+1313 Given that D= 0.48 sin (5.6x +84t) mm, and standard notation given is, x = Acos (Omega*t + Fi), Omega is 2pi*frequency, and fi is the displacement of it's start point along the x-axis, please elaborate on the logic behind identifying and using this equation and its parts. Such that I can easily know in the first equation how to find wavelength, frequency and velocity (magnitude and dirrection). Also, what is the mm specifying/how can I tell or is it the standard unit measurement here?
I can identify velocity as the dx/dt. Amplitude as the constant in front of sin. I didn't clearly identify how 5.6 is lambda or its relationship to the variable x. Frequency given 2pi*f where f is = to 84.So is that always an assumption I can make that 84 is per unit of time (t) of 1 second? I mean t could be anything right but frequency is set, so 84 times where t is .5 seconds is not the same frequency as where t has changed. Or is it that t and x are not unkown and they are just place markers or something? Ok Thanks for who can just give me some logical steps regarding this.
+33661 You are probably thinking of the approximation that sin(θ) ≈ θ when θ is very small. But this only holds when the angle is in radians. For example suppose θ = 0.1 radians (=5.73°)
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{5.73}}^\circ\right)} = {\mathtt{0.099\: \!840\: \!745\: \!983}}$$
So sin(0.1radians) = sin(5.73°) ≈ 0.1
Or sin(0.1radians) ≈ 0.1
sin(5.73°) ≠ 5.73
This is true whenever you have the sin of a small angle, no matter the context.
+1313 Um I was doing calculations and my unit notation in meters for waves. Is that not correct? Or where in lies the difference. Thanks agian for a great reply and the time to answer. I have to absorb all that in case it pops up in the exam. Cheers.
+33661 What you wrote was mm. This means millimetres; metres would just be m (1mm = 0.001m).
+1313 Regarding the information of this post: I recall in lectures that for angles less than (a small number of aproximation 13 degrees, the calculator should be in radians, of the top of my head. Was that regarding this wave formula or another area, or in general and can you elaborate please. Thanks.
+33661 You are probably thinking of the approximation that sin(θ) ≈ θ when θ is very small. But this only holds when the angle is in radians. For example suppose θ = 0.1 radians (=5.73°)
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{5.73}}^\circ\right)} = {\mathtt{0.099\: \!840\: \!745\: \!983}}$$
So sin(0.1radians) = sin(5.73°) ≈ 0.1
Or sin(0.1radians) ≈ 0.1
sin(5.73°) ≠ 5.73
This is true whenever you have the sin of a small angle, no matter the context.
+8262 Wo is giving you too many negative points, Stu? I am trying to replace them.
+1313 Great answer again Alan.
Not sure your meaning DragonSlayer554 unless there were troll comments again.
I really appreciate that you answered the entire question Alan. As you know there is a lot to absorb and mostly its in details such as these. The stuff that is not only hard to hold on to in the memory, but is not common but is important and which also is not covered in depth. This is also compounded by a few general knowledge gaps or logical though processes for understanding the material on the first attempt, for most people.
Thanks.
