A mass inside a horizontal spring

[tecnica]

Homework Statement

We have a spring of length l₀ tied to two vertical non moving sticks. We place a mass m at 0,45l₀ and let it oscillate. If we measure the period of an oscillation, we can find the angular frequency and calculate k. The question is, how can I calculate k₁ and k₂, the constants of each one of the pieces on both sides of the mass?

Homework Equations

w=sqrt (k/m)

The Attempt at a Solution

So the thing is, I don't know if k=k₁+k₂ or 1/k=1/k₁+1/k₂

 

[Orodruin]

Which one do you think is correct and why?
We require students to make attempts, the reason for this is that it helps you better in the long run if we help you think rather than simply provide you with an answer.

 

[tecnica]

Which one do you think is correct and why?
We require students to make attempts, the reason for this is that it helps you better in the long run if we help you think rather than simply provide you with an answer.

I guess it is k=k₁+k₂, because when you move it Δx, the force acting on it is -(k₁+k₂)Δx, right?

 

[Orodruin]

Each of the parts of the spring give a force proportional to their spring constants so what you have given is the spring constant relevant for computing the angular frequency. But you should stop and think for a moment if this is actually the spring constant k of the full spring.

 

[tecnica]

@Orodruin So the spring constant relevant for computing the angular frequency is the one that obeys k=k1+k2, and the spring constant of the full spring obeys 1/k=1/k1+1/k₂, right?

 

[Orodruin]

Yes. In order to solve this you will also need to think about an additional condition for the relationship between k₁ and k₂ (a hint is that it will depend on where the mass was connected onto the spring). I also suggest calling the spring constant of the original spring k₀ and the spring constant relevant for the angular frequency k in order not to mix them up.

 

[Chestermiller]

tecnica,

If you have two identical springs, except that one spring is twice as long as the other, which one exhibits a higher spring constant?

Chet

 

[tecnica]

tecnica,

If you have two identical springs, except that one spring is twice as long as the other, which one exhibits a higher spring constant?

Chet

I'm guessing the shorter one has a higher spring constant.

Yes. In order to solve this you will also need to think about an additional condition for the relationship between k₁ and k₂ (a hint is that it will depend on where the mass was connected onto the spring). I also suggest calling the spring constant of the original spring k₀ and the spring constant relevant for the angular frequency k in order not to mix them up.

Could it be l₁k₁+l₂k₂=l₀k (where this k is the one used for the angular frequency)? l₁ is the distance between the left vertical stick and the mass, and l₂ is the distance between the mass and the right stick.

 

[haruspex]

I'm guessing the shorter one has a higher spring constant.
Could it be l₁k₁+l₂k₂=l₀k (where this k is the one used for the angular frequency)? l₁ is the distance between the left vertical stick and the mass, and l₂ is the distance between the mass and the right stick.

It might help if you concentrate on one side, l₁ say. How does k₁ relate to k?

 

[tecnica]

It might help if you concentrate on one side, l₁ say. How does k₁ relate to k?

Could it be (l₁/l₀)k₁=k ?

 

[haruspex]

Could it be (l₁/l₀)k₁=k ?

Yes.

 

[Orodruin]

Yes, but with k₀, the spring constant of the full spring, not k, the spring constant appearing in the angular frequency.

 

[haruspex]

Yes, but with k₀, the spring constant of the full spring, not k, the spring constant appearing in the angular frequency.

Glad you caught that - I hadn't checked back with the OP to see exactly what k meant here.

 

[Orodruin]

Glad you caught that - I hadn't checked back with the OP to see exactly what k meant here.

Well, part of the problem was that it was not defined in the OP but the same symbol k was used for both ... I defined it in post #6.

 

[tecnica]

Thank you guys, I understand it all now