Pendulum experiment systematic errors

[Shyanne]

Homework Statement

So we had to the simple pendulum experiment and were measuring the effect of the length of the pendulum on its period of motion. However, our results produced a line of best fit that was significantly higher than the expected line of best fit (with length vs period squared). This suggests that there's systematic error involved but the error percentage decreases with increasing length. The lengths we used were: 5cm, 10cm, 15cm, 20cm, 25cm, 30cm.

Homework Equations

T=2pi*sqrt(L/g)

The Attempt at a Solution

As we measured the period using PASCO (velocity sensor) and a stopwatch, I don't think there's any significant systematic error associated with the devices as both graphs were similar. But, perhaps there error associated with the length of the pendulum -consistently measuring higher lengths but I don't know how we could've managed that. :(

 

[mfb]

How large are the deviations, and how did you measure the height? How large (size) was the test mass, and which point on it did you use for the length measurement? Did you take its moment of inertia into account?
Not length-related, but what was your angle of the oscillations, and did you consider deviations from the harmonic motion?

The length measurement is typically the largest source of uncertainty in such a lab experiment.

 

[Shyanne]

How large are the deviations, and how did you measure the height? How large (size) was the test mass, and which point on it did you use for the length measurement? Did you take its moment of inertia into account?
Not length-related, but what was your angle of the oscillations, and did you consider deviations from the harmonic motion?

The length measurement is typically the largest source of uncertainty in such a lab experiment.

The percentage errors were :
5cm- 73.8%
10cm- 25.26%
15cm- 10.64%
20cm- 4.724%
25cm- 2.092%
30cm- 0.5460%

I measured using a ruler and the size of the mass had a diameter of approximately maybe 5cm and we measured from the top of it (I realize that's a mistake but shouldn't the period of lower because of a lower length?).
We ensured that the angle was always the same as it was around 45 degrees.

 

[haruspex]

the angle was always the same as it was around 45 degrees.

That is really too much. SHM is only an approximate behaviour, valid for small angles. I don't know whether that accounts for your results.
The failure to consider the position of the mass centre of the mass sounds quite significant, but I'm not sure how to reverse engineer your readings from the errors you quote. Please post your actual measurements (and not as an image... something I can cut and paste from).

 

[mfb]

A mass with a diameter of 5 cm? That is ... large.

The 45 degree angle will lead to a constant factor between expected and actual time, and the wrong length measurement looks like it can explain the length-effect, but the sources values would be necessary to clarify this. My guess: adding 5 cm to the length of everything and taking the second order into account for the large angles will lead to a good agreement.

 

[haruspex]

adding 5 cm to the length of everything

5cm was the diameter, so likely a bit less to add. As against that, there is the rotational inertia of the mass, which adds a little.

 

[Shyanne]

That is really too much. SHM is only an approximate behaviour, valid for small angles. I don't know whether that accounts for your results.
The failure to consider the position of the mass centre of the mass sounds quite significant, but I'm not sure how to reverse engineer your readings from the errors you quote. Please post your actual measurements (and not as an image... something I can cut and paste from).

Sure thing! Here are the period values I obtained:
5cm: 0.780
10cm: 0.795
15cm: 0.860
20cm: 0.940
25cm: 1.025
30cm: 1.105

I thought that perhaps it could have something to do with the lengths -do you think it's reasonable to say that there'd be more error involved with smaller lengths?

 

[Shyanne]

A mass with a diameter of 5 cm? That is ... large.

The 45 degree angle will lead to a constant factor between expected and actual time, and the wrong length measurement looks like it can explain the length-effect, but the sources values would be necessary to clarify this. My guess: adding 5 cm to the length of everything and taking the second order into account for the large angles will lead to a good agreement.

I thought that too (error in length could explain the length-effect) but we measured to the top of the mass so wouldn't our length be smaller than the actual length and thus, our period should've been smaller than the expected values BUT it was actually larger. :(

 

[haruspex]

I thought that too (error in length could explain the length-effect) but we measured to the top of the mass so wouldn't our length be smaller than the actual length and thus, our period should've been smaller than the expected values BUT it was actually larger. :(

A shorter pendulum is faster. If you underestimated the length of the pendulum then you predicted a faster rate than you will get. That is, the observed period will come out too long.

 

[haruspex]

Sure thing! Here are the period values I obtained:
5cm: 0.780
10cm: 0.795
15cm: 0.860
20cm: 0.940
25cm: 1.025
30cm: 1.105

I thought that perhaps it could have something to do with the lengths -do you think it's reasonable to say that there'd be more error involved with smaller lengths?

It cannot be explained by the undermeasuring of the pendulum length. The errors are far too great.
That you saw almost the same period for 5cm and 10cm is just crazy. How did you measure the periods? I assume you counted some number of oscillations. How many? Was it still a fairly vigorous swing at the end?

 

[mfb]

At an angle of 45°, the period is 4% longer, which leads to an 8% deviation in the measurement of g - the same deviation for all measurements.

5cm was the diameter, so likely a bit less to add. As against that, there is the rotational inertia of the mass, which adds a little.

The 5 cm were an estimate based on the quoted errors, but that doesn't work out with the actual measurements now.

The finite moment of inertia of the object acts in the same way as a length offset. But no length offset can make those values agree with the expectation, neither with the large-angle correction nor without. You need a length offset of about 15 cm until the values look similar, but then the measured g is about 15 m/s² and still with large fluctuations.

It is possible to measure g with a precision better than 1% in such a lab experiment. 10% deviations are way too large, and 50% does not make sense.

Do you have a picture of the setup? How did your ridiculously huge mass look like and what was its shape?

