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Chapter 17: Problem 38

Standing Wave A standing wave results from the sum of two transverse waves traveling in opposite directions given by $$ y_{1}=(0.05 \mathrm{~m}) \cos ((\pi \mathrm{rad} / \mathrm{m}) x-(4 \pi \mathrm{rad} / \mathrm{s}) t) $$ and $$ y_{2}=(0.05 \mathrm{~m}) \cos ((\pi \mathrm{rad} / \mathrm{m}) x+(4 \pi \mathrm{rad} / \mathrm{s}) t) $$ (a) What is the smallest positive value of \(x\) that corresponds to a node? (b) At what times during the interval \(0 \leq t \leq 0.50 \mathrm{~s}\) will the particle at \(x=0.00 \mathrm{~m}\) have zero velocity?

Short Answer

Expert verified
For part (a), x = 0.5 m. For part (b), the particle at x = 0 m has zero velocity at t = 0 s, 0.25 s, and 0.5 s.

Step by step solution

01

- Understand the given waves

The two given waves are: 1. \(y_{1} = (0.05 \text{ m}) \cos ((\backslashpi \text{ rad/m}) x - (4 \backslashpi \text{ rad/s}) t)\)2. \(y_{2} = (0.05 \text{ m}) \cos ((\backslashpi \text{ rad/m}) x + (4 \backslashpi \text{ rad/s}) t)\)
02

- Sum of the waves

Add the two waves to form the standing wave: \(y = y_{1} + y_{2}\)Using the trigonometric identity: \[\text{cos}(A) + \text{cos}(B) = 2 \text{cos} \left( \frac{A+B}{2} \right) \text{cos} \left( \frac{A-B}{2} \right)\]We have: \(y = 2(0.05 \text{ m}) \cos (\pi x) \cos (-(4 \backslashpi t)) \)Since \(\cos (-(4 \backslashpi t)) = \cos (4 \backslashpi t)\), the standing wave equation becomes:\[y = 0.1 \text{ m} \cos (\pi x) \cos (4 \backslashpi t)\]
03

- Find nodes from the standing wave

Nodes occur where the displacement is always zero. Thus: \(\cos (\pi x) = 0 \)Solving gives: \(\pi x = (2n + 1) \frac{\pi}{2} \)Thus: \(x = (2n + 1) \frac{1}{2}\)
04

- Smallest positive value of x for a node

The smallest positive \(x\) is when \(n = 0\). Thus: \[x = \frac{1}{2} \text{ m}\]
05

- particle velocity at x = 0

The particle鈥檚 velocity is found by differentiating the displacement with respect to time: \(v = \frac{dy}{dt}\)Substitute \(x = 0\) into the standing wave equation to get: \[y = 0.1 \cos (4 \pi t)\]Differentiate it: \[v = -0.4 \pi \sin (4 \pi t)\]For zero velocity, we have: \(\sin (4 \pi t) = 0\)Solving for time, \(t\), gives: \(4 \pi t = n \pi \Rightarrow t = n \frac{1}{4}\)
06

- Interval [0, 0.50 s)

For \(0 \leq t \leq 0.50 \text{ s}\), the valid solutions are found by letting \(n = 0, 1, 2\), giving:\[t = 0, 0.25 \text{ s}, 0.5 \text{ s}\]

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Key Concepts

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

Nodes in Standing Waves
In standing waves, nodes are points where the displacement remains zero at all times. They occur due to destructive interference between two waves traveling in opposite directions. For the waves given in the exercise, the standing wave equation can be found as: Since nodes are where the displacement is zero, we can set the equation equal to zero and solve for x. By doing this, we find that nodes happen at: This means that nodes occur at positions where the product of the cosine term is zero. The smallest positive value of x that corresponds to a node is when n = 0, giving us:
Particle Velocity in Waves
In a standing wave, the particle velocity indicates how fast a point on the wave is moving up and down. To find the velocity of a particle at any point in the wave, we differentiate the displacement function with respect to time. For the standing wave obtained from the sum of the two traveling waves, we focus on the point where x = 0. By substituting x = 0 into our standing wave equation and differentiating with respect to time, we get: Next, solving for times when this velocity is zero, we set the sine term in our equation equal to zero and solve for t. These times are the moments when the particle at x = 0 momentarily stops moving before reversing direction. Within the interval 0 鈮� t 鈮� 0.50 s, the times for zero velocity occur at:
Wave Superposition
Wave superposition is the principle that when two or more waves traverse the same medium, their displacements add up at every point. In this exercise, two transverse waves with identical amplitudes and frequencies are traveling in opposite directions. By using the superposition principle, we can sum these waves as: This results in a standing wave where certain points (nodes) always have zero displacement. The combined wave is characterized by the interference pattern of the two original waves. The phenomenon of standing waves is an important application of wave superposition, demonstrating both constructive and destructive interference throughout the medium.
  • Constructive Interference: where the waves align with each other, creating points of maximum amplitude.

  • Destructive Interference: where the waves cancel each other out, creating nodes.

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Most popular questions from this chapter

A String Under Tension A string under tension \(F^{\text {tension }}\) oscillates in the third harmonic at frequency \(f_{3}\), and the waves on the string have wavelength \(\lambda_{3} .\) If the tension is increased to \(\tau_{f}=4 \tau_{i}\) and the string is again made to oscillate in the third harmonic, what then are (a) the frequency of oscillation in terms of \(f_{3}\) and (b) the wavelength of the waves in terms of \(\lambda_{3} ?\)

Wire Clamped The tension in a wire clamped at both ends is doubled without appreciably changing the wire's length between the clamps. What is the ratio of the new to the old wave speed for transverse waves traveling along this wire?

Amplitude of the Resultant Determine the amplitude of the resultant wave when two sinusoidal string waves having the same frequency and traveling in the same direction on the same string are combined, if their amplitudes are \(3.0 \mathrm{~cm}\) and \(4.0 \mathrm{~cm}\) and they have initial phases of \(0.0\) and \(\pi / 2 \mathrm{rad}\), respectively.

Violin Strings The heaviest and lightest strings on a certain violin have linear densities of \(3.0\) and \(0.29 \mathrm{~g} / \mathrm{m} .\) (a) What is the ratio of the diameter of the heaviest string to that of the lightest string, assuming that the strings are of the same material? (b) What is the ratio of speeds if the strings have the same tension?

String Fixed at Both Ends A string fixed at both ends is \(8.40 \mathrm{~m}\) long and has a mass of \(0.120 \mathrm{~kg} .\) It is subjected to a tension of \(96.0 \mathrm{~N}\) and set oscillating. (a) What is the speed of the waves on the string? (b) What is the longest possible wavelength for standing wave? (c) Give the frequency of that wave.
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