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Two waves of the same frequency have amplitudes \(1.00\) and \(2.00 .\) They interfere at a point where their phase difference is \(60.0^{\circ} .\) What is the resultant amplitude?

Short Answer

Expert verified
The resultant amplitude is approximately 2.65.

Step by step solution

01

Convert Phase Difference to Radians

The phase difference is given in degrees, so we first convert it to radians. The conversion factor from degrees to radians is \( \pi/180 \). Thus, \( 60.0^{\circ} \) becomes \( \frac{60 \times \pi}{180} = \frac{\pi}{3} \) radians.
02

Formula for Resultant Amplitude

For two interfering waves, the resultant amplitude \( A_r \) can be calculated using the formula: \[ A_r = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} \]Here, \( A_1 = 1.00 \), \( A_2 = 2.00 \), and \( \phi = \frac{\pi}{3} \).
03

Calculate \( \cos(\phi) \)

Calculate the cosine of the phase difference. Since \( \phi = \frac{\pi}{3} \), we find \( \cos(\frac{\pi}{3}) = \frac{1}{2} \).
04

Substitute Values into the Formula

Substitute all known values into the formula from Step 2:\[ A_r = \sqrt{1.00^2 + 2.00^2 + 2 \times 1.00 \times 2.00 \times \frac{1}{2}} \]Simplify the equation:\[ A_r = \sqrt{1 + 4 + 2} \]
05

Simplify and Calculate Resultant Amplitude

Continue simplifying the equation:\[ A_r = \sqrt{7} \]Calculate the square root of 7 to find the resultant amplitude:\[ A_r \approx 2.65 \]

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Key Concepts

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

Amplitude
Amplitude is a key concept when discussing wave interference. It refers to the height or strength of a wave. Imagine a wave on the ocean; its amplitude would be the height from the calm water to the top of the wave.
This height tells us how much energy the wave carries.
In physics, amplitude is typically measured in units, such as meters, for mechanical waves or volts for electrical waves. For the problem at hand, we are dealing with two different amplitudes, 1.00 and 2.00. These values indicate the individual strengths of the two waves before they interact.
  • The larger the amplitude, the more energetic the wave.
  • Even if waves have the same frequency, differing amplitudes will affect their interaction outcome.
The concept becomes more important when waves interface since the combined or resultant wave's amplitude depends on both individual amplitudes and how they interact.
Phase Difference
Phase difference is like the timing between two repeating events. When dealing with waves, this timing or phase difference (phi \( \phi \)) helps determine how two or more waves interact. Consider two people starting to jump a skipping rope. If they start at the same time, they "in phase." If one starts jumping slightly later than the other, they are "out of phase."
In our scenario, the phase difference is given as 60 degrees, which is equivalent to \( \frac{\pi}{3} \) radians. Here's what to know about phase difference:
  • The greater the phase difference, the more the waves are desynchronized.
  • A phase difference of zero leads to maximum constructive interference, where waves amplify each other.
  • A phase difference of 180 degrees leads to complete cancellation, known as destructive interference.
The particular phase difference in our exercise affects how the strength or amplitude of the resultant wave is calculated.
Resultant Amplitude
Resultant amplitude is the outcome of wave interference. It's the amplitude of the new wave formed by combining two or more waves. This new wave is influenced by both the individual amplitudes and the phase difference between them. The formula to calculate the resultant amplitude for our situation is: \[ A_r = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} \]
Here, \( A_r \) is the resultant amplitude, while \( A_1 \) and \( A_2 \) are the amplitudes of the individual waves. \( \phi \) is the phase difference.
  • Resultant amplitude can be larger or smaller than individual amplitudes, depending on how waves interfere.
  • Constructive interference (small \( \phi \)) usually increases amplitude, while destructive interference (large \( \phi \)) reduces it.
In this exercise, using the formula, we determine that the resultant amplitude is approximately 2.65 when simplified. This provides a clear understanding of how two interacting waves combine in the given context.

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Most popular questions from this chapter

In a double-slit experiment, the fourth-order maximum for a wavelength of \(450 \mathrm{~nm}\) occurs at an angle of \(\theta=90^{\circ} .\) (a) What range of wavelengths in the visible range \((400 \mathrm{~nm}\) to \(700 \mathrm{~nm})\) are not present in the third-order maxima? To eliminate all visible light in the fourth-order maximum, (b) should the slit separation be increased or decreased and (c) what least change is needed?

Light of wavelength \(700.0 \mathrm{~nm}\) is sent along a route of length \(2000 \mathrm{~nm}\). The route is then filled with a medium having an index of refraction of \(1.400 .\) In degrees, by how much does the medium phase-shift the light? Give (a) the full shift and (b) the equivalent shift that has a value less than \(360^{\circ}\).

In a double-slit arrangement the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits. (a) What is the angular separation in radians between the central maximum and an adjacent maximum? (b) What is the distance between these maxima on a screen \(50.0 \mathrm{~cm}\) from the slits?

A disabled tanker leaks kerosene \((n=1.20)\) into the Persian Gulf, creating a large slick on top of the water \((n=1.30) .\) (a) If you are looking straight down from an airplane, while the Sun is overhead, at a region of the slick where its thickness is \(460 \mathrm{~nm}\), for which wavelength(s) of visible light is the reflection brightest because of constructive interference? (b) If you are. scuba diving directly under this same region of the slick, for which wavelength(s) of visible light is the transmitted intensity strongest?

If the distance between the first and tenth minima of a double-slit pattern is \(18.0 \mathrm{~mm}\) and the slits are separated by \(0.150\) \(\mathrm{mm}\) with the screen \(50.0 \mathrm{~cm}\) from the slits, what is the wavelength of the light used?

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