Spring constant and oscillation expression? Help.

[nukeman]

Homework Statement

Here is the question:

Homework Equations

The Attempt at a Solution

I know that SHM is: accel = -(constant) (displacement)

Linear from my book says: Ax = Ftotal/m (dont quite get this)

Any help? THanks!

 

[technician]

I would say they are identical for a given displacement.
F = 2kX in each case

 

[nukeman]

Can you show me how?

I need to be able to write an expression for the net force on block 1, and use it to derive an expression for the period of oscilation, in terms of k and m

 

[technician]

In each case I think the force on the block is 2kx (x is the displacement)
I would use this to write an expression for the acceleration (= F/m)
then acc =ω^2x to get an expression for ω^2 then get the time period expression.
get back to me if you cannot do this

 

[nukeman]

In each case I think the force on the block is 2kx (x is the displacement)
I would use this to write an expression for the acceleration (= F/m)
then acc =ω^2x to get an expression for ω^2 then get the time period expression.
get back to me if you cannot do this

Having issues with this!

Ok FIRST for a) Write an expression for the net force on block 1 - Let's start with this ok?

Would it not be just: 2kx --- is that correct? Its because there is 2 spring constants.

---->NOW if that is right, let's move onto " Use it to derive an expression for the period of oscillation, in terms of k and m"

I don't know how to get this...?

 

[technician]

So if the force = 2kx then acceleration = F/m = 2kx/m
Acceleration = ω^2 x
so ω^2 = 2k/m
Can you get an expression for frequency or time period from this?
ask if you are not sure... but you know the 'rules' of this place... I prefer not to give you the answer directly... you are almost there.

 

[nukeman]

So if the force = 2kx then acceleration = F/m = 2kx/m
Acceleration = ω^2 x
so ω^2 = 2k/m
Can you get an expression for frequency or time period from this?
ask if you are not sure... but you know the 'rules' of this place... I prefer not to give you the answer directly... you are almost there.

K wait... acceleration is 2kx/m

Where did "Acceleration = w^2x, so w^2 = 2k/m come from? Can you explain this to me?

 

[nukeman]

Umm: Period = 2pi sqrt(m/2k)

Is that what you mean?

and the frequency would be 1/Period

 

[technician]

No problem... be patient
SHM is when F is proportional to displacement,x,
that means F = kx... should be F = -kx because displacement is a vector but forget that for now
This means that F = ma = kx so a = kx/m... looks better like this ... (k/m)x
By analysing SHM and its links with circular motion you get a = ω^2 x, so ω^2 = k/m
In this problem, because there are 2 springs k = 2k
so ω^2 = 2k/m... are you OK now?
I will explain in more detail if you do not get it ! Don't panic ... it is tricky

 

[technician]

you have got it
well done

 

[nukeman]

you have got it
well done

Oh cool. So,

Umm: Period = 2pi sqrt(m/2k)

Is that what you mean?

and the frequency would be 1/Period

This answers a) completely?

 

[technician]

I think so

 

[technician]

it would be good to have confirmation from another contribution but I am happy with my answer