SHM graphing (1 Viewer)

[kimmeh]

i know this is an easy question.. but how do i do it ? lol



graph: x = 6 sin (2t + 45 degrees)

and can someone tell me how to graph in general :

x = a sin (nt + alpha)

or x = a cos (nt + alpha)

i get stuck on the ' alpha ' bit..

like... how is it shifted up or down?

 

[...]

hmm, max and min pt will be +6 and -6 as max of value of sin = -1 and +1

then u would want to know where the graph will hit in the x axis

so:
x=0
sin(2t+pi/4) =0 <---pi/4 = 45 degree
2t+pi/4= sin^-1 0
sin^-1 0 = 0, pi, 2pi

therefore:
2t+pi/4= 0 and pi and 2pi

and solving for t

then u would get t= -pi/8, 3pi/8, 7pi/8

so this sin curve would not exactly start at (0,0) but rather at -pi/8

hope thats correct

 

[kimmeh]

yeah it does start at -pi/8
the LHS of the axis... ummm
i meant.. where does it start on the y axis...
like u know how a normal sine curve starts at 0 and a normal cosine curve starts at 1...

 

[...]

well, this is a sin curve, so it will start at y=0, if its cos, then starts at y=0(or whatever amp given)

that equation will not affect where sine would start in terms of y-axis...that equation only effects the amplitude, how much its been shifted compare to y=sinx and the frequency

 

[kimmeh]

yeah i think i'm stuck in the 'shifted' part...

wait...

it starts on the y axis [in this case, the x axis] at 6/[root 2]

hmm lemme find it in a textbook first

 

[...]

woah? 6/sqr(2) ???

edit:sorry...

x=6 * sin(2t+pi/4)

sub t=0

sin=1/[root2]

therefore, 6* 1/[root2]

 

[kimmeh]

LOL i was about to say... ehe
"u could let t = 0"

 

[...]

so u get it now?

 

[kimmeh]

Originally posted by ...
so:
x=0
sin(2t+pi/4) =0 <---pi/4 = 45 degree
2t+pi/4= sin^-1 0
sin^-1 0 = 0, pi, 2pi

i ddint know u could sin^-1 both sides..

and how did u get pi? ASTC? but its equal to 0..

 

[...]

i wrote sin^-1 to show how i got those pi's...(ie what pi needed to get x=0)

and u do it from calculator and it gives u 45 degree, hence pi/4

so 2t+pi/4=0 hence t= pi/8
" 2t+pi/4=pi <----as sin(pi)=0, hence t=.......
and so on...

and what do u mean ASTC?? ASTC was to show the -ve and +ve of a trig function isn't it?
in this case u rather want a sine curve and know when sinx=0 so using ur calculator u would type sin^-1 0 (sin^-1 is above sin usually)
i inverse sin one side so 2t+pi/4 dun have sin function with it, hence solving for 't'

 

[kimmeh]

ahh i get it now.. i remeber my teacher telling me 'if u use astc.. ur going to get missing values.. thus always use the graph to find answers..'


in other words.. u did sin^-1 for both sides to get rid of the sin on the LHS.. right? .. lol sorry about this..

 

[...]

yea, i sin^-1 so i can get rid of sin on LHS and this answer tells me what digit will hit 0 for sinx

 

[kimmeh]

ohhh pokays ehehehee thanks alot !

now i wonder if i can cough this up in the exam

 

[...]

meh, just practise a few more and u'll know what to do...