# Homework Help: Interesting harmonic motion lab

1. Jan 27, 2014

### rhino1000

1. The problem statement, all variables and given/known data
This is a required analysis for my Physics II lab. We recorded the motion of an object oscillating on spring, and are asked to use the slopes of different graphs that we plotted using the data collected in lab in order to find the spring constant (k).

Both graphs were derived from the same data, which was amplitude of the objects motion, period of its motion, and maximum velocity of the object. We collected all of this information for a mass of 25 grams, then 30 grams, etc. all the way up to 75 grams.

The first graph is a graph of Period^2 (s^2) vs. mass of object (kg). The line of best fit for this graph was f(x)=19.699x + 0.0652. In other words, f(x) should be Period^2, and x should represent mass of the object (s).

The second graph is a graph of Vmax vs A/sqrt(m). The line of best fit for this graph was f(x) = 1.169x + 0.0396. In other words, f(x) should be Vmax, and x should represent the mass of the object.

Both of the mentioned graphs should be linear. They were, and also had a R^2 value of .98 or higher each!

2. Relevant equations

T = 2∏/ω = 2∏*sqrt(m/k)

Vmax = Aω = A*sqrt(k/m)

3. The attempt at a solution
We graphed T^2 vs m. So, using the relevant equation, we get:
T^2 = 4∏^2 *m/k
This means that if we graphed T^2 vs m, we should get
Slope of graph (the derivative of T^2 with respect to m) = 4∏^2 / k
Therefore k should = 4∏^2 / (slope of graph) ... right?
If we plug in the slope that we obtained from our line of best fit, we get:
k = 4∏^2 / 19.699 = 2.004 N/m

The other graph was Vmax vs A/sqrt(m). Using the relevant equation, we get
Vmax = A/sqrt(m) * sqrt(k) ... so, the slope of the graph should be the derivative of Vmax with respect to [A/sqrt(m)]. So...
Slope of second graph (dVm/d(A/sqrt(m)) = sqrt(k) ...... right?
So then, k = slope^2
If we plug in the slope that we obtained from our line of best fit, we get:
k=1.169^2 = 1.367N/m.

Note that 2.004 does NOT agree with 1.367! Why do you guys think this is? Do you guys see a flaw in my reasoning? I would not be surprised at all, because it seems like my derivative techniques were questionable at best!

Last edited: Jan 27, 2014
2. Jan 27, 2014

### vanhees71

Your formulas are correct, but for the max velocity you get
$$v_{\text{max}}^2=A^2\omega^2=A^2k/m.$$
So from the slope of the corresponding fit depends also on the amplitude.

3. Jan 27, 2014

### rhino1000

Keep in mind the graph is Vmax (not V^2max) vs A/sqrtm. We don't need to square anything until we actually plug in the experimentally-determined slope in order to find k (or so I would think). On another note, In my work, I - think - I took into account Amplitude, when I wrote that the slope should be d(Vmax) / d(A/sqrt(m)) . Isn't that true? I don't see where I forgot to take into account Amplitude... Expand on this?

Last edited: Jan 27, 2014
4. Jan 27, 2014

### rhino1000

Thanks for the reply, by the way!

5. Jan 27, 2014

### lightgrav

method 1 comes from timing an entire oscillation top-bottom-top (T) with different masses hanging on it (m);
your horizontal intercept (not the vertical one) should be -1/3 the spring's mass (9 gram is a bit light)

method 2 must be from sonic ranger v(t) and x(t) data traces;
vmax is the top of the v(t) curve (was that about -vmin at curve bottom?)
A is 1/2 the top-bottom displacement, and for √m you converted to kg before taking the root, right?

6. Jan 27, 2014

### rhino1000

Sorry - my internet wasn't working well for a bit there! Both graphs were formed from the same data. The data we collected was the amplitude of the object's motion, maximum velocity of the object, and period of the oscillation. We collected all of this information with 5 gram increments of mass, every increment from 25 grams to 75 grams.

