Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
anishsaini10
Group Title
hii everyone...i urgently need a solution to my question due to my exams. My question is:
Show that two simple harmonic motions with frequency P and 2P when added will result in a periodic function of frequency P. Generalize the above for a number of harmonic functions with freqyencies P, 2P , nP, etc.
 3 years ago
 3 years ago
anishsaini10 Group Title
hii everyone...i urgently need a solution to my question due to my exams. My question is: Show that two simple harmonic motions with frequency P and 2P when added will result in a periodic function of frequency P. Generalize the above for a number of harmonic functions with freqyencies P, 2P , nP, etc.
 3 years ago
 3 years ago

This Question is Closed

occaam Group TitleBest ResponseYou've already chosen the best response.1
Hi, I realize that you might be look for a more mathematical answer and that what I will be offering is a more graphical answer, but I hope the proof will suffice. 1. Imagine a sine curve with a frequency P that has no phase value (the phase value is largely irrelevant given your question, but we will consider it later). The period is 1/P. The f(x)=0 points of the sign function are at 0, 1/2(P), P, 3/2(P), etc. At these points the function, if added to another function has no additive value. In between these points the function has, alternatively, an additive value, then a subtractive value, then an additive value, etc. 2. Now imagine a sine curve with a frequency 2P. Here the period is 1/2(P) and the f(x)=0 points are at 0, 1/4P, 1/2P, 3/4P, P, etc. Again, the additive value of f(x) at these points is 0 and between these points is +, , + alternatively. 3. Combine the two graphs in your mind, but take it step at a time. Notice something about the 0 points. Every time your original equation was at an f(x)=0 point, so was the second equation. At the point that you have gone through the original period of the original equation, this is extremely intuitive. The second equation has gone through exactly 2 periods and will come back to exactly the same point at which it started. For half the original period, using this easy example, the point is also probably quite obvious, but to answer the third part of your question (generalizing to nP), the key is that harmonic motion is symmetrical, so that any given function over its period can be divided into two parts, the positive and negative. (But we will leave this part of the discussion for a little later.) When the original function, is at a 0 point, the derived function (2P) will also be at a 0 point. Which means that you will have an additive value of 0 and the combined value will be 0. 4. Now to either side of those 0 points, 2 things are happening. 1) Your original function is adding a given value to whatever your derived function is or 2) it is subtracting that same value from your derived function. Your derived function is going to behave in one of two ways on either side of the zero function of your original function: a) it will be a repeat of itself if the coefficient is even (2P, 4P, 6P, etc.) or it will be a symmetrical reflection of itself if the coefficient is odd (3P, 5P, 7P....). For derived functions, the addition of the original function will have no impact at any given 0, then a positive impact, then no impact, then a negative impact, all of which is periodic across P. For derived functions in which the coefficient is even, that impact will create a pattern that has one particular type of reflection and for odd, a very different type, but for either, the combined function will have a 0 value at x=0, a 0 value at x=1/(2P), a 0 value at x=1/(P) at which point the combined function will repeat itself. The point, perhaps quite simple point, is that any function that has a period of nP where n is an integer, will have a 0 point at 1/P (because this function completes its cycle n times by 1/P), and a 0 point at 1/(2P) ( for odd n functions because harmonic functions are reflected at 1/2 their periods and for even n functions because 1/2n is still and integer). The impact of the adding the original function will not be the same between 0 points but rather will be the same every other 0 point and will either impact a function that looks exactly the same between original function 0 points (even n functions) or is the reflection of itself across original function 0 points (odd n functions). Furthermore, the rate of change that the original function gives to the derived function is always either the same value or the negative of that value at any given value d off one of its zero points (the nature of the harmonic motion). Ergo, you have created a situation in which a function repeats itself at N*(1/P) and reflects itself at (N*(1/P) + 1/(2P)), where N is any given integer. I hope this helps. I look back at my answer and realize that the language is nightmarish, but the picture of it in my head is beautiful. Warm regards, Occaam
 3 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.