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

step 1 Okay, this is a wave problem. For part one, we have this formula for our wave y1 of x is equal t

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

A wave has the form $y=A \cos (2 \pi x / \lambda+\pi / 3)$ when $x<0$. For $x>0,$ the wavelength is $\lambda / 2 .$ By applying continuity conditions at $x=0,$ find the amplitude (in terms of $A$ ) and phase of the wave in the region $x>0 .$ Sketch the wave, showing both $x<0$ and $x>0$

Key Concepts

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.

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.

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.

Transcript

00:01 Okay, this is a wave problem.

00:04 For part one, we have this formula for our wave y1 of x is equal to a cosine 2 pi x over lambda plus pi over 3 which is the face so this is for negative x and and for positive values for x, we have to find a and face for that.

00:44 So it should have the form of a, let's say two, cosine.

00:53 Everything is same except we have a new lambda.

00:57 We're going to call it lambda 2 plus a new phase, which we don't know what is that.

01:05 And the goal is finding that so we're going to call it five so by continuity conditions you're going to figure out what is a 2 and uh 5 for the positive values of x greater than 0 okay so what else is given in the question uh it said lambda 2 is equal to lambda over two so the frequency is twice so our waves are denser in the extraction okay so we have this it's good now what are the continuity conditions we have conditions for y so for positive and negative values of x y should be equal to each other and we have derivative from this guy which is d y d x and we have to solve them at x equal to zero so it means we have to have y1 equal to y2 at x equal to 0 and we have dy1 d x equal to d y2 d x at x equal to uh 0 now i'm gonna substitute these guys y1 and y2 here and solve these two equations and find a 2 and 5 okay let's do this so y1 equal to y2 is equivalent of saying a cosine pi over 3 equal to a 2 cosine a 2 cosine 5 and d y and y d y 1 over d x equal to d y to d x is equivalent of saying i'm going to write it here, minus 2 pi a over lambda.

03:39 I'm taking derivative of the first equation and putting x equal to 0.

03:47 It will always sign pi over 3 equal to minus 4 pi a 2 over lambda 2.

04:02 By phi.

04:05 So i'm going to divide the second of these equations by the first...

Please give Ace some feedback