Can You Solve This Four Spring Oscillation Problem?

[Tanya Sharma]

Homework Statement

Figure shows a particle of mass m attached with four identical springs each of length L₀. Initial tension in each spring is F₀.Neglect gravity calculate the period of small oscillations of the particle along a line perpendicular to the plane of figure. (Take F₀ = (0.01 π²)N, m = 100 gm, L₀ = 10cm)

Homework Equations

The Attempt at a Solution

In the equilibrium state length of the spring L = L₀ + z ,where L₀ is the original length and z is the initial extension in the spring when in equilibrium.

Initial tension in each spring when in equilibrium = kz = F₀

Suppose the mass is displaced by x amount upwards .In the attachment L' represents the new length.

L'=√(L²+x²)

Force by a spring = k(L'-L₀) = = k(L'-L+z) = k(L'-L)+kz

This force will have a a horizontal component and a a vertical component .The horizontal component will be canceled by a similar component from the opposite string .

The net force from all four springs will be F = 4k(L'-L₀ )sinθ = 4k(L'-L)sinθ + 4kzsinθ

sinθ = x/L'

So net force =4kx(L'-L)/L' + 4kxz/L’

=4kx(1-L/L') + 4kxz/L’

I somehow feel I am not approaching the problem correctly .

I would be grateful if somebody could help me with the problem.

 

[ehild]

Homework Statement

Figure shows a particle of mass m attached with four identical springs each of length L₀. Initial tension in each spring is F₀.Neglect gravity calculate the period of small oscillations of the particle along a line perpendicular to the plane of figure. (Take F₀ = (0.01 π²)N, m = 100 gm, L₀ = 10cm)

Homework Equations

The Attempt at a Solution

In the equilibrium state length of the spring L = L₀ + z ,where L₀ is the original length and z is the initial extension in the spring when in equilibrium.

Initial tension in each spring when in equilibrium = kz = F₀

Suppose the mass is displaced by x amount upwards .In the attachment L' represents the new length.

L'=√(L²+x²)

Force by a spring = k(L'-L₀) = = k(L'-L+z) = k(L'-L)+kz

This force will have a a horizontal component and a a vertical component .The horizontal component will be canceled by a similar component from the opposite string .

The net force from all four springs will be F = 4k(L'-L₀ )sinθ = 4k(L'-L)sinθ + 4kzsinθ

sinθ = x/L'

So net force =4kx(L'-L)/L' + 4kxz/L’

=4kx(1-L/L') + 4kxz/L’

I somehow feel I am not approaching the problem correctly .

I would be grateful if somebody could help me with the problem.

It is correct so far. Now you consider the case when x is small with respect to L. Approximate the square root accordingly.


ehild

 

[Tanya Sharma]

Hello ehild...

L'=√(L²+x²)
L'=L√(1+x²/L²)

L' ≈ L(1+x²/2L²) when x<<L

Is it correct ?

I think it is not right .If we had x/L instead of x²/L² ,then it would have been alright . x²/L² term will be too small .But then if we don't keep it then L' ≈ L ,which again looks incorrect .Not sure when to keep which terms while making approximations.

 

[ehild]

It is correct. You have to keep terms linear in x/L. Note that the square root is in the denominator. What is the force?

ehild

 

[Tanya Sharma]

F = 4kx³/2L²(1+x²/2L²) + 4kxz/(1+x²/2L²)

Right ?

 

[ehild]

You made it very complicated...First: the force is opposite to x. Factor out 4kx. $$F=-4kx\left(1-\frac{L-z}{L \sqrt{1+(x/L)^2}}\right)$$

ehild

 

[Tanya Sharma]

You made it very complicated...First: the force is opposite to x. Factor out 4kx. $$F=-4kx\left(1-\frac{L-z}{L \sqrt{1+(x/L)^2}}\right)$$

ehild

Sorry for the mess up.

Is $L \sqrt{1+(x/L)^2}$ ≈ $L$ ?

If correct then $F = \frac{-4kxz}{L}$ or $F = \frac{-4F_{0}x}{L}$

Right?

 

[ehild]

Sorry for the mess up.

Is $L \sqrt{1+(x/L)^2}$ ≈ $L$ ?

If correct then $F = \frac{-4kxz}{L}$ or $F = \frac{-4F_{0}x}{L}$

Right?

Yes, it is correct, but write L in terms of Fo and Lo, as they are given.

ehild

 

[Tanya Sharma]

Yes, it is correct, but write L in terms of Fo and Lo, as they are given.

ehild

L = L₀+z and kz=F₀

But ,neither z nor k is given in the problem ?

I think there is something wrong in the question .The length given should be the stretched length of the spring in equilibrium not the natural length .

What do you say ?

 

[ehild]

Yes, Lo might be the initial length instead of the unstretched length.

ehild

 

[Tanya Sharma]

Thanks...

There is a similar problem of four springs just as in the OP.But in this case the springs are in their natural length i.e there is no initial tension.Here the oscillations are occurring in the plane of the figure , not perpendicular to it.We need to find the time period when the mass is displaced slightly towards the top vertical spring i.e the oscillations are occurring in the plane of the figure .Neglect gravity.

My approach - For calculating forces by the left and the right spring ,we will approach exactly the way we have done in this question .Force due to left spring = Force due to right spring ≈ 0

Force due to top spring + Force due to bottom spring = -2kx

Net force F = -2kx

Do I make sense ?

 

[ehild]

The forces of the left and right springs are not zero if there is an initial tension in the springs. But the springs have their natural length this time, so the tension is zero. F=-2kx.

ehild

 

[Tanya Sharma]

Thank you so much :)

 

[Tanya Sharma]

ehild...I found a few things not so obvious in the Original problem ?

1) L' ≈ L , even when the springs are getting stretched .

2) If the springs are not under initial tension in the OP ,then the mass would not be performing SHM , as net force will be zero .

What do you think ?

 

[ehild]

ehild...I found a few things not so obvious in the Original problem ?

1) L' ≈ L , even when the springs are getting stretched .

Not necessarily... Those springs can be stressed quite much. Think of rubber bands.

2) If the springs are not under initial tension in the OP ,then the mass would not be performing SHM , as net force will be zero .

What do you think ?

The new force would not be zero, but it would not be linear in the displacement. The mass will move somehow, it would be interesting to see, how.

ehild