I did indeed convert to kilograms, and yes the A is 1/2 the top-down displacement. "Luckily for us," the computer that the motion sensor was hooked up to spat out the amplitude for us, and we didn't have to use such complicated reasoning. Lolz.

Last edited: Jan 27, 2014
7. Jan 27, 2014

### rhino1000

Clause 1: correct.
Clause 2: Is this important? How do you figure?

8. Jan 27, 2014

### lightgrav

Your graph analysis looks ok to me ... so there's a mistake earlier, before graphing.
You're looking for a factor of 2: the time from top-to-bottom is ½ the oscillation T.
How did you MEASURE the Amplitude as the object moved past it? bottom-to-top is 2A.
did you include the spring's mass with the additional mass? did you have a "mass hanger" that was overlooked?

9. Jan 27, 2014

### rhino1000

I'm beginning to think that my reasoning was correct, and my numbers are as good as can be expected with the data that we have. This means that I am starting to think that faulty data is the cause of this 48% discrepancy between the two k values. The mass of the spring was not taken into account, I think that it is possible that this negligence could be the harbinger of bad results. Any thoughts?

10. Jan 27, 2014

### rhino1000

Lol! I didn't see this post before my latest one. You have to manually refresh on this site! We have similar thoughts, then! Awesome.

The computer literally gave us the period, so I am pretty confident our period isn't off by a factor of two. Same thing goes for amplitude. This leaves the spring's mass as the culprit. Also, we had a "mass hanger," but the hanger had a disk-weight welded onto it, that said a certain mass. I am pretty sure the hanger/welded disk combo is already accounted for in the varying masses of the object complex (unless my partner screwed up, but I SERIOUSLY doubt that this is the problem). In other words, I think that the 25 gram starting mass includes the mass of the hanger - and I am confident in this.

Last edited: Jan 27, 2014
11. Jan 27, 2014

### lightgrav

hmm. did the computer also give you vmax as a single value for each mass? called vmax, or vrms?
how big do you think the spring's mass was?

12. Jan 27, 2014

### rhino1000

So I guess the question is, would the neglected mass of the spring affect the one or the other adversely - and to what degree?

13. Jan 27, 2014

### rhino1000

The computer gave us the amplitude of the sine function that fit the velocity data. So the amplitude of the velocity's value should be the Vmax. I do not really remember how big the spring's mass was, but I suppose like 10-20g - it would seem to small to vary the number by 50% like this. It might have been even greater.

Perhaps it would be a good idea to calculate the mass of the spring that would cause a discrepancy this large? Also, do you think a spring's mass of, say, 20g, would have the same effect on the objects mass as if the object was 20g more massive?

14. Jan 27, 2014

### lightgrav

let's suppose you ignored the mass hanger's 25g.
method 1 multiplies by mass, so you're multiplying by a value that's too small, only ½ m at smallest mass.
method 2 divides by √m , so it essentially multiplies by too BIG a value, 1.4 times m at the smallest mass.

15. Jan 27, 2014

### rhino1000

Let me think about this. I don't see how you can simply multiply the result by some factor of m, rather than add m to the mass of the actual object, and seeing what effect such an addition would have on the data.

Edit: unless you actually did mean that m represents the mass of the object + the mass of the spring.

16. Jan 27, 2014

### rhino1000

I see that you think that the mass of the spring matters in both situations, any thoughts on which one would affect the results more?

Also, are you sure/ do you think that the mass of the spring can be modeled as an addition to the mass of the hanging object?

17. Jan 27, 2014

### lightgrav

yes, adding the same amount to each mass value ... would tweak them all by a different factor.
I was just showing that it pushes the two "k" results in opposite directions.

How did the computer display the period for the sine-wave?

18. Jan 27, 2014

### rhino1000

It just said Period = x s
Along with Amplitude = y m
In the velocity information box, it showed Velocities amplitude = z m/s
Along with Period = x s (same as displacement period) etc.

19. Jan 27, 2014

### rhino1000

Thanks for all the help, lightgrav, I really appreciate it!