# Homework Help: Error source of SHM experiment

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1. Apr 3, 2015

### smokedvanilla

1. The problem statement, all variables and given/known data
Investigating the effect of mass on the period of oscillation.

This experiment is about SHM of a floating cylinder, and the theory is explained in this website:
http://physics.stackexchange.com/questions/64154/shm-of-floating-objects

Also, I'm attaching a diagram of my experimental setup. (x is the displacement of the cylinder under water)

2. Relevant equations

Buoyant force acting on the cylinder, F=Vρg=Axρg
As Newton's second law states that F=ma, a=-Axρg/m
Since a is directly proportional to -x, this fulfils the condition for SHM.

a=ω^2x, and ω=2∏/T

therefore a=(4∏^2)x/T^2

-Axρg/m=(4∏^2)x/T^2

From the derivation above, it can be concluded that 1/m is directly proportional to 1/T^2, and therefore m is proportional to the square of period.

3. The attempt at a solution
I tried to perform the above experiment. The cylinder does bob up and down in the water, but it does not perform SHM. First of all, the amplitude keeps decreasing, but I have no idea what is the source of energy loss. Besides, the correlation between m and T^2 is also not directly proportional.

Does anyone have an idea what's the error in this experiment?

#### Attached Files:

• ###### Photo on 4-3-15 at 6.44 PM.jpg
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Last edited: Apr 3, 2015
2. Apr 3, 2015

### Staff: Mentor

While it does not change the period of the oscillation, you forgot the gravitational force on the bob in the equations.
I agree.
You are moving water around all the time. Could this lead to losses?

3. Apr 3, 2015

### smokedvanilla

Does this mean that the net force acting on the oscillating cylinder is (buoyant force)-(weight of cylinder) or vise versa depending on the direction of its motion? (Which means ma=Fb-mg, is this eqn correct?)

This is the graph of square of period against mass.

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4. Apr 4, 2015

### Staff: Mentor

Right (and the sign just depends on the definition of "direction of force"). As the weight of the cylinder does not change this is just an offset to the equilibrium position, the derivatives don't change.
The straight line looks like a very nice fit. It has an offset that needs an explanation, but I see where it can come from.
I wonder how you got frequency measurements with that precision.

5. Apr 5, 2015

### smokedvanilla

I'm sorry I didn't quite get what you meant by this.

As for the graph, I took about 7 measurements (time taken for 3 oscillations) for each mass and calculated the average value. I just realized that it is actually a very nice fit, coz I was too carried away by the fact that it doesn't pass thru the origin..