Object Movement
A 500 gram object was attached to the end of a coiled string, which in turn was latched onto the top of a faucet. Once steady, the mass of 500g was let go of, and it proceeded to steadily drop down, then back up, and this repeats for a while until the system slowly runs out of energy, therefore coming to a complete rest.
Graphs of Position, Velocity, and Acceleration vs. Time
This Position vs. Time Graph is for one period, or a single cycle from 'Point X' in the air going down, and returning to 'Point X.' It shows, over the course of a couple of seconds, the relative location of the object in direct correlation to the time. If this line were to be scrunched up, one would notice that it would look just like the spring mass in motion, going up and down, and restart... |
This Velocity vs. Time Graph is also for one period, and it denotes the overall speed and direction of the object mass in proportionality to the time covered. Barring the few lapses in the Tracker program that caused a few discrepancies, the general consensus of this graph would be that the object, over the given period of time, maintains a constant direction of going up and then back down, but loses speed eventually until it stops. The velocity is at a maximum point when going up or down, but not yet at the top. Velocity will be found at it's lowest point when it begins to slow down when it reaches the top or bottom. Zero velocity is discovered at the point where mass is at it's highest or lowest possible position.
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Again, for a lone period, we have an Acceleration vs. Time Graph that displays the slightly inconsistent speeds (Tracker errors) at which the object is going at, when coming down hard, or going back up slowly. Generally speaking, the common bystander would expect the mass to, as aforementioned, gradually lose enough speed so it could come to a complete stop. Acceleration is positive when the mass is located near the end of it's fall downwards. The acceleration will be negative when the mass is approaching the peak of it's jump upwards. There will be zero acceleration when the mass is found at the exact middle of it's journey.
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Spring Constant Details
The spring constant is simply a number that exemplifies an object's resistance to compression, and the higher the value, the stiffer it becomes, and thus, it becomes increasingly difficult to be stretched any further. For example, a fully expanded Slinky toy shows the maximum spring constant, given that you can exert that much force on it. Therefore, this can also be referred to as 'elasticity.'
Below, there is a completely solved out equation for this particular lab and it's values. 'T' is representative to the time period of oscillation of the object, 'm' is the mass of the object in kilograms, and 'K' is the spring constant that is to be solved for.
Below, there is a completely solved out equation for this particular lab and it's values. 'T' is representative to the time period of oscillation of the object, 'm' is the mass of the object in kilograms, and 'K' is the spring constant that is to be solved for.
\[T=2\pi\sqrt{\frac{m}{K}}\]
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This is the equation to solve for the spring constant of this harmonic motion lab. Hooke's Law also proposes another equation; F = -kX, however, this is a much more difficult formula to solve for.
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\[\frac{T}{2\pi}=\sqrt{\frac{m}{K}}\]
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To start off, I simply divided '2π' from both sides, and this gets us one inch closer to isolating 'K.' |
\[({\frac{T}{2\pi}})^2={\frac{m}{K}}\]
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Here, I squared both sides to remove the root from the right side. |
\[({\frac{T}{2\pi}})^2\times{\frac{1}{m}}={\frac{1}{K}}\]
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All I have done for this part is multiply both sides by 'm' to get 'K' as the only variable on the right side. |
\[K=1/({\frac{T}{2\pi}})^2\times m\]
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This step may seem a bit more confusing than the previous ones, and rightfully so. Here, I switched the sides to put K off to the left, and then I multiplied each term so that it is a reciprocal of itself.
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\[K={\frac{4\pi^2m}{T^2}}\]
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At this point, I just simplified the left side of this equation, first solving the bottom of the first term and flipping it, since it was below the '1,' and then I merely multiplied this by 'm.'
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\[K={\frac{4\pi^2(0.5)}{(0.77)^2}}
\]
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Substitution now comes into effect, as I replace 'm' with '0.5,' since that is the mass of the 500 gram object in kilograms, and 'T' with the period of oscillation for the object, which was about 0.77 seconds.
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\[K≈33.29\space{\frac{kg}{s^2}}
\]
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In one step, I solved the entire aforementioned formula, and I rounded the answer to two significant figures. |
\[K≈33.3\space{\frac{N}{m}}
\]
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However, we are not done just yet, as the proper units for spring constant equations are Newtons/meters, not kilograms/seconds squared, because this involves force.
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