t = 4.00\) s and position \(x = 2.16\) m is given by \(y_2 = (35.0 \ \text{mm}) \sin (2.88 \pi (4.00) + 0.400 \pi (2.16))\). Calculate the phase angle: \(2.88 \pi \times 4.00 + 0.400 \pi \times 2.16 = 11.52 \pi + 0.864 \pi = 12.384 \pi\). Now calculate \(y_2 = 35.0 \ \text{mm} \times \sin(12.384 \pi)\). Since \(\sin(12.384\pi) \approx 0\), \(y_2 = 0\).
03

Net Displacement at t = 4.00 s, x = 2.16 m

The net displacement \(y\) at time \(t = 4.00\) s and position \(x = 2.16\) m is the sum of \(y_1\) and \(y_2\). Therefore, \(y = y_1 + y_2 = 0 + 0 = 0\) mm.
04

Evaluate Displacement of First Wave at t = 4.00 s, x = 2.56 m

The displacement for the first wave at time \(t = 4.00\) s and position \(x = 2.56\) m is given by \(y_1 = (24.0 \ \text{mm}) \sin (9.00 \pi (4.00) - 1.25 \pi (2.56))\). Calculate the phase angle: \(9.00 \pi \times 4.00 - 1.25 \pi \times 2.56 = 36.00 \pi - 3.20 \pi = 32.80 \pi\). Now calculate \(y_1 = 24.0 \ \text{mm} \times \sin(32.80 \pi)\). Since \(\sin(32.80\pi) \approx 0\), \(y_1 = 0\).
05

Evaluate Displacement of Second Wave at t = 4.00 s, x = 2.56 m

The displacement for the second wave at time \(t = 4.00\) s and position \(x = 2.56\) m is given by \(y_2 = (35.0 \ \text{mm}) \sin (2.88 \pi (4.00) + 0.400 \pi (2.56))\). Calculate the phase angle: \(2.88 \pi \times 4.00 + 0.400 \pi \times 2.56 = 11.52 \pi + 1.024 \pi = 12.544 \pi\). Now calculate \(y_2 = 35.0 \ \text{mm} \times \sin(12.544 \pi)\). Since \(\sin(12.544\pi) \approx 0\), \(y_2 = 0\).
06

Net Displacement at t = 4.00 s, x = 2.56 m

The net displacement \(y\) at time \(t = 4.00\) s and position \(x = 2.56\) m is the sum of \(y_1\) and \(y_2\). Therefore, \(y = y_1 + y_2 = 0 + 0 = 0\) mm.

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Key Concepts

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

Wave Displacement
Wave displacement refers to how far a point on the wave moves from its equilibrium position. For any wave traveling through a medium, the displacement at any given time can be affected by factors like wave amplitude, frequency, and phase. In the given exercise where waves travel in opposite directions on the same string, we determine the displacement caused by each wave individually, and then sum them to find the net displacement.

For example, the displacement of the first wave is calculated by taking the amplitude (24.0 mm) and multiplying it by the sine of its phase angle. Similarly, the displacement of the second wave depends on its respective amplitude (35.0 mm) and phase angle. The displacement's sign, positive or negative, further tells us the direction of movement, ensuring wave interference's correct modeling. Understanding these calculations offers insights into real-world phenomena like vibration and wave interactions.
Phase Angle
The phase angle in a wave determines the wave's position at a particular point in time and space. These angles are given by the terms within the trigonometric function's argument, such as \(9.00 \pi t - 1.25 \pi x\) for the first wave and \(2.88 \pi t + 0.400 \pi x\) for the second wave. Phase angles are crucial because they tell us whether waves are in phase, out of phase, or anywhere in between.

  • An in-phase condition might reinforce the wave's energy, leading to larger displacements.
  • Out-of-phase waves can cancel each other, reducing the displacement amplitude.

In real-world applications, understanding phase relationships is essential for technologies like noise-canceling headphones, which use phase differences to eliminate unwanted sound.
Trigonometric Functions
Trigonometric functions, especially sine and cosine, are fundamental in describing waveforms. In wave mechanics, sine functions help depict oscillatory behavior, encapsulating information about wave displacement, amplitude, and phase. For example, the exercise uses sine functions to model both waves traveling in opposite directions along a string.

The sine of a phase angle yields values between -1 and 1, determining the wave's influence on displacement. Given the periodic nature of these functions, waves can exhibit repetitive patterns, such as crests and troughs, crucial for analyzing sound waves, light, or any oscillations. Deeper understanding of trigonometric functions aids in control and manipulation of waves, enabling advancements in various fields from telecommunications to music production.

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Most popular questions from this chapter

The E string on an electric bass guitar has a length of \(0.628\) \(\mathrm{m}\) and, when producing the note \(\mathrm{E},\) vibrates at a fundamental frequency of \(41.2\) \(\mathrm{Hz}\) Players sometimes add to their instruments a device called a "D-tuner." This device allows the \(\mathrm{E}\) string to be used to produce the note \(\mathrm{D},\) which has a fundamental frequency of \(36.7\) \(\mathrm{Hz} .\) The D-tuner works by extending the length of the string, keeping all other factors the same. By how much does a D-tuner extend the length of the E string?

A person hums into the top of a well and finds that standing waves are established at frequencies of \(42,70.0,\) and 98 Hz. The frequency of \(42\) \(\mathrm{Hz}\) is not necessarily the fundamental frequency. The speed of sound is \(343\) \(\mathrm{m} / \mathrm{s} .\) How deep is the well?

A string has a linear density of \(8.5 \times 10^{-3}\) \(\mathrm{kg} / \mathrm{m}\) and is under a tension of \(280\) \(\mathrm{N}\). The string is \(1.8\) \(\mathrm{m}\) long, is fixed at both ends, and is vibrating in the standing wave pattern shown in the drawing. Determine the (a) speed, (b) wavelength, and (c) frequency of the traveling waves that make up the standing wave.

A tube with a cap on one end, but open at the other end, has a fundamental frequency of \(130.8\) \(\mathrm{Hz} .\) The speed of sound is \(343\) \(\mathrm{m} / \mathrm{s}\) (a) If the cap is removed, what is the new fundamental frequency of the tube? (b) How long is the tube?

Divers working in underwater chambers at great depths must deal with the danger of nitrogen narcosis (the "bends"), in which nitrogen dissolves into the blood at toxic levels. One way to avoid this danger is for divers to breathe a mixture containing only helium and oxygen. Helium, however, has the effect of giving the voice a high-pitched quality, like that of Donald Duck's voice. To see why this occurs, assume for simplicity that the voice is generated by the vocal cords vibrating above a gas-filled cylindrical tube that is open only at one end. The quality of the voice depends on the harmonic frequencies generated by the tube; larger frequencies lead to higher-pitched voices. Consider two such tubes at \(20^{\circ} \mathrm{C} .\) One is filled with air, in which the speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) The other is filled with helium, in which the speed of sound is \(1.00 \times 10^{3} \mathrm{m} / \mathrm{s}\) To see the effect of helium on voice quality, calculate the ratio of the \(n\) th natural frequency of the helium-filled tube to the \(n\) th natural frequency of the air-filled tube.
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