# Spring constant and SHM

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1. Mar 3, 2015

### squashen

The goal is to measure the spring constant of a spring and then calculate a theoretical period of the oscillation and compare the results to a real life measurement.

Extension of spring A:
Neutral: 52 cm
1 N: 41 cm
2 N: 30 cm
F = kx
k = 9.1 N/m

10 oscilliations = 9.26 seconds

Calculating the theoretical period for 2 newton:
t=2pi*sqrt(m/k)
t=2pi*sqrt(2/9.1)=2.9 which is completely wrong compared to the real life result.
If i set k to 91 instead of 9.1 the resulting period is 0.93, much closer to reality.

Have I botched up the units along the way or what have I done wrong?

I'd be very greatful for help.

2. Mar 3, 2015

### Quantum Defect

On the calculation of the natural frequency, what units does "m" have? What were the units of the "2" that you put into the numerator in side the square root?

3. Mar 3, 2015

### squashen

the unit M is in newtons, and its for measuring the oscilliation of the spring with a spring constant of 9.2 when its weighed down with 2 newtons

4. Mar 3, 2015

### squashen

I belive I´ve found a solution, by converting the unit m from newtons to kg it looks more correct. Sorry for the inconvenience!

5. Mar 3, 2015

### Quantum Defect

Are you sure?

t has what units? sec ? right?

What units do you get for "t" when you use M with units of Newtons?

6. Mar 3, 2015

### Quantum Defect

Yup! "m" is the mass of the bob in kg. Newtons are units for the force that the bob exerts. F = m*g

This is why it is always a good idea to work with units on your numbers. You would have seen that you got nonsense units for the period of oscillation if you had kept units of "m" and "k".