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(II) The velocity of waves on a string is \(96 \mathrm{~m} / \mathrm{s}\). If the frequency of standing waves is \(445 \mathrm{~Hz}\), how far apart are the two adjacent nodes?

Short Answer

Expert verified
The nodes are approximately 0.108 m apart.

Step by step solution

01

Understand the relationship between speed, frequency, and wavelength

The velocity of a wave on a string is given as \( v = 96 \, \text{m/s} \). The frequency of the waves is \( f = 445 \, \text{Hz} \). We need to find how far apart the nodes are, which is half of the wavelength (\( \lambda \)). The formula connecting speed \( v \), frequency \( f \), and wavelength \( \lambda \) is: \[ v = f \lambda \]
02

Calculate the Wavelength

Rearrange the formula for velocity to solve for wavelength \( \lambda \): \[ \lambda = \frac{v}{f} \] Substituting in the given values, \( v = 96 \ \text{m/s} \) and \( f = 445 \ \text{Hz} \), we get: \[ \lambda = \frac{96}{445} \approx 0.2157 \ \text{m} \]
03

Determine Node Separation

The distance between two adjacent nodes in a standing wave is half the wavelength. Therefore, the separation between nodes is: \[ \text{Node Separation} = \frac{\lambda}{2} = \frac{0.2157}{2} \approx 0.10785 \ \text{m} \]

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Key Concepts

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

Understanding Standing Waves
Standing waves are a fascinating phenomenon in physics that occurs when waves of the same frequency travel in opposite directions and interfere with each other. This results in a distinct wave pattern that appears to be "standing still" rather than traveling. They are common in musical instruments, where strings vibrate in this manner to produce sound. The most important feature of standing waves is the creation of nodes and antinodes. Nodes are points along the medium where there is no movement. At these points, the interfering waves cancel each other out. Meanwhile, antinodes are regions where the medium reaches maximum displacement because the interfering waves reinforce each other.
  • Standing waves form when incident and reflected waves superimpose.
  • Each harmonic pattern consists of nodes and antinodes.
  • Nodes remain stationary over time; however, energy is continuously passing through these points.
Understanding the formation and characteristics of standing waves is crucial when discussing node separation and wavelength calculation.
Node Separation in Standing Waves
Node separation refers to the distance between two consecutive nodes in a standing wave pattern. In practical terms, this is essentially half of the wavelength of the wave. Nodes are critical points in standing waves distinguished by the absence of motion, indicating that the wave’s destructive interference occurs here. This particular separation helps define the wave's characteristics.
  • Node separation = Half the wavelength.
  • Plays a role in defining the sound or pitch in musical instruments, linked to harmonic frequencies.
  • Aids in effective wave analysis since different musical or physical properties depend on node placement.
By simply understanding node separation, one learns more about the medium and the conditions in which standing waves arise.
How to Calculate Wavelength
Calculating the wavelength is a fundamental step in any wave-related problem. For a wave traveling on a string, such as in the given exercise, the wavelength can be calculated using the formula that relates the velocity of the wave, the frequency, and the wavelength itself. The formula is: \[ \lambda = \frac{v}{f} \]Here, \(v\) is the velocity of the wave, and \(f\) is the frequency. When you rearrange the formula to solve for wavelength, you find out how long one complete cycle of the wave is from crest to crest or trough to trough. In the example exercise provided, the values we plug into this formula were 96 meters per second for velocity and 445 Hz for frequency, leading to a wavelength of approximately 0.2157 meters.
  • The wavelength, symbolized by \( \lambda \), represents the length of one complete wave cycle.
  • Essential for understanding wave properties such as frequency and speed.
  • Directly affects node separation and subsequently, the physical characteristics of the medium.
Knowing how to compute the wavelength can help in determining other important properties like frequency, velocity, and node separation in wave phenomena.

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Most popular questions from this chapter

(I) Water waves approach an underwater "shelf" where the velocity changes from \(2.8 \mathrm{~m} / \mathrm{s}\) to \(2.5 \mathrm{~m} / \mathrm{s}\). If the incident wave crests make a \(35^{\circ}\) angle with the shelf, what will be the angle of refraction?

Two strings on a musical instrument are tuned to play at \(392 \mathrm{~Hz}(\mathrm{G})\) and \(494 \mathrm{~Hz}(\mathrm{~B}),(a)\) What are the frequencies of the first two overtones for each string? \((b)\) If the two strings have the same length and are under the same tension, what must be the ratio of their masses \(\left(m_{\mathrm{G}} / m_{\mathrm{A}}\right) ?(c)\) If the strings, instead, have the same mass per unit length and are under the same tension, what is the ratio of their lengths \(\left(\ell_{\mathrm{G}} / \ell_{\mathrm{A}}\right) ?(d)\) If their masses and lengths are the same, what must be the ratio of the tensions in the two strings?

(II) When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end, remaining relatively calm at the center. Suppose the frequency to produce such a standing wave in a \(45-\mathrm{cm}\) -wide tub is \(0.85 \mathrm{~Hz}\). What is the speed of the water wave?

(II) Determine if the function \(D=A \sin k x \cos \omega t\) is a solution of the wave equation.

(II) A transverse wave pulse travels to the right along a string with a speed \(v=2.0 \mathrm{~m} / \mathrm{s}\). At \(t=0\) the shape of the pulse is given by the function $$ D=0.45 \cos (2.6 x+1.2) $$ where \(D\) and \(x\) are in meters. (a) Plot \(D\) vs. \(x\) at \(t=0\). (b) Determine a formula for the wave pulse at any time \(t\) assuming there are no frictional losses. (c) Plot \(D(x, t)\) vs. \(x\) at \(t=1.0 \mathrm{~s} .\) (d) Repeat parts \((b)\) and \((c)\) assuming the pulse is traveling to the left. Plot all 3 graphs on the same axes for easy comparison.

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