Simple Harmonic Motion. (1 Viewer)

[mazza_728]

Do we need to know how to derive the equations?
If so how do u derive the equations.. especially in reverse i.e i have a question that says that a=-n²x show that x=acos(nt+@)??
How do i do it?
Thanks xoxo

 

[mojako]

>> i have a question that says that a=-n2x show that x=acos(nt+@)??
How do i do it? <<
I don't think you can do this one ^_^

 

[Li0n]

why cant you just go

x = acos(nt+@)
x'= -nasin(nt+@)
x"= -n²[acos(nt+@)]
x"=a= -n²x

 

[mojako]

He / she was asking it the other way around.
Without knowledge or assumption that the x has the form of acos(nt+@), I don't think it can be done.
But maybe it can

 

[mazza_728]

This is the question:
1986 HSC Question 5
ii) A particle is oscilatting in simple harmonic motion such that its displacement x metres from a given origin O satisifies the equation
d²x/dt²=-4x
where t is the time in seconds

(a) show that x=acos(2t+@) is a possible equation of motion for this particle, where a and @ are constants.


I also cant do this part:

(b) The particle is observed at time t=0 to have a velocity of 2 metres per second and a displacement from the origin of 4 metres. Show that the amplitude of oscillation is (sqrt)17 metres.

I tried solving this using v²=n²(a²-x²) but got a=4 (i.e. (sqrt)16) ???

 

[mojako]

it says: "a possible solution"
so you do what Mr Li0n said
(it's like when proving the natural / exponential growth formula... do it from the possible solution)

you have
4=4(a^2-16)
1=a^2-16
17=a^2

and just an alternative answer without that formula, you can do
x=acos(2t+@) , i.e. 4=acos(@) [since it's when t=0]
differentiating,
v=-2asin(2t+@), i.e. 2=-2asin(@)

squaring 4=acos(@): 16=a^2 cos^2(@)
squaring 2=-2asin(@): 4=4a^2 sin^2(@), i.e. 1=a^2 sin^2(@)
Adding the two: 16 + 1=a^2 cos^2(@) + a^2 sin^2(@)
17 = a^2 [ cos^2(@) + sin^2(@) ]
The thing in [...] = 1 (Pythagorean identity)

 

[mazza_728]

Thankyou muchly. I used the v^2=n^2(a^2-x^2) but just made a terribly stupid mistake which i hope i never repeat again. As for the proof, can you prove it the other way i mean can u go from a=-n^2x to x=acos(nt+@)?? Or can we always do it by integrating?

 

[mojako]

>> i mean can u go from a=-n^2x to x=acos(nt+@)?? <<
I'm quite sure that I said you cannot
So if you ask me then you'll get that answer
And I have a big faith that other people will say the same
EDIT: hmm.. I withdraw all those statements

>> Or can we always do it by integrating? <<
Don't you mean differentiating?
Yes you can always differentiate.

Oh, you can integrate x''=-n^2x using x'' = d(1/2 v^2)/dx
but you'll get v^2=n^2(a^2-x^2), using the fact that at endpoints v is zero, and the endpoints are x=a and x=-a
EDIT:wait.. if you keep going using v=dx/dt and then inverting the equation, I think you can get x in terms of t.

 

[Li0n]

why cant you just go

x = acos(nt+@)
x'= -nasin(nt+@)
x"= -n²[acos(nt+@)]
x"=a= -n²x

i knew i was right

 

[cj_bridle]

lol how good is quoting yourself

 

[Li0n]

i knew i was right

my precioussssssssssssssssssssssssssssssssssssssss

 

[cj_bridle]

lol how good is quoting yourself

i agree with cj

 

[mojako]

>> i mean can u go from a=-n^2x to x=acos(nt+@)?? <<
I'm quite sure that I said you cannot
So if you ask me then you'll get that answer
And I have a big faith that other people will say the same
EDIT: hmm.. I withdraw all those statements

>> Or can we always do it by integrating? <<
Don't you mean differentiating?
Yes you can always differentiate.

Oh, you can integrate x''=-n^2x using x'' = d(1/2 v^2)/dx
but you'll get v^2=n^2(a^2-x^2), using the fact that at endpoints v is zero, and the endpoints are x=a and x=-a
EDIT:wait.. if you keep going using v=dx/dt and then inverting the equation, I think you can get x in terms of t.

I completely agree with mojako.
He always has the same thought with me.
We are true friends !!

 

[CrashOveride]

mojako's only friend is himself.

 

[Li0n]

awww only a true friend can be that trully honset