#$&*
course Phy 241
1/9 8
You suspended a candy bar from a rubber band chain and counted its oscillations for a minute. You did the same for a stack of 2 dominoes, and for a stack of four dominoes.
Insert a copy of your data here, along with any previously submitted work you wish to include:
****
Candy Bar Oscillation
Here I had a rubber band chain of 31cm long unstretched. One Snickers candy bar (regular size) with mass of 97 grams. and 8 dominoes, with total mass 126.5 grams
Here I dont measure each dominoes mass individually because again we had a couple or three of those thin ones. so I'm just using 126.5 overall mass for the 8 dominoes.
I measured the oscillations of the candy bar by setting it in motion with my arm holding it in front of me. (candy bar moving up and down with a displacement of no more than say 10cm.)
During that 1 minute, I counted 90 oscillations. . so 90 oscillations / minute
I also have here that 4 dominoes stretch the 31cm long rubber band chain to 39cm, and 8 dom stretch it to 59cm long.
Oscillating Candy Bar
#$&*
For each number of dominoes, determine the force constant k, based on your observations of the number of oscillations in a minute (which you can use to find the angular frequency omega) and the mass of the domino stack.
****
F_net = -kx
ma = -kx
k= -ma/x
a= x""
k= - (mx"")/x
using 97grams as mass, change in positition as 10cm. and 2pi rad/s^2 as acceleration. My force constant is -143.6 g/s^2.
@&
Acceleration is not constant for a simple harmonic oscillator. 2 pi rad/s^2 might be an angular acceleration, but it is not a linear acceleration and wouldn't be multiplied by m to get a force.
Your length data below allows you to graph tension force vs. chain length. The slope of this graph is your value of k. It looks like that would be omething in the neighborhood of .4 N / cm.
Combining this with the angular frequency omega would allow you to determine the mass.
*@
I got 2pi rad/s^2 because it had a frequency of 1.5 oscillations/sec. . one revolution or frequency can be measured in radians, one complete radian being 2pi. 1.5*2pi is 3pi radians / second.
to get acceleration from there, divide by our seconds, (1) to get 3 pi rad/s^2. -97grams * 3pi rad/s^2 / 20pi = -143.6 g / s^2
20pi in the denominator came from the 10cm travel. because 10cm oscilation is one revolution is 2pi.... 10cm *2pi is 20pi.
initial length of 31cm
# Dom | Length(cm)
1 32
2 33.5
3 36
4 39
5 43
6 48
7 54
8 59
#$&*
Are your values of the force constant reasonably consistent?
****
I think they are, almost twice the force to turn the candy bar around to return up, and vice verca.
#$&*
Based on your best estimate of the force constant and your observation of the frequency of the oscillations of the candy bar, what is the mass of the candy bar?
****
based on this data, the mass is 323.1 grams. get this by multiplying the time twice to take care of the s^2 in the denominator. This makes sense because such a force downward on the candy bar
makes it feel heavier, briefly, when actually its just turning around. This means this has to be the force on the return up the oscilation. else the candy bar would feel weightless at the top when gravity takes over momentarily.
#$&*
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
@&
I actually get something close to 400 grams when I calculate based on your data, but that has to be off. That's nearly a pound.
Check my notes and be sure you know how to calculate m from this data. Your value of k just isn't done correctly.
*@