Velocity Derivative of a Sinusoidal Wave (Counter-Intuitive)

In summary, the problem the author is getting hung up on is that he can't seem to justify the derivative of the position function through the logic that one uses when manipulating mathematics to fit physical interpretation. He finds a solution to the problem repeated a couple different times, so he thinks he made a mistake.
  • #1
Alpha Scope
4
0
What's the matter:
So, I think I have some skills when it comes to differentiation after taking calculus 2 last semester, but when it starts to intertwine with physics, and interpreting physical phenomenon through equations, It appears I could use some help. Anyway, the problem that I got hung up on is admittedly simple by anyone's standards; introductory classical physics, more specifically dealing with waves.

If you take the dy/dt of position function, in this case a sinusoidal wave function, you find it's velocity function.
y(x,t)=xmsin(xk(0)-ωt) you should get:
U=-xmωcos(-ωt)
However, that's not what happens.
BTW. Two different sources attest to this being the right answer.
http://[URL=http://s172.photobucket.com/user/Alpha_Scope/media/Derivative%20Question_zpsryocyscm.jpg.html][PLAIN]http://i172.photobucket.com/albums/w15/Alpha_Scope/Derivative%20Question_zpsryocyscm.jpg
Derivative%20Question_zpsryocyscm.jpg


It's counter-intuitive to me since this seems like a simple derivative, and I'm unable justify it through the logic that one uses when manipulating mathematics to fit physical interpretation. As I mentioned previously, I'm still getting the hang of all this. Perhaps a little sooner with your help.

Anyone care to shed some light?
 
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  • #2
Your image does not show, you cannot link to things which are in your yahoo account and expect them to be visible to others.

Furthermore, your derivative is missing the kx of the cosine argument.
 
  • #3
Okay, should I expect it to appear now? I did so intentionally because x is said to equal 0.
 
  • #4
Alpha Scope said:
y(x,t)=xmsin(xk(0)-ωt) you should get:
U=-xmωcos(-ωt)
This is the correct derivative if x = 0 (or a multiple of pi/k).
 
  • #5
Apparently this is a typo, can't have ##\pi/t## there but the correct is ##\pi/5##. I am not sure though what you don't understand. Why dy/dt is the velocity of the point of matter at x=0? It is by the definition of velocity, we know that the point of matter at x=0 does an oscillation given by ##y_0=f(0,t)=-4sin(-\pi t/5)##, hence the velocity of this oscillation is ##\frac{dy_0}{dt}=\frac{df(0,t)}{dt}=-4(-\frac{\pi}{5})cos(-\frac{\pi t}{5})##.
 
  • #6
Okay, maybe there ain't a problem here after all. I found this solution (if I should even call it one) to this problem reiterated a couple different times, so I thought I made a mistake. I guess nothing's the matter, since this ain't my bad.

THANKS GUYS!
 
  • #7
Ok i see now the term ##\pi/t## if it was correct it would mean that the velocity goes to zero as time grows larger (goes to infinity) which we know it isn't correct for an oscillating point of matter. I wonder how can it be that many different sources reproduce this obvious(both mathematically and physically) mistake.
 
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Related to Velocity Derivative of a Sinusoidal Wave (Counter-Intuitive)

1. What is the velocity derivative of a sinusoidal wave?

The velocity derivative of a sinusoidal wave refers to the rate of change of the wave's velocity at a specific point in time. It is a measure of how quickly the wave's velocity is changing.

2. Why is the velocity derivative of a sinusoidal wave considered counter-intuitive?

The velocity derivative of a sinusoidal wave is considered counter-intuitive because it can be difficult to conceptualize how the velocity of a wave can change even though the wave itself is not changing. It may seem counter-intuitive that the velocity of a wave can change without any external force acting upon it.

3. How is the velocity derivative of a sinusoidal wave calculated?

The velocity derivative of a sinusoidal wave can be calculated by taking the derivative of the wave's velocity function, which is the first derivative of the wave's displacement function. This can be done using calculus principles.

4. What factors can affect the velocity derivative of a sinusoidal wave?

The velocity derivative of a sinusoidal wave can be affected by various factors, such as the amplitude, frequency, and wavelength of the wave, as well as the medium through which the wave is traveling. Changes in these factors can result in changes in the velocity derivative of the wave.

5. What is the significance of understanding the velocity derivative of a sinusoidal wave?

Understanding the velocity derivative of a sinusoidal wave is important in various fields of science and engineering, such as acoustics, optics, and electromagnetics. It can help us better understand and predict the behavior of waves in different mediums and can also be used to design and optimize technologies that rely on wave phenomena.

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