Find Antinode Closest to x=0.25m in Standing Wave

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In summary, the question asks to locate the antinode closest to x=0.25 m for the standing wave 0.005sin(30x)cos(420t) m. The formula for the distance between antinodes is λ/2, and the given solution uses the formula kXn=(2n+1)π/2 to find the value of Xn when n=2, which is 0.261 m. The conversation then moves on to discussing the spatial factor sin(kx) and the values of x that give the maximum amplitude, which are π/2, 3π/2, 5π/2, 7π/2, and so on. It is noted that
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lha08
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Homework Statement


Consider the standing wave: 0.005sin(30x)cos(420t) m. Locate the antinode closest to x=0.25 m.

Homework Equations





The Attempt at a Solution


I'm not really sure how to tackle this problem except that i know that antinodes occur at the crests of the standing wave and that the distance between antinodes is lamda/2. In the answers it says that: kXn=(2n+1)pi/2... Xn=(2n+1)pi/(2k) and when you plug in n=2 it is 0.261 m...does anyone know how they got the initial formula?
 
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  • #2
Consider the spatial factor sin(kx). What values of x give the maximum amplitude?
 
  • #3
Doc Al said:
Consider the spatial factor sin(kx). What values of x give the maximum amplitude?

aren't they pi/2, 3pi/2, 5pi/2, 7pi/2...etc?
 
  • #4
lha08 said:
aren't they pi/2, 3pi/2, 5pi/2, 7pi/2...etc?
Good. But that's the value for kx, not x. (In this problem, k = 30.) How would you write that series using n, where n = 0, 1, 2, and so on? Compare that to the formula given.
 

1. What is a standing wave?

A standing wave is a type of wave that forms when two waves with the same frequency and amplitude travel in opposite directions and interfere with each other. This results in a wave pattern that appears to be standing still, hence the name "standing wave".

2. What is an antinode?

An antinode is a point on a standing wave where the amplitude of the wave is at its maximum. In other words, it is the point on the wave where the particles are moving with the greatest displacement from their resting position.

3. How do you find the antinode closest to x=0.25m in a standing wave?

To find the antinode closest to x=0.25m in a standing wave, you will need to first determine the wavelength of the wave. This can be done by measuring the distance between two adjacent nodes (points of zero amplitude) on the standing wave. Once you have the wavelength, you can use the formula λ=nL/2, where n is the number of antinodes and L is the length of the standing wave, to find the distance between each antinode. From there, you can determine which antinode is closest to x=0.25m.

4. Why is finding the antinode closest to x=0.25m important?

Finding the antinode closest to x=0.25m is important because it allows us to accurately measure the wavelength of the standing wave. This information can then be used to calculate other important parameters such as the frequency and velocity of the wave.

5. What are some practical applications of understanding standing waves and antinodes?

Understanding standing waves and antinodes has many practical applications, including in the fields of acoustics and musical instruments. The concept of standing waves is used in the design of musical instruments, such as guitars and pianos, to produce specific sounds and notes. Antinodes are also important in the study of resonance, which is used in fields such as engineering and medicine. Additionally, standing waves are used in various scientific experiments, such as in the study of quantum mechanics and atomic structure.

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