applicable? (1 Viewer)

[JamiL]

in 2001 Q4 they say,
a particle, who displacement is x, moves in simple harmonic motion.
find x as a function of t, if a=-4x and if x=3 v=6root3 when t=o

i assumed x=acos (nt+@) since it was S.H.M and when on from there 2 find that
x=6cos(2t + 60*) which was the right answer, althou my principle seems 2 not believe that it is a valid proof, althou my teacher who has been teaching 4 30 years seems 2 fink it is, although he has neva applyed 2 do hsc marking, and wouldn no. is it valid to simply assume that x=acos(nt + @).
P.S in sucess 1 they do it my principles way, its just it is so long i dont get the point of doin it that way when i can easly getthe same result in a few lines??
how who any of us go about it?

 

[FinalFantasy]

pretty sure u can assume stuff like x=asin(nt+@), x=acos(nt+@), x=bsin (nt)+c cos (nt).... ect

 

[acmilan]

From syllabus:

An alternative treatment of simple harmonic motion begins with the differential equation a = –n²x and then uses the inverse trigonometric functions to derive a solution of the form x = a cos (nt + a). A rigorous treatment along these lines is difficult because x is not an invertible function of t (and v is not a function of x) unless the range of x is restricted. It can be observed that x = x₀ cos(n(t – t₀)) + (v₀/n)sin(n(t – t₀)) is a solution of the equation, which satisfies the additional conditions x (t₀) = x₀, v(t₀) = v₀. Questions of uniqueness of solution are outside the scope of the syllabus.

My interpretation of that is that you dont need to know how to do it.

 

[JamiL]

From syllabus:

An alternative treatment of simple harmonic motion begins with the differential equation a = –n²x and then uses the inverse trigonometric functions to derive a solution of the form x = a cos (nt + a). A rigorous treatment along these lines is difficult because x is not an invertible function of t (and v is not a function of x) unless the range of x is restricted. It can be observed that x = x₀ cos(n(t – t₀)) + (v₀/n)sin(n(t – t₀)) is a solution of the equation, which satisfies the additional conditions x (t₀) = x₀, v(t₀) = v₀. Questions of uniqueness of solution are outside the scope of the syllabus.

My interpretation of that is that you dont need to know how to do it.

so wot ur telling me is that the question on the 2001 HSC is not on the syllabus, wow... ill let my him, no that
and on that it isn hard, because a=d([V^2]/2)/dx so u can find v easly and then flip dv/dx to get dx/dv and intergrate... not that hard

 

[acmilan]

so wot ur telling me is that the question on the 2001 HSC is not on the syllabus, wow... ill let my him, no that
and on that it isn hard, because a=d([V^2]/2)/dx so u can find v easly and then flip dv/dx to get dx/dv and intergrate... not that hard

What they're are saying is when you use that method, you are working with inverses and thus have to start worrying about restrictions on domain because, as you have learnt in inverse functions, not all equations are invertible

 

[Idyll]

so wot ur telling me is that the question on the 2001 HSC is not on the syllabus, wow... ill let my him, no that

The question is definitely still part of the syllabus. Your method of assuming x=acos(nt+alpha) is valid. It doesn't really constitute a proof that your equation is correct, since the definition of SHM is that:

acceleration = -n^2*x

To prove it you could show that x=acos(nt+alpha) is a solution to the differential equation above, or derive x=acos(nt+alpha) from the equation. However, none of this is at all necessary.

The question simply asked you to find the equation, which means you just write down x=acos(nt+alpha), then use the information given to find a, n and alpha (if you're in any doubt, that's also how the MANSW solutions solve the questions).

 

[JamiL]

The question is definitely still part of the syllabus. Your method of assuming x=acos(nt+alpha) is valid. It doesn't really constitute a proof that your equation is correct, since the definition of SHM is that:

acceleration = -n^2*x

To prove it you could show that x=acos(nt+alpha) is a solution to the differential equation above, or derive x=acos(nt+alpha) from the equation. However, none of this is at all necessary.

The question simply asked you to find the equation, which means you just write down x=acos(nt+alpha), then use the information given to find a, n and alpha (if you're in any doubt, that's also how the MANSW solutions solve the questions).

yer thx, also my teacher cheaked with some bored of studies solutions and they seem 2 agree wif me... stupid principle doesn even no maths :S mind u he wants 2 teach our class... fukn idiot. i should be typing this cos if he reads it he will no its me and then ill prob get fukd
lol