Guest Posted February 9, 2015 Report Share Posted February 9, 2015 (edited) A question.If s = 2 sin( pi * t ) + sin (2 * pi * t ), t is bigger or equals to 0. When is the period of the function s? and why? Edited April 20, 2015 by Guest Reply Link to post Share on other sites More sharing options...
Vioh Posted February 9, 2015 Report Share Posted February 9, 2015 (edited) A really quite question.If s = 2 sin( pi * t ) + sin (2 * pi * t ), t is bigger or equals to 0. When is the period of the function s? and why? I'm not really sure what the best way is to explain this question. I suspect there must be a way to re-write function 's' into the standard form (from which we can derive the period); however, I haven't found it yet. But anyway, I'll to explain this the best I can using words: It's clear that 's' is just the superposition of 2 separate functions, and (we will forget about the number '2' in front of the sine, because the amplitude doesn't play a role in finding the period). In other words, if you put the graphs of these 2 functions on top of one another, and then superpose them, you'll get the function 's'. Now, it's pretty obvious that function has the period of 2, and function has the period of 1. Notice that these 2 periods are multiples of each other. So by the time has finished one period, has done 2 periods. And by that time, they will repeat this process once again. Hence, the period of function 's' is nothing more than the period of the function itself. Conclusion, 's' has the period of 2. I've already checked this with the calculator. Feel free to ask if you have further questions Edited February 9, 2015 by Vioh Reply Link to post Share on other sites More sharing options...
Guest Posted February 9, 2015 Report Share Posted February 9, 2015 (edited) A really quite question.If s = 2 sin( pi * t ) + sin (2 * pi * t ), t is bigger or equals to 0. When is the period of the function s? and why? I'm not really sure what the best way is to explain this question. I suspect there must be a way to re-write function 's' into the standard form (from which we can derive the period); however, I haven't found it yet. But anyway, I'll to explain this the best I can using words: It's clear that 's' is just the superposition of 2 separate functions, and (we will forget about the number '2' in front of the sine, because the amplitude doesn't play a role in finding the period). In other words, if you put the graphs of these 2 functions on top of one another, and then superpose them, you'll get the function 's'. Now, it's pretty obvious that function has the period of 2, and function has the period of 1. Notice that these 2 periods are multiples of each other. So by the time has finished one period, has done 2 periods. And by that time, they will repeat this process once again. Hence, the period of function 's' is nothing more than the period of the function itself. Conclusion, 's' has the period of 2. I've already checked this with the calculator. Feel free to ask if you have further questions I might need some time to digest it. Haha . BUT thank you so so much. Edited February 9, 2015 by Guest Reply Link to post Share on other sites More sharing options...
Guest Posted February 10, 2015 Report Share Posted February 10, 2015 now I get it. Thank you soso much. Reply Link to post Share on other sites More sharing options...
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