simple harmonic motion proof help (1 Viewer)

[Steven12]

how do you proof

v'2=36-6x-2x'2

is a simple harmonic motion?


'2 means To the square of 2

 

[Xayma]

v²=36-6x-2x²
d/dx(1/2 v²)=a
∴
1/2 v²=18-3x-x²
d/dx (1/2 v²)=-3-2x
a=-3-2x

Therefore acceleration is directly proportional to displacement. Therefore it undergoes simple harmonic motion.

 

[Heinz]

Originally posted by Xayma
v²=36-6x-2x²
d/dx(1/2 v²)=a
∴
1/2 v²=18-3x-x²
d/dx (1/2 v²)=-3-2x
a=-3-2x

Therefore acceleration is directly proportional to displacement. Therefore it undergoes simple harmonic motion.

You can go on further to say that its a = -1(2x+3) and so is in the form -n²x

 

[kpq_sniper017]

Originally posted by Heinz
You can go on further to say that its a = -1(2x+3) and so is in the form -n²x

just on that note:
if u get x"=-n²(x-a), can u leave it like that and simply say it undergoes SHM?

or do u have to go further and say
let X=x-a
.'. X"=x"
.'. x"=-n²X
which is of the form -n²x
.'. it undergoes SHM?

 

[George W. Bush]

shm, by definition, is x'' in the form of -n^2(x-b)

 

[kpq_sniper017]

Originally posted by George W. Bush
shm, by definition, is x'' in the form of -n^2(x-b)

so it wouldn't be necessary to do the extra 4 steps? coz they do that in fitzpatrick and i wans't too sure whether it was required.

 

[CM_Tutor]

I would show the X = x - a substitution - the usual HSC definition of SHM is x'' = -n²x, not x'' = -n²(x - a).

Originally posted by Xayma
a=-3-2x

Therefore acceleration is directly proportional to displacement. Therefore it undergoes simple harmonic motion.

Be careful saying things like this - acceleration is directly proportional to displacement means a = kx for some k, not a = kx + C for some constants k and C.

Originally posted by Heinz
You can go on further to say that its a = -1(2x+3) and so is in the form -n²x

Heinz, this is not in the form -n²x, and n is not 1 in this case. If you follow the substitution method mentioned, you'll find that n is actually sqrt(2).

 

[Heinz]

Originally posted by CM_Tutor

Heinz, this is not in the form -n²x, and n is not 1 in this case. If you follow the substitution method mentioned, you'll find that n is actually sqrt(2).

Thank god your back. Actually, i have no idea what the substitution method is, i never learnt it. The reason why i said n² = 1 was because for some unknown reason, i always assumed n was an integer. Maybe i should go learn this method. Thanks for that

 

[kpq_sniper017]

thanks for clearing that up cm.
haven't seen u here in awhile.....still working on that PhD thesis?

in the exam today, i did that X=x-a method. This is what I did, and I hope it's right:

x"=-4(x-3)
Let X=x-3
.'. X"=x"
.'. X"=-4X, which is S.H.M.

 

[Steven12]

Originally posted by Xayma

i dont understand how you can say that

a=-3-2x
is proportional to the displacement when displacement equation wasnt given

please explain

 

[kpq_sniper017]

Originally posted by Steven12
i dont understand how you can say that

a=-3-2x
is proportional to the displacement when displacement equation wasnt given

please explain

you're given v² in terms of x (i.e. displacement). so when u differentiate, u also get acceleration in terms of x. there's no need to have a displacement equation in terms of t because velocity is already given in terms of x.

 

[CM_Tutor]

Actually, a = -3 - 2x does not have acceleration proportional to displacement, as this would require a = kx.

What you can do is an X = x + (3 / 2) substitution, which gives X'' = -2X, where acceleration is proportional to displacement.

 

[kpq_sniper017]

Originally posted by CM_Tutor
Actually, a = -3 - 2x does not have acceleration proportional to displacement, as this would require a = kx.

What you can do is an X = x + (3 / 2) substitution, which gives X'' = -2X, where acceleration is proportional to displacement.

yea.
that was what i was referring to earlier on i.e. whether its necessary to use the substitution.
at least i know now that it's a necessity.

 

[DcM]

Originally posted by CM_Tutor
Actually, a = -3 - 2x does not have acceleration proportional to displacement, as this would require a = kx.

What you can do is an X = x + (3 / 2) substitution, which gives X'' = -2X, where acceleration is proportional to displacement.

how do u know wat to sub in??

 

[Xayma]

Originally posted by DcM
how do u know wat to sub in??

Solve -3-2x=0
If the answer is b then X will be x-b

 

[kpq_sniper017]

Originally posted by DcM
how do u know wat to sub in??

transform into the form: -n²(x-b)

 

[CM_Tutor]

Originally posted by DcM
how do u know wat to sub in??

Originally posted by pcx_demolition017
transform into the form: -n²(x-b)

So, you can take out the coefficient of x as the factor (-2 in the question at issue here).

 

[kpq_sniper017]

Originally posted by CM_Tutor
So, you can take out the coefficient of x as the factor (-2 in the question at issue here).

yep

 

[Steven12]

alrite alrite

I know you people are talking about the substitution method
but this is as far as i got


a=-3-2x
a=-1(3+2x)

since a=-n'2x
let x=3+2x

right? now what?
I am a bit slow so please bear with me

 

[mojako]

alrite alrite

I know you people are talking about the substitution method
but this is as far as i got


a=-3-2x
a=-1(3+2x)

since a=-n'2x
let x=3+2x

right? now what?
I am a bit slow so please bear with me

a=-3-2x
a=-1(3+2x)
a=-1(2x+3)

pcx_demolition in post #16 said that the condition for something to be SHM is:
a = -n'2(x-b)
where b is any constant (any number basically)
[if the motion fulfils this condition then it's automatically an SHM]

You have a=-1(2x+3)
a=-2 (x+1.5)
a=-2 (x-[-1.5])
so, the n'2 part is "2" and the b part is "-1.5"

And while we are on this equation, the SHM will have its centre of motion at x=1.5.
this kind of thing is something that always happens... if a=-n'2(x-b) then it's simple harmonic and its centre of motion is x=b.

Also, the standard way of writing x'2 on computers is x^2