# Homework Help: Simple Harmonic Motion of springs

1. Oct 7, 2007

### aurora14421

[SOLVED] Simple Harmonic Motion

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

A mass rests on a frictionless horizontal table and is connected to rigid supports via 2 identical springs (which obey Hooke's law). Each spring is stretched to a length (l) considerably greater than its rest length. At rest, the mass is located at the origin.

What is the magnitude and direction of the net force exerted by the springs on the mass when it is displaced in a perpendicular direction to the springs (i.e. along the y axis) for the special case of a small displacement.

Hence, write down the equation of motion governing small oscillations in the y direction and give the period (using small angle approximation for sin theta).

2. Relevant equations

(y/l)<<1

Hooke's law: F=-kx

3. The attempt at a solution

2. Oct 7, 2007

### Hootenanny

Staff Emeritus
Okay, so suppose we displace the mass a small distance $\delta y$ as the question said, can you write down a vector equation representing the resultant force expericend by the mass?

3. Oct 7, 2007

### captain.joco

I have sketched a diagram, I can write thy equation, but still stucked..

4. Oct 7, 2007

### Hootenanny

Staff Emeritus
I'm assuming you two are in the same class, not the same person with schizophrenia...

It might be helpful if you posted the equations you've got...

5. Oct 7, 2007

### captain.joco

the x components are equal, thy will cancel out, so we're left only with the y components.
I've got for one spring only ( Since y/l << 1, ):

F-k*y*cos(thetha) I might be terribly wrong though

I don't know thy other person, this is just a problem similar to mine..

6. Oct 7, 2007

### aurora14421

I think the force is given by:

$$F=-k(l-l_{0}) sin \Theta$$

where $$sin \Theta= \frac{y}{l}$$

Is that right?

7. Oct 7, 2007

### aurora14421

Sorry, forgot to say F is the vertical component of the force.

8. Oct 7, 2007

### Hootenanny

Staff Emeritus
I'm assuming you mean F = kycos(theta) and the angle your taking is that between the y-axis and the spring. If that is the case, then your solution is correct, the net force would be the sum of the two springs thus; F = 2kycos(theta)

However, the question drops the hint that you should use the small angle approximation for sin theta; which leads me to believe you should define theta to be the angle between the spring and the x-axis.

9. Oct 7, 2007

### Hootenanny

Staff Emeritus
This is correct, but be careful with your notation, the question states that l is the natural length of the spring so you should consider changing your equation to something like;

$$F=-k(l-l') \sin \theta$$

You can now apply the small angle approximation ($\sin\theta \approx \theta$) to your equation.

10. Oct 7, 2007

### captain.joco

Yeah, I have used the angle between the y axis and the spring.

Using sin...
If I use approximation saying that for very small displacements, l will remain the same, the force ( y component ) will probably be
F = - k*l*sin(theta ).
can I do this?

And for the last part of the question about the period of oscillation.. I have w = 2pi / T hence T= 2pi/w, but what does it mean in the y direction, can you please clarify

11. Oct 7, 2007

### Hootenanny

Staff Emeritus
From aurora's post you can see that;

$$\sin\theta = \frac{y}{l'}$$

Where l' is the extended length. And given the small angle approximation we can say that;

$$\sin\theta \approx \theta \Rightarrow \sin\theta \approx \frac{y}{l'}$$

12. Oct 7, 2007

### captain.joco

Alright, I get it now! I got confused with notation a bit, but thanks. Do you have any hint about the period?

13. Oct 7, 2007

### Hootenanny

Staff Emeritus
Well, the period depends on the spring constant and the mass and the spring constant is the constant of proportionality in Hooke's law. So in our equation;

$$F = -2kl'\sin\theta = -2kl'\frac{y}{l'} = -2ky$$

So if we now say that $F\propto -y$, what would be the constant of proportionality?

14. Oct 7, 2007

### aurora14421

Ok, so just to make sure I understand, the magnitude of the net force is:

$$F=-k(l-l\prime)\frac{y}{l}$$

Or does it get multiplied by two since there are 2 springs?

The direction of the force is downwards.

The equation of motion is just the net force divided by the mass of the object attached to the spring and this will be equal to $$-\omega^2 y$$ .

Never mind, your previous post just answered me I think. I'm to slow...

15. Oct 7, 2007

### captain.joco

well, i think 2k ?

I know that w = ( k/m )^1/2

16. Oct 7, 2007

### captain.joco

What happened with the "2" in F = -2ky ? I think it should be included in the equation of motion somehow, and that's the trick with the period..

17. Oct 7, 2007

### Hootenanny

Staff Emeritus
That's the vertical force from a single spring, it is multiplied by 2 to account for the second spring.
Correct
Yes, the new spring constant is twice that of the original.

18. Oct 7, 2007

### captain.joco

If I am right, the period T = pi*m/k ?

19. Oct 7, 2007

### Hootenanny

Staff Emeritus

20. Oct 7, 2007

### captain.joco

Oh, sorry, too hasty...

T ={ (2)^1/2 * Pi } / w or T = Pi * ( 2k/m )^1/2

21. Oct 7, 2007

### Hootenanny

Staff Emeritus
Noope, try again, your fraction is upside down, I'm sure there's a $2/pi$ in there somewhere as well....

22. Oct 7, 2007

### captain.joco

maybe I start from the wrong thing.. I start from :

( 2k/m )^1/2 = 2Pi/T from there I got the thing written above

23. Oct 7, 2007

### Hootenanny

Staff Emeritus
This is correct, your algebra must be tripping you up;

$$\sqrt{\frac{2k}{m}} = \frac{2\pi}{T}$$
$$\frac{T}{2\pi} = \sqrt{\frac{m}{2k}}$$

$$\boxed{T = 2\pi\sqrt{\frac{m}{2k}}}$$

24. Oct 7, 2007

### captain.joco

I checked, my working is correct, just multiply my equation by 2, you get the same answer. Algebra is all i have..

Thanks for everything!

25. Oct 7, 2007

### aurora14421

Yeah, thanks for your help. I'm pretty sure I understand everything now.