# What is wrong with this k constant?

I used two methods the find k constant from the SAME spring. I used f=k(y-yo) and for oscillation T=2pie*sqr.root mass/k. I got really OFF answers for the two k constants, what did I do wrong?

First method:

(kg) initial position(m) final position(m) force(N) total elongation K(N/m)

0.05 0.059 0.719 0.49 0.660 1.24

Method two: oscillation

Mass(kg) 0.050

period 1 0.3129

period 2 0.3020

period 3 0.3090

average 0.3080

K constant(N/m) 21.2

square root of mass(kg) 0.224

Standard deviation 0.00552

Its 1.24 vs 21.2, please help!

### 1 Answer

- wy125Lv 49 years agoFavorite Answer
First, you might want to double check the spring constant you get from the first method. I get:

k = F/elongation = (0.05)(9.8)/0.660 = 0.74 N/m

now when you're using the oscillation period it's not just the mass of the object that you need to take into account but the equivalent mass of the spring. This isn't the total mass of the spring because not all the spring's own mass is pulling itself down. So your equation for period becomes:

T = 2 * Pi * sqrt( (m + m_es) / k )

where m_es is the equivalent mass of the spring (most use 1/3 mass of the spring as this value).

Here's how we can figure this out. Let's square both sides:

T^2 = 4*PI^2 * (m + m_es) / k

T^2 = 4*PI^2 * m/k + 4*PI^2*m_es/k

now you can see that if you made a plot of T^2 vs m you would get line with slope 4*Pi^2*/k which means you can get k from there.

Now the problem is that you didn't vary the mass so you only have data for one value of m. You need at least 2 different values of m to get a line but I would recommend 4 or more.

Hope that wasn't too confusing...

Edit:

Was just thinking that if you didn't use the equivalent mass of the spring you will get a k value that's smaller than the actual so that doesn't explain the discrepancy. But you still need to make a plot to figure out k when your using the oscillation period.

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