Welcome back everyone. So in this problem, we're given a wave function. It's given as y of x and t is equal to 6.5 times the cosine. Then we got some numbers in this equation. We've got 2π/28 x and then 2π/0.36 t. In this problem, we want to calculate the wavelength and frequency of the wave, and then eventually we want to figure out what direction the wave travels in. So let's go ahead and get started here.
Part A, we want to calculate the wavelength. We know that the general form of an equation or a wave function is going to be a sine of kx plus or minus omega t or cosine. In this case, we have an amplitude, we have a cosine, and the value that goes in front of the x is k. Our omega over here is the 2π/0.36. We want to calculate what the wavelength is. That's not k or omega, but we can get the wavelength by looking at those values. Remember that k is related to the wavelength, and omega is related to the frequency or the period.
So to find the wavelength, we really just focus on that wave number given to us in our wave function. We know that k=2π/λ and in our problem here, in our wave function, it's given as 2π/28. So lambda is equal to 28 meters. It's straightforward to extract the wavelength when it's written in this form.
Now, for Part B, we want to calculate the frequency. Instead of looking for the wavelength by looking at k, if we want the linear frequency f, we're going to have to look at omega. Omega is represented as 2π×f or 2π/T. In our equation, this 2πf was equal to just 2π/0.36. So f=1/0.36, and this ends up being 2.78 Hz.
For the last part of the problem, we determine the direction the wave travels in. We just have to look at the sign between x and t. If the form is kx plus omega t, it means that you are traveling to the left. So, the plus sign indicates that the direction this wave travels in is left.
That's really all there is to it, folks. Let me know if you have any questions. Thanks for watching, and I'll see you in the next video.