A wave has the form y=A cos(2 πx / λ+π/ 3) when x 0, the wavelength is λ/ 2 . By apply

A wave has the form y=A cos(2 πx / λ+π/ 3) when x<0 . For...

Key Concepts

Phase Shift and Amplitude Adjustment at Boundaries

When a wave encounters a boundary where its propagation properties change (such as a change in wavelength), the phase of the wave can shift and the amplitude may adjust in order to maintain continuity. By applying the continuity conditions at the boundary, one can determine the new amplitude and phase in the second region in terms of those in the first.

Wave Number and Wavelength Relationship

The wave number is defined as k = 2?/? and describes the spatial variation of the wave. When the wavelength changes between regions, the wave number changes accordingly, altering the phase factor in the wave's mathematical form. Understanding this relationship is key to analyzing how the wave’s phase and amplitude adjust at boundaries.

Wave Function Continuity

In wave mechanics, the wave function must be continuous at boundaries, meaning that the value of the function at the interface of two regions must be the same when approached from either side. This condition ensures that there are no sudden jumps in displacement and is used to find relationships between parameters (such as amplitude and phase) of waves in different regions.

Transcript

00:01 Hello, and in this question here, we're going to give an example of how to match boundary conditions.

00:08 So we have a wave, and for x less than zero, it is described by y1, and one refers to the fact that this wave is, that this is the description of the wave for x less than zero, and this is equal to a1, so the amplitude times the cosine of 2 pi divided by lambda 1, which is the wavelength when x is less than 0, times x plus a phase of pi over 3.

00:40 For x greater than 0, it is described by y2, and the subscript of 2 refers to the fact that this is when x is greater than, so this describes the wave when x is greater than 0, and this is equal to a wavelength, to an amplitude a 2 times the cosine of 2 pi divided by lambda 2 times x plus an unknown phase.

01:08 Okay? and we want to match boundary conditions to find a2 and phi 2 in terms of a1.

01:18 Okay? well, to do this, what are the boundary conditions we need to have? well, the first boundary condition is that when we approach x equals zero from the left, we'd want that to be equal to x equals zero from the right.

01:33 So what that actually means is that when we have y1, when x goes to zero, and we'd want that to be equal to y2 at x equals zero.

01:46 So that that is to say that this is continuous.

01:51 We also want the first derivative to be continuous, and that means that we want this condition to be true.

02:01 Well, technically speaking, it should be the limit as x goes to zero, but for this example here, we can just fill an x equal zero.

02:09 And that the limit of the first derivative as x goes to zero from the right hand side should be equal to the left hand side, which is what this condition here is.

02:19 Okay, so from these two conditions here, we hope to be able to find a2 and phi 2 in terms of a1.

02:29 Okay, so let's fill into the first condition.

02:32 So we have y1 of 0, and that equals a1 times the cosine of pi over 3.

02:43 And we also have y2 of 0, and that equals a2 times the cosine of the unknown phase.

02:52 5x2.

02:53 So setting y1 at x equals 0 equal to y2 at x equals 0 we can work out that a 2 is equal to a1 times the cause pi over 3 all divided by the cause of the unknown phase.

03:12 Now the cause of pi over 3 is equal to a half so this gives that a 2 is equal to a 1 sorry, there should have been an a1 there, all divided by 2 times the cosine of the unknown phase, phi 2.

03:29 Now, filling into the second boundary condition, which is this one here.

03:37 So first, we'll calculate the derivatives.

03:40 So, dy1, dx, is equal to minus a1 times 2 pi divided by lambda 1, times the sign of 2 pi divided by lambda 1 times x plus pi over 3 and the derivative of y2 with respect to x is equal to minus a 2 times 2 pi divided by lambda 2 times the sign of 2 pi divided by lambda 2 times x plus 52 which is the unknown phase.

04:18 So we want in the limit as x approaches zero from the left to be equal to x as equal to x approaching zero from the right.

04:30 So that means here that we set, we can just set this condition that i mentioned above.

04:42 So filling this in for x equals zero, we get minus a1 times 2 pi lambda 1 times sign of pi over 3 is equal to minus a 2 times 2 pi lambda 2 sign of 5 2 okay well how do we progress from here well from the question we're told that lambda 2 is equal to lambda 1 divided by 2 so that means we can do some simplification to get my so we can get a1 is equal to sign pi over three times equal to two times a two sign of five two okay now we can use the fact that sign of pi over three is equal to three root two okay so this equals to the sorry, square root of 3 divided by 2.

05:50 So we can use this to work out that a1, a1 divided by a2, is equal to 4 divided by the square root of 3 times the sign of 5 .2.

06:05 We also have, from here, that a1 divided by a2, well, that is equal to, we have a1 divided by a2, is also equal to twice the cause of phi 2...

Please give Ace some feedback