how to i solve the amplitude 3sin(2x)+4sin(x)
The amplitude of one component is 3; the amplitude of the other is 4. When two sine waves are added together, regardless of their frequencies or phases, they will periodically be in phase and add together constructively. They will add to an amplitude of 7 at those points.
thx , is amplitude find through max/min pts?
Far be it from me to disagree with Chroot, but I don't see how sin(x) and sin(2x) will "constructively interfere" to give an amplitude of 3+ 4. 4sin(x)= 4 only for x= pi/2+ 2npi while 3sin(2x)= 3 only for 2x= pi/2+ 2mpi or x= pi/4+ mpi. That means we would have to have (m-2n)pi= pi/4 for some integers m and n.
The way I would do this problem is to find the maximum value of y= 3sin(2x)+ 4sin(x) as expscv suggested:
If y= 3sin(2x)+ 4sin(x) then y'= 6cos(2x)+ 4cos(x). cos(2x)= 2cos2(x)- 1 so this is y'= 12 cos2(x)+ 4cos(x)- 6= 0.
Solving that quadratic equation for cos(x), then converting that to sin(x) and sin(2x) (sin(x)= sqrt(1- cos2(x)) of course, and sin(2x)= 2sin(x)cos(x)), I get that 3sin(2x)+ 4sin(x) has a maximum value of approximately 6.10, the same thing I get by graphing y= 3sin(2x)+ 4sin(x).
Whoops. Integer multiple frequencies. :sheepish grin:
thx both in reply, espically HallsofIvy ,
I got too curious.
The red and blue are the two components. The black one the sum.
cool wat software is that?
Nice looking functions. You can do that with Excel as well, although it would be a bit less convenient.
It's called ROOT. You can download it from here. It is mainly used for high energy physics.
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