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Two out-of-phase radio antennas at \(x=\pm 300 \mathrm{m}\) on the \(x\) axis are emitting \(3.0 \mathrm{MHz}\) radio waves. Is the point \((x, y)=\) \((300 \mathrm{m}, 800 \mathrm{m})\) a point of maximum constructive interference, perfect destructive interference, or something in between?

Short Answer

Expert verified
The interference is neither maximum constructive nor perfect destructive. The interference is somewhere in between.

Step by step solution

01

Calculate Distance to Antennas

Begin by calculating the distance from the given point to each antenna. The distances \(d_1\) and \(d_2\) from the point \((300 m,800 m)\) to the antennas at \(x=\pm 300\) m are found using the Pythagorean theorem \(d=\sqrt{x^2+y^2}\). This results in \(d_1=\sqrt{(-300m)^2+(800m)^2}\) and \(d_2=\sqrt{(300m)^2+(800m)^2}\).
02

Calculate Phase Difference

The phase difference between the signals at the point in question depends on the difference in distances the waves have traveled from the two antennas. It is calculated using the wave properties. The formula to calculate phase difference is \(\Delta\phi=2\pi* \frac{\Delta d}{\lambda}\)
03

Calculate the Difference Between the Distances

Calculate \(\Delta d=D_2-D_1\) where \(D_2\) and \(D_1\) are the distances, substituting the values obtained from step 1.
04

Calculate Wavelength

Calculate the wavelength of the radio waves \(\lambda\). Use the formula \(\nu=c/\lambda\) where \(\nu=3.0MHz\) is the frequency and \(c=3x10^8m/s\) is the speed of light.
05

Compute the Phase difference

Compute the phase difference by substituting \(Delta d\) and \(\lambda\) in phase difference formula calculated in step 2.
06

Determine Interference

If the phase difference is a multiple of \(2pi\), then it indicates maximum constructive interference. If the phase difference is an odd multiple of \(pi\), this suggests perfect destructive interference. In other cases, the interference will be somewhere in between.

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Key Concepts

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

Pythagorean Theorem
The Pythagorean Theorem is a fundamental principle in geometry that relates to the lengths of the sides of a right-angled triangle. It's a valuable tool when you need to find the distance between two points in a coordinate system. The theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Essentially, if you have a triangle with sides labeled as "a," "b," and "c鈥 (where 鈥渃" represents the hypotenuse), the relationship can be expressed as:
  • \( c^2 = a^2 + b^2 \)
In the case of the radio antennas, we use the Pythagorean Theorem to calculate the straight-line distances from each antenna to a specific point in the coordinate plane. These distances (\(d_1\) and \(d_2\) for antennas at \(x = -300 \ \text{m} \) and \(x = +300 \ \text{m} \)) help in understanding how the radio waves travel and interact at the target point.
Phase Difference
Phase difference is a key concept in wave interference, representing the difference in the phase of two points on separate waveforms. When dealing with interference from two sources like antennas, phase difference tells us whether the waves will reinforce or cancel each other. It's calculated based on the difference in the paths traveled by the waves from each source. The formula used to find phase difference \( \Delta \phi \) is:
  • \( \Delta \phi = 2 \pi \frac{\Delta d}{\lambda} \)
Here \( \Delta d \) is the path difference, and \( \lambda \) is the wavelength. Understanding the phase difference helps predict whether the waves at a specific point result in constructive interference (amplified waves) or destructive interference (cancelled waves). If \( \Delta \phi \) equals a multiple of \( 2\pi \), it indicates full alignment, resulting in constructive interference. Conversely, if \( \Delta \phi \) is an odd multiple of \( \pi \), complete cancellation occurs, leading to destructive interference.
Wavelength Calculation
Wavelength calculation is a fundamental step in understanding wave behavior. The wavelength \( \lambda \) is the distance between two consecutive points of a wave in phase and is inversely proportional to the frequency of the wave. For electromagnetic waves like radio waves, the relationship between speed, frequency, and wavelength is given by the formula:
  • \( c = u \cdot \lambda \)
Where \( c \) is the speed of light (\(3 \times 10^8 \ \text{m/s}\)), \( u \) is the frequency of the wave, and \( \lambda \) is the wavelength. Solving this formula for \( \lambda \) allows us to determine the wavelength when the frequency is known:
  • \( \lambda = \frac{c}{u} \)
In the context of the antennas, this calculation is vital for determining the precise phase difference between the wave sources, which ultimately decides the type of interference occurring at a given point.

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Most popular questions from this chapter

A 121 -cm-long, \(4.0 \mathrm{g}\) string oscillates in its \(m=3\) mode with a frequency of \(180 \mathrm{Hz}\) and a maximum amplitude of \(5.0 \mathrm{mm}\) What are (a) the wavelength and (b) the tension in the string?

A soap bubble is essentially a very thin film of water \((n=\) 1.33) surrounded by air. The colors that you see in soap bubbles are produced by interference, much like the colors of dichroic glass. a. Derive an expression for the wavelengths \(\lambda_{\mathrm{c}}\) for which constructive interference causes a strong reflection from a soap bubble of thickness \(d\) Hint: Think about the reflection phase shifts at both boundaries. b. What visible wavelengths of light are strongly reflected from a 390 -nm- thick soap bubble? What color would such a soap bubble appear to be?

An open-open organ pipe is \(78.0 \mathrm{cm}\) long. An open-closed pipe has a fundamental frequency equal to the third harmonic of the open-open pipe. How long is the open-closed pipe?

Il A \(40-\mathrm{cm}-\) long tube has a \(40-\mathrm{cm}-\) long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As the insert is slowly pulled out, the sound from the tuning fork creates standing waves in the tube when the total length \(L\) is \(42.5 \mathrm{cm}, 56.7 \mathrm{cm},\) and \(70.9 \mathrm{cm} .\) What is the frequency of the tuning fork? Assume \(v_{\text {sound }}=343 \mathrm{m} / \mathrm{s}\).

What is the thinnest film of \(\mathrm{MgF}_{2}(n=1.39)\) on glass that produces a strong reflection for orange light with a wavelength of \(600 \mathrm{nm} ?\)

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