For the simple harmonic motion equation , what is the period?
unaltered, sin and cos have a period of 2pi
if we alter it by say sin(kx) , then we know when kx=2pi, we begin a new cycle x = 2pi/k is therefore the period of an altered function of sin or cos
so how do we apply this information to your problem?
did you see the pic? I wasn't able to send the equation
i see the pic yes
wouldn't it be sin, because that is what is in the problem?
sin is the name of a function, its not a value of a period.
sin and cos have the same period ... you have sin in the problem yes from what ive posted, how do we apply this to our problem?
I'm not sure
ok, then lets approach it this way. lets say we have a function sin(u) what is the period of this function?
u?
u is the input variable ... it has no value. what is the normal period of the sine function? sin(u) has a period of what? ive already posted it in about my first posting if you need to reread it.
2pi
good that means that when u=2pi we start all over again u = 2pi, now lets redefine u as the arguement of your problem pi/3 x = u so lets replace u by this: pi/3 x = 2pi what is the value of x that makes this true?
sin(u) = sin(pi/3 x) sin(2pi) = sin(pi/3 x) when x = ......
-1 or 1?
we have reduced the problem to some pretty basic math here. pi/3 x = 2pi , how do we solve for x? to test your values: -pi/3 is not equal to 2 pi and: pi/3 is not equal to 2pi so x = 1 or -1 is not a solution pi/3 x = 2pi , how do we solve for x?
can we minus pi/3 from each side?
technically we can, but that doesnt help us out in the long run how about we start my multiplying both sides by 3? what do we get after that?
x=6pi, right?
oooh, close (3)pi/3 x = (3)2 pi pi x = 6 pi divide by pi and we get x=6, right?
yeah
then that is the value that starts the process over again; 6 is our period of sin(pi/3 x)
ok, thank you
youre welcome
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