<p>if the wavelength given was 2m… and it moved half a period,</p>
<p>where y is landa for wavelength:
v = fy = y times 1/T = y/T, T would equal 1/10 since it’s half a period (1/20 * 2 for a full period)</p>
<p>thus v = 2/ (1/(1/10)) = 20 m/s… i answered 40 m/s but i want to why if you said it moved half a wavelength, you used 2… did the drawing only show 2 to be half the wavelength?</p>
<p>Also, for the 20Hz, was the 0.4m not the wavelength? Because if it was, v = fy = 20 times 0.4 = 8 m/s… confused with what you did again, why is your equation solving for f if it’s known? j/c about ur thought process</p>
<p>also,
the image of ur last comment:</p>
<p>positive conductive rod is held near a neutral conducting metal with two ends R and B (R on the side of nearer the rod)</p>
<p>and then it provided options such as R is positive, B is positive, RB is positive, RB is neutral etc. All I remember saying is that I said R was negative because the e-'s would go as close to the rod as possible…?</p>
<p>Does anyone remember the question concerning quasars? It was something about what the age of a quasar, 6 billion years if I recall correctly, implied about quasars in general. I believe that one of the answers had something to do with quasars existing in the milky way, one had to do with quasars having been around since near the beginning of the universe, and one answer had something to do with red shift. I don’t remember what the fourth choice was, and I’m not sure what the correct answer actually was. Anyone know?</p>
<p>Sorry if my post confused you let me elaborate:</p>
<p>For the sinusoidal graph, the graph was not distance versus time - it was distance versus distance. From the first graph, it started like a sin graph. The intersection points were at 0m, 2m, 4m, etc. From the second graph (looked like cos ), the intersections were at 0m, 2m, 4m, etc. Remember that this is distance versus distance (position y vs. position x). Also note that the time change between the two graphs was 1/20 seconds. So since the first graph was sin and the second graph was cosine, there was no way the object could have traveled 4m (or else the second graph would look exactly like the first). So it could be reasoned from the graph that the object traveled only 2 meters. It asked for the velocity - 2m / (1/20 seconds) = 40 meters / second.</p>
<p>For the standing wave, the LENGTH (not wavelength) was .4. There is an equation that states f = nv / 2L. Solving for v, v = f2L / n. N = 1 because it was a standing wave so therefore v = 2<em>f</em>L = 2* 20 * .4 = 16 m/s</p>
<p>For the RB, i believe that was the correct answer</p>
<p>For the wavefront at P, the waves are 1m in difference when looking at incident paths taken, and they are in phase with a wavelength at .5m. This means that at P, there will be a maximum. Since it is “leading” to the maximum at P, the amplitude to the left of P will be increasing towards this maximum values. Just because this value is lesser than the amplitude does not mean that the amplitude is decreasing. Since the waves leave each other afterward, the amplitude decreases to the right of P.</p>