Shm (1 Viewer)

[George W. Bush]

your book is wrong.

many SHM questions have centres of motion that aren't x=0 e.g. every tide problem

if the particle is moving with SHM between x=3 and x=5, then x'' = -n^2(x-4)

 

[CrashOveride]

ok so is 'x' relative to origin or the centre of motion?
coz for a tide question i did, for it to work i have to take it from centre of motion...but in that example i just posted up x=cos^2(3t) you would have to take it from the origin ?

 

[CrashOveride]

Surely Bush(and Heinz) arnt the only people to have done SHM, come on you lot =p

 

[CrashOveride]

Ok say i had a tide that at high tide was 10m, low tide 6m. if one morning it was at low tide at 6am, when does it first reach 9m

Couldn't i "translate" this as being x=-2 to x=2, with the centre of motion as the origin? So i let t=0 x=-2...then later on find out what the time is to take to reach x=1 ...that would correspond to the time for it to go 6m --> 9m ?
x=acos(nt+&)
its going in the positive direction (even with sin) at t=0, x=-2 so it should work? i end up getting -ve time

i asked some ppl and they said u use something like x = b + acos(nt) to do these questions? its apparently outlined in the Excel book but its not listed in mine so any comments would be appreciated

 

[George W. Bush]

Originally posted by CrashOveride
Couldn't i "translate" this as being x=-2 to x=2, with the centre of motion as the origin? So i let t=0 x=-2...then later on find out what the time is to take to reach x=1 ...that would correspond to the time for it to go 6m --> 9m ?

Yes - but it will require a line or so writing what you are doing...personal preference i guess.

i asked some ppl and they said u use something like x = b + acos(nt) to do these questions? its apparently outlined in the Excel book but its not listed in mine so any comments would be appreciated

Tide starts at maximum distance from the origin - implies the use of cosine. Since it starts at its lowest point:

x= 8 - 2cos(nt)

 

[CrashOveride]

Also how do u just exclude the PHI from cos(nt) ?

Ok doing it this way now
x = 8 - 2cos(nt)

-1/2 = cos(nt)
nt = 2pi/3
since n = 4pi / 25
t = 4 +1/6
which is correct
and if i took the other angle, i get t = 8+1/3 ...so this means when the 2nd time it reaches 9 is 8hrs20min after 6am...so that would mean it goes from 9m to 10m and back to 9m again in the same time it took from 6m to 9m (4+1/6 hrs)

but also, wat if i said -1/2 = cos(nt)
1/2 = -cos(nt)
1/2 = cos(pi - nt)
pi - nt = pi/3
then i get t = 25/24 ?

and some -ve time if i took the angle as 5pi/3

Originally posted by George W. Bush
Yes - but it will require a line or so writing what you are doing...personal preference i guess.

i did it this way using cos...and i ended up with a -ve time

 

[George W. Bush]

Originally posted by CrashOveride
Also how do u just exclude the PHI from cos(nt) ?

you can sub t=0 if you insist, im pretty sure you can just look at it and say theta=0, this isn't general maths

Ok doing it this way now
x = 8 - 2cos(nt)

-1/2 = cos(nt)
nt = 2pi/3
since n = 4pi / 25
t = 4 +1/6
which is correct
and if i took the other angle, i get t = 8+1/3 ...so this means when the 2nd time it reaches 9 is 8hrs20min after 6am...so that would mean it goes from 9m to 10m and back to 9m again in the same time it took from 6m to 9m (4+1/6 hrs)

what's the other angle? n>0, t>0

but also, wat if i said -1/2 = cos(nt)
1/2 = -cos(nt)
1/2 = cos(pi - nt)
pi - nt = pi/3
then i get t = 25/24 ?

i don't know how you go that, since you have -nt = -2pi/3, which is the same thing as above.....

 

[CrashOveride]

Originally posted by George W. Bush
you can sub t=0 if you insist, im pretty sure you can just look at it and say theta=0, this isn't general maths

ah crap, sorry wasnt thinking

the other angle i got was 4pi/3...which gives me time as 100/12
because cos4pi/3 = -1/2

 

[CrashOveride]

Originally posted by George W. Bush
what's the other angle? n>0, t>0

The other angle was 4pi/3....and in getting it n>0...t>0
t turns out to be 100/12

If someone could just point out why this is is happening and im taking the other answer right...i think ive got my head around SHM

 

[CrashOveride]

^bump

 

[Harimau]

Originally posted by CrashOveride
^bump

You don't necessarily need to have an angle in the bracket, you can always manipulate or define the equation such that it would always start at the maximum or minimum.

This might sound weird, but i couldn't find the question you were asking. Could you please quote it again or something? Sorry...

 

[prusso]

just started shm - i got no fuckin idea, class is spending a long time learning it :S

 

[George W. Bush]

your second t will be the latter time when the tide is at that height - if you think about your graph of the displacement function, it starts at the bottom, goes up to the top and the time lies somewhere on that curve, it then goes back down to low tide again, your nt=4pi/3 time is on that part of the curve. So, if you were asked to find the range of times, it would be between those.

Originally posted by prusso
just started shm - i got no fuckin idea, class is spending a long time learning it :S

OH PLEASE
SHARE MORE DETAILS OF YOUR PERSONAL LIFE

jk

 

[CrashOveride]

Originally posted by CrashOveride
Ok doing it this way now
x = 8 - 2cos(nt)

-1/2 = cos(nt)
nt = 2pi/3
since n = 4pi / 25
t = 4 +1/6
which is correct
and if i took the other angle, i get t = 8+1/3 ...so this means when the 2nd time it reaches 9 is 8hrs20min after 6am...so that would mean it goes from 9m to 10m and back to 9m again in the same time it took from 6m to 9m (4+1/6 hrs)

So that's pretty much what ur saying? so the nt = 2pi/3 is the first time ? (you said it was the time on the "return" journey) and the other time (which was exactly double...like 8+1/3 hours) must actually be the time it comes backs to 9m on its return journey?

Also..if i did it the other way using x=aCos(nt + &) letting x=-2 when to = 0 etc etc.... i get nt + pi = pi/3, 5pi/3
Now if i take the 2nd angle, all is well. The first angle gives me -ve time...is it because since on the cos displacement graph we're starting at the lowest point...the lowest angle is actually giving me the time it was 9m because its going "backwards" on the graph so to speak (as cos starts from max positive displacement) ? and thus the -ve time ?

 

[George W. Bush]

from what you're saying i think you got it

if you draw the graph it should reveal all