Video: Finding the Resultant of the Superposition of Two Identical Waves with a Phase Difference between Them

Two sinusoidal waves are moving through a medium in the positive π‘₯-direction, both having an amplitude of 7.00 cm, a wave number of 3.00 m⁻¹, an angular frequency of 2.50 s⁻¹, and a period of 6.00 s, but one has a phase shift of an angle πœ‹/12 rad. What is the displacement of the resultant of these waves at a time 𝑑 = 2.00 s and a position π‘₯ = 0.530 m?

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Video Transcript

Two sinusoidal waves are moving through a medium in the positive π‘₯-direction, both having an amplitude of 7.00 centimeters, a wave number of 3.00 inverse meters, an angular frequency of 2.50 inverse seconds, and a period of 6.00 seconds, but one has a phase shift of an angle πœ‹ divided by 12 radians. What is the displacement of the resultant of these waves at a time 𝑑 equals 2.00 seconds and a position π‘₯ equals 0.530 meters?

Let’s begin by highlighting some of the vital information given to us. We’re told that the amplitude of both of these sinusoidal waves is 7.00 centimeters. We’ll call that capital 𝐴. We’re also told that each wave has a wave number of 3.00 inverse meters. We’ll call that value π‘˜. Each wave also has an angular frequency of 2.50 inverse seconds. We’ll call that πœ”. And they also each have a period of 6.00 seconds, which we’ll call capital 𝑇. And finally, one of the two waves has a phase shift of an angle of πœ‹ over 12 radians. We’ll call that phase shift πœ™.

What we want to solve for is the displacement of the resulting wave at a time when 𝑇 is equal to 2.00 seconds and π‘₯, the position, is 0.530 meters. We’ll call that displacement 𝑦, which itself is a function of position in time. With these terms defined, we can now move to solving for 𝑦 where π‘₯ is 0.530 meters and time is 2.00 seconds.

To solve for the amplitude of our resultant wave, we’re going to use the principle of superposition. This principle says that when waves overlap, they add or subtract from one another. When they add together, that’s called constructive interference. And when they subtract one from the other, that’s called destructive interference.

Our resultant wave amplitude 𝑦, which is a function of position in time, is equal to the combination or the sum of the two other waves mentioned in the problem statement. We’ll call these waves 𝑦 sub one of π‘₯ and 𝑇 and 𝑦 sub two of π‘₯ and 𝑇.

Before we begin to create 𝑦 sub one and 𝑦 sub two from the values given in the problem statement, let’s recall the general equation that describes a wave. The displacement of the material through which a wave travels, 𝑦 as a function of wave position in time, is equal to the wave amplitude, 𝐴, multiplied by the sine of the wave number π‘˜ times π‘₯, position, minus the angular frequency πœ” times time, 𝑇, plus πœ™, the phase angle of the wave.

We can apply this general equation to 𝑦 sub one and 𝑦 sub two, so that we know what terms to fill in for each. Let’s rewrite 𝑦 sub one and 𝑦 sub two as general wave equations. When we do this, we see that 𝑦 sub one of π‘₯ sub 𝑇 is equal to 𝐴 sub one times the sine of π‘˜ sub one π‘₯ minus πœ” sub one 𝑇 plus πœ™ sub one. And 𝑦 sub two of π‘₯ and 𝑇 is equal to 𝐴 sub two times the sine of π‘˜ sub two π‘₯ minus πœ” sub two 𝑇 plus πœ™ sub two.

As we look through the different terms of these equations and then to the right-hand side showing the given values from our problem statement, we recall that capital 𝐴, which is 7.00 centimeters, is the amplitude for both of our waves, 𝑦 sub one and 𝑦 sub two.

Likewise, the wave number π‘˜ of 3.00 inverse meters is the wave number for both of our waves. And πœ”, the angular frequency of 2.50 inverse seconds, is the angular frequency for both waves. The one difference comes with πœ™.

We’re told that the phase angle of πœ‹ divided by 12 radians is true for one of these two waves but not the other. So let’s choose that πœ™ one is equal to zero and πœ™ sub two is equal to πœ™, that is πœ‹ divided by 12 radians. We’re now ready to substitute in for all the terms of 𝑦 sub one and 𝑦 sub two. Let’s do that now.

We now, with this substitution, have an expression that says that the resulting amplitude of our combined waves is equal to the sum of two general waves with values given in the problem statement and written here. Now this equation gives us a general solution for any position, π‘₯, in any time, 𝑇. In the problem though, we’re asked for a solution at a specific position and specific time.

We want to know the resulting wave amplitude when π‘₯, the position, is 0.530 meters and when 𝑇, the time, is 2.00 seconds. This means that for each instance of the position π‘₯ we find in our equation, we substitute in 0.530 meters for that π‘₯. And for each instance of 𝑇 that we see, we substitute in the time value of 2.00 seconds.

When we make this substitution, we find a resulting wave amplitude from the combination of these two separate waves of 1.90 centimeters. That’s the amplitude that results from mixing these two waves when the position is 0.530 meters and the time is 2.00 seconds.

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