# Classical mechanics, mass on a spring

1. Jun 11, 2013

### Cogswell

1. The problem statement, all variables and given/known data
Consider the classical system of a mass of one kg attacked to the ceiling with a spring constant k=50N/m.
The mass is held at rest such that the spring hangs vertically but is not extended. The mass is then released and falls under gravity. Neglect air resistance.

1. What is the maximum extension of the spring?
2. What is the maximum speed of the mass?
3. What is the frequency of oscillation of the mass?

2. Relevant equations

Undamped, undriven oscillator:

$\ddot{x} = -kx$

$x(t) = B_1 \cos (\omega t) + B_2 \sin (\omega t)$

$t = 2 \pi \sqrt{\dfrac{m}{k}}$

3. The attempt at a solution

1. For the first one, I used the fact that:
$F = mg$
$F = -kx$
$mg = -kx$

$x = -\dfrac{mg}{k}$

2. I don't quite know how to do this.
I think the maximum velocity is when it's going downwards, at its equilibrium position, but I don't really know how to find it...

3. f = 1/t
$t = 2 \pi \sqrt{\dfrac{m}{k}}$
$f = \dfrac{1}{2 \pi \sqrt{\dfrac{m}{k}}}$

Is 1 and 3 right, and can someone help me with 2?

2. Jun 11, 2013

### Saitama

1 is incorrect. You find the extension at the equilibrium position but the question asks the maximum extension. Use conservation of energy.

For 2nd, you are correct that maximum velocity is at equilibrium. You can use conservation of energy again.

And your answer for third is correct. :)

3. Jun 11, 2013

### Cogswell

Oh okay, so using the conservation of energy...

$E_{\text{total}} = T + U$

$E_{\text{total}} = \dfrac{1}{2}m \dot{x}^2 + \dfrac{1}{2}k x^2$

I know E has to be constant, but how would I go about solving for either of them?

When the kinetic energy equals zero, that's the maximum displacement from equilibrium, and so

$E_{\text{total}} = \dfrac{1}{2}k x^2$

$\sqrt{\dfrac{2 E_{\text{total}}}{k}} = x$

Now how do I know what E(total) is?

4. Jun 12, 2013

### Saitama

You are missing the gravitational potential energy. ;)

5. Jun 12, 2013

### Cogswell

So adding in gravity...

$E_{\text{total}} = T + U$

$E_{\text{total}} = \dfrac{1}{2}m \dot{x}^2 + \dfrac{1}{2}k x^2 + mgx$

(x is positive when the spring is going up)

When I solve for x, for when the kinetic energy is zero, I will get 2 answers. But it'll still have the E(total) term in it. What is the E(total) term, or how can I find it?

6. Jun 12, 2013

### Saitama

Where have you defined the gravitational potential energy to be zero?

Consider the gravitational potential energy equal to zero at the initial position. So, initial energy is zero. When the mass moves down a distance x, compute the final energy.

7. Jun 12, 2013

### Cogswell

What do you mean?

Okay, so the gravitational potential energy is zero at when the spring is at its starting position.

And so the initial total energy is zero.

The final energy will be when is moves a distance of x.
(x is positive when it goes up, and so negative when it goes down)

When the mass is released, the mass will go downwards.

$E_i = 0$

$E_f = \dfrac{1}{2} m ( -\dot{x})^2 + \dfrac{1}{2} k (-x)^2 - mgx$

So the change in energy is just $E_f$

$\Delta E = \dfrac{1}{2} m ( -\dot{x})^2 + \dfrac{1}{2} k (-x)^2 - mgx$

And then...?

8. Jun 12, 2013

### Saitama

Why you think about the change? You simply need to conserve energy i.e. initial energy=final energy. And what is x in terms of other variables in $E_f$? (you calculated it in OP but there's a sign error.)

9. Jun 12, 2013

### Cogswell

Haha I don't know why I was thinking change in energy...

So,

$\dfrac{1}{2} m \dot{x}^2 + \dfrac{1}{2} k (x)^2 - mgx = 0$

And from the OP:

F = -mg (going downwards)
F = -kx

$x = \dfrac{mg}{k}$

I don't see the point in that though...

Carrying on, I know the kinetic energy is zero when the displacement is at its maximum:

$\dfrac{1}{2} k (x)^2 = mgx$

$k x = 2mg$

$x = 2 \dfrac{mg}{k}$

??

And then similarly for kinetic energy (but it doesn't work...)
The kinetic energy is at its highest at the point x=0 (equilibrium when potential energy = 0)

$\dfrac{1}{2} m \dot{x}^2 = 0$

...

10. Jun 12, 2013

### Saitama

Well, I think you were trying to calculate the velocity at the equilibrium position, don't you think the above relation gives the answer after you substitute x?

Also, which direction you are considering to be positive? Upwards or downwards?
Let the upward direction be positive, then $kx-mg=0$ at equilibrium position.
You just calculated the maximum extension which was required for the part 1.

Sorry, I can't follow this.

11. Jun 12, 2013

### voko

Since it is already known that the max velocity is at the equilibrium, you just need to recall the equilibrium really is.

12. Jun 12, 2013

### Cogswell

Oh wait, I get it now...

So the maximum velocity is at equilibrium, where the force of the spring is balanced out with the force of gravity, and that's what the $x = \dfrac{mg}{k}$ means.

Putting that into the energy equation, I get that

$\dfrac{1}{2} m \dot{x}^2 + \dfrac{1}{2} k \dfrac{m^2 g^2}{k^2} - mg \dfrac{mg}{k} = 0$

$\dfrac{1}{2} m \dot{x}^2 = \dfrac{1}{2} \dfrac{m^2 g^2}{k}$

$\dot{x}^2 = \dfrac{m g^2}{k}$

$v_{\text{max}} = g \sqrt{\dfrac{m}{k}}$

Last edited: Jun 12, 2013
13. Jun 12, 2013

### voko

A slightly more extravagant way to obtain max velocity is by maximizing kinetic energy using the energy equation.