# Springs Double

1. Dec 19, 2013

### oneplusone

1. The problem statement, all variables and given/known data
You have two springs with spring constants k_1 = 10 and k_2 = 20 vertically attached to a wall, and a mass of mass 3.00kg is hung from it. Find the period of oscillation.

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||
|| <-- first spring, k_1 = 10
||
--
||
|| <== second spring, k_2 = 20
||
{} <== mass 3.00kg

2. Relevant equations

F=-kx
T = 2pi/omega

3. The attempt at a solution

Consider the second spring. We have:
$F=-kx \implies 3(-9.8) = -(20)(x) \implies x = 1.47$

Now the second spring undergoes a displacement of:

$F=-kx \implies 3(-9.8) = -(10)x \implies x = 2.94$.

So k effective is:

$F = -k (x_1+x_2) \implies 3(9.8) = k(2.94+1.47) \implies k = 4.41$

From here, i just found $\omega$ by using $\omega =\sqrt{k}{m}$. and the period was easy from there.

Is the first part of my solution correct? Im having trouble visualizing it.

2. Dec 19, 2013

### Staff: Mentor

So far so good.

4.41 is the total displacement, not the spring constant. Solve for the effective spring constant.

3. Dec 19, 2013

### oneplusone

oops i meant 6.666. Thank you.

A follow up question to this has the same springs, except oriented like this:

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|| ||
|| ||
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| {} |
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With the springs parallel. How would you solve this? I am still not convinced that there is a solution (to me, the "block" will be tilted).

4. Dec 19, 2013

### Staff: Mentor

Assume the system is constrained so that the springs get the same displacement.