FIGURE P16.57 shows a snapshot graph of a wave traveling to the right along a string at 45 m/s. At this instant, what is the velocity of points 1, 2, and 3 on the string?

Question

Graph showing a wave function with points a, b, c, and wave traveling at 60 m/s.

Accepted Answer

Welcome back, everyone. We are making observations about this segment of a wave here and we are called a couple of different things. We are told that the velocity of our wave is 60 m per second. And we are tasked with finding what is the velocity of our wave at the positions of A B and see here. Now, I want to be very specific here. The velocity here of 60 m per second is the wave traveling in the positive X direction where our velocities at our different points are going to depend on something a little bit different here. Now, first and foremost, here's what I want to point out. I wanna point out or look at our point B here. Now, our point B is a local minimum. And so what we can say is that the velocity or the instantaneous velocity at our point B is going to be zero m per second. But what about A and C here? Well, A and C are zeros of our waves. So what does this mean? Well, what this means? Let me actually change colors here real quick that we have V of A is going to equal V of C where the magnitude of V fa is going to equal the magnitude of V F C which is just going to equal the magnitude of our maximum velocity here. Now how do we go about calculating that? Well, we have a formula for this, we have that V max is equal to omega times our amplitude. Now, we know that our angular velocity is equal to two pi times our frequency And we have that frequency is equal to our wave speed divided by our wave length. So substituting our equation for frequency into our equation for angular velocity. What we get is that angular velocity is equal to two pi times V divided by our wavelength. So what we can say is that V max is equal to the amplitude of our wave times two pi times V all divided by our wave length. Now, what is going to be our amplitude and our wave length? Well, we can see that our amplitude is given by the maximum height or maximum displacement from our D equals zero line here. So this is going to be our amplitude which we are told is eight centimeters. So really 0.8 m. So we have 0.8 times two pi times our velocity of 60 divided by our wavelength. Well, what is our wavelength going to be? Well, our wavelength comes from looking at the distance from one peak to another peak. And what we are told is that it is 80 centimeters from peak to peak. So that makes our wavelength 80 centimeters or 800.8 m. So then plugging this into a calculator, what we get is 38 m per second. So that's the magnitude for V of A and V of C. But what about the direction here as a reminder we have that V of B is zero meters per second. Let's look at V of A, we are going from positive to negative. So the V of A is going to be negative whereas for C we are going from negative to positive. So V of C will be positive. So now we have found our V of A V F B and V F C which corresponds to our final answer. Choice of C. Thank you all so much for watching. I hope this video helped. We will see you all in the next one.

FIGURE P16.57 shows a snapshot graph of a wave traveling to the right along a string at 45 m/s. At this instant, what is the velocity of points 1, 2, and 3 on the string?

Verified Solution

Key Concepts

Wave Velocity

Transverse Waves

Phase of a Wave