Constructive and Destructive Interference of Waves at pi/50 Time - Find Points

[Tom McCurdy]

Homework Statement

y1(x,t)=4*cos(20*t-30*x)
y2(x,t)=4*cos(20*t+30*x)
set t=pi/50 and find the points where the wave interferes constructively and destructively

The Attempt at a Solution

So I tried to take the derivative of y1 and y2 added together and set it equal to zero.

$$120*sin(\frac{2 \pi}{5}-30x) = 120*sin(\frac{2 \pi}{5}+30x)$$

then I took out the 120 on both sides, but I am not sure what to do from their exactly. I tried to take the arcsin a few different ways but I couldn't get anything concrete.

I know the answer is $$x= \frac{\pi}{60}+\frac{2 \pi n}{30}$$ for constructive
and $$x= \frac{n \pi}{30}$$ for destructive

this makes sense to me since the period would be 2pi/30 but I don't get how they actually got there. Could someone point me in the right direction

 

[Mindscrape]

Since both of these waves have the same amplitude and the same angular frequency when they interfere they will give 0 for destructive interference and wave*2 for constructive.

wave1 + wave2 = 0 happens when?

how does the constructive go?

think about what a phase really means

 

[m2003]

?Firstly, it's important to note that the given equations represent waves traveling in opposite directions with the same frequency and amplitude. This means that when they intersect, they will either add constructively (creating a larger amplitude) or destructively (canceling each other out).

To find the points where the waves interfere constructively or destructively, we need to set the two equations equal to each other and solve for x. This is because at these points, the waves will have the same displacement (y) and therefore will either add or cancel each other out.

So, setting y1(x,t) = y2(x,t) and substituting t=pi/50, we get:

4*cos(20*(pi/50)-30*x) = 4*cos(20*(pi/50)+30*x)

Simplifying and using the cosine addition formula, we get:

cos(4pi/5 - 30x) = cos(4pi/5 + 30x)

For constructive interference, cos(4pi/5 - 30x) = cos(4pi/5 + 30x) = 1, which means that 4pi/5 - 30x = 4pi/5 + 30x + 2n*pi (where n is an integer). Solving for x, we get:

x = (4pi/5 - 4pi/5 - 2n*pi)/60 = (2n*pi)/60 = n*pi/30

So, the points where the waves interfere constructively are x = n*pi/30, where n is an integer.

Similarly, for destructive interference, cos(4pi/5 - 30x) = cos(4pi/5 + 30x) = -1, which means that 4pi/5 - 30x = -(4pi/5 + 30x) + 2n*pi. Solving for x, we get:

x = (4pi/5 - 4pi/5 - 2n*pi)/60 = (-2n*pi)/60 = -n*pi/30

So, the points where the waves interfere destructively are x = -n*pi/30, where n is an integer.

In summary, the points where the waves interfere constructively are x = n*pi/30 and the points where they interfere destructively are x = -n*pi/30, where n is an