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GR8677 #4
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Wave Phenomena$\Rightarrow$}Wave Equation

Perhaps this formula is more familiar: $y=A\sin(\omega t - k x)$. But then again, they define the familiar quantities $k$ and $\omega$ for you in theirs...

(A) The amplitude is A. (Recall that the amplitude of $y=sin(x)$ is just 1, not 2.)

(B) How exactly does the argument of $\sin$ make it a traveling wave? Well, a traveling wave keeps the same waveform at all times. Thus, the overall argument has to be constant. $\omega t - kx = constant\Rightarrow x \propto - constant +\omega t$. This looks like the old high school kinematics equation $x=-constant + vt$. In fact, it's basically the same thing. That high school equation describes a particle going in the positive x-direction. So does this equation.

(C) Dimensions don't match. Period has units of time, not units of time/meter.

(D) The speed of the wave is $v=\lambda/T$. From the kinematics explanation explained in (B), the velocity of the wave is obviously $v=\frac{\omega}{k}=\frac{2\pi\lambda}{2\pi T}$

(E) True. See (D) and (B).

Alternate Solutions
 syreen2013-09-23 20:27:12 the argument looks like a traveling wave sin(wt-kx) where w=1/T, k=1/lamda. Know that vphase=w/k and so w/k=lamda/TReply to this comment
syreen
2013-09-23 20:27:12
the argument looks like a traveling wave sin(wt-kx) where w=1/T, k=1/lamda. Know that vphase=w/k and so w/k=lamda/T
Prologue
2009-10-12 08:38:10
I think the explanation for B needs to be spelled out a little better. The wave looks the same in 'some reference frame traveling through space' no matter what time it is. This means you need to find a relationship between x and t such that the argument for sine gives a constant output through all time for the same x' - where x' is not the velocity but the space coordinate in the new moving frame. We need to find a $\frac{dx}{dt}$ such that $\frac{d[sin(x')]}{dt}=0$. If you want this to be independent of time then that means the argument has to be constant with respect to time. So $2\pi(\frac{t}{T}-\frac{x}{\lambda})=constant=C$. Now you can solve for x to get $x=(\frac{\lambda t}{T}-\frac{\lambda C}{2 \pi})$. Now you have the time dependence for x and you can just take the time derivative to get $\frac{dx}{dt}=v=\frac{\lambda}{T}$.

I know this is long winded but these little details might be nice for someone that isn't comfortable with shifting.
 Prologue2009-10-12 10:54:45 That $\frac{d[sin(x')]}{dt}$ should logically be a $\frac{d[sin(\frac{2 \pi}{\lambda}x')]}{dt}$ but it doesn't change the rest of the post at all.
anmuhich
2009-03-19 08:39:59
Or you just know that for a sine wave the velocity is just the wavelength over the period.
icelevistus
2007-09-27 16:29:19
The only variables/constants we can assume the units of are x and t. All others are unspecified, despite what convention suggests.

Thus, the application of dimensional analysis for eliminating answers only works for D.
 icelevistus2007-09-27 16:30:20 Never mind, the units are implicit for the sin argument to be unitless.

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