# Homework Help: Spring constant and oscillation expression? Help.

1. Jan 17, 2012

### nukeman

1. The problem statement, all variables and given/known data

Here is the question:

2. Relevant equations

3. 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!

2. Jan 17, 2012

### technician

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

3. Jan 17, 2012

### 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

4. Jan 17, 2012

### 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

5. Jan 17, 2012

### nukeman

Having issues with this!

Ok FIRST for a) Write an expression for the net force on block 1 - Lets 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, lets move onto " Use it to derive an expression for the period of oscillation, in terms of k and m"

I dont know how to get this...?

6. Jan 17, 2012

### 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.

7. Jan 17, 2012

### nukeman

K wait... acceleration is 2kx/m

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

8. Jan 17, 2012

### nukeman

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

Is that what you mean?

and the frequency would be 1/Period

9. Jan 17, 2012

### 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

10. Jan 17, 2012

### technician

you have got it
well done

11. Jan 17, 2012

### nukeman

Oh cool. So,

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

Is that what you mean?

and the frequency would be 1/Period

12. Jan 17, 2012

### technician

I think so

13. Jan 17, 2012

### technician

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