 

[Shyanne]

It cannot be explained by the undermeasuring of the pendulum length. The errors are far too great.
That you saw almost the same period for 5cm and 10cm is just crazy. How did you measure the periods? I assume you counted some number of oscillations. How many? Was it still a fairly vigorous swing at the end?

It cannot be explained by the undermeasuring of the pendulum length. The errors are far too great.
That you saw almost the same period for 5cm and 10cm is just crazy. How did you measure the periods? I assume you counted some number of oscillations. How many? Was it still a fairly vigorous swing at the end?

I used both, a PASCO velocity sensor (which is known to be accurate) AND we counted the time taken for 20 oscillations and obtained the period from that. Both graphs were very similar which is why I thought the only factor that could produce some results is the length?

 

[Shyanne]

At an angle of 45°, the period is 4% longer, which leads to an 8% deviation in the measurement of g - the same deviation for all measurements.The 5 cm were an estimate based on the quoted errors, but that doesn't work out with the actual measurements now.

The finite moment of inertia of the object acts in the same way as a length offset. But no length offset can make those values agree with the expectation, neither with the large-angle correction nor without. You need a length offset of about 15 cm until the values look similar, but then the measured g is about 15 m/s² and still with large fluctuations.

It is possible to measure g with a precision better than 1% in such a lab experiment. 10% deviations are way too large, and 50% does not make sense.

Do you have a picture of the setup? How did your ridiculously huge mass look like and what was its shape?

It's weird because the greater the length of the pendulum, the more accurate the g value became. For the 30cm pendulum it was 9.7ms^2 whereas for the 5cm one it was 3.24m/s^2 which is just ridiculous. Is it possible for the accuracy to increase with increasing lengths? I looked at other experiments and sample sizes tended to begin at 30cm.

Sure, I've attached a picture -it's similar to that one except we only used one of the masses (one golden mass whereas in the picture they used two).

 

[mfb]

Is it possible for the accuracy to increase with increasing lengths?

Any fixed length error will have a smaller effect if the pendulum is longer.

Your picture leads to so many questions...

- was the pendulum tilted as shown in the picture?
- did you take the mass of the rod into account?
- did you just shift the golden mass along the rod, did you shift the rod, or how did you change the length?

If you just shifted the mass along the rod I can perfectly see why you got nearly the same period for 5 and 10 cm. The pendulum was mainly the rod then, with just a small effect from the golden mass. In that case there is hope to get the values in agreement with expectations.

 

[Shyanne]

Any fixed length error will have a smaller effect if the pendulum is longer.

Your picture leads to so many questions...

- was the pendulum tilted as shown in the picture?
- did you take the mass of the rod into account?
- did you just shift the golden mass along the rod, did you shift the rod, or how did you change the length?

If you just shifted the mass along the rod I can perfectly see why you got nearly the same period for 5 and 10 cm. The pendulum was mainly the rod then, with just a small effect from the golden mass. In that case there is hope to get the values in agreement with expectations.

No the pendulum wasn't tilted, it was just straight and was instead attached to a retort stand. I didn't realize that the mass of the pendulum would affect its period -what formula would that relate to? Yes, we did shift the golden mass along the rod and the length of our pendulum was essentially from the position that the rod is attached to the retort stand to the top of the golden mass.

 

[mfb]

Why do you expect the golden mass to play a role but the rod to be irrelevant?

I didn't realize that the mass of the pendulum would affect its period -what formula would that relate to?

Not the mass itself, but the mass distribution. If it is not a point-mass at the end, you get something called a physical pendulum: you have to take the mass distribution into account via the moment of inertia of the pendulum.

 

[Ray Vickson]

It's weird because the greater the length of the pendulum, the more accurate the g value became. For the 30cm pendulum it was 9.7ms^2 whereas for the 5cm one it was 3.24m/s^2 which is just ridiculous. Is it possible for the accuracy to increase with increasing lengths? I looked at other experiments and sample sizes tended to begin at 30cm.

Sure, I've attached a picture -it's similar to that one except we only used one of the masses (one golden mass whereas in the picture they used two).

Does the black box also swing up and down as the pendulum swings back and forth? If so, the black box acts a second pendulum attached at right angles to the first one, and unless it (the box) is feather-light, its mass and motion must be taken into account. So, it looks like you have a two-armed pendulum, with one end fixed and the other (longer) end at variable length.

 

[Shyanne]

Why do you expect the golden mass to play a role but the rod to be irrelevant?Not the mass itself, but the mass distribution. If it is not a point-mass at the end, you get something called a physical pendulum: you have to take the mass distribution into account via the moment of inertia of the pendulum.

So I did some more research and conducted more measurements. The mass of the weight at the end of the pendulum is 0.076kg and the pendulum rod itself is 0.030kg. I think the equation for the physical pendulum refers to the mass of the rod however how could I incorporate the mass of the weight as well? I thought maybe T=2pi*sqrt(I/mgL) where I = (1/2 x mass of rod x radius of rod squared) + (mass of weight x radius of weight squared) but I'm not too sure if that's correct.

 

[haruspex]

T=2pi*sqrt(I/mgL) where I = (1/2 x mass of rod x radius of rod squared) + (mass of weight x radius of weight squared)

Nearly right. The MoI of a uniform rod about one end is 1/3 x mass of rod x length of rod squared.

How long is the rod, 30cm?

Edit: by taking the rod as 40cm, the mass as a square block of width 5cm, the measured positions of the mass as being to its top, I got an average over the 6 readings of 9.90m/s². The range was 9.04 to 10.32.

 

[Shyanne]

Nearly right. The MoI of a uniform rod about one end is 1/3 x mass of rod x length of rod squared.

How long is the rod, 30cm?

Edit: by taking the rod as 40cm, the mass as a square block of width 5cm, the measured positions of the mass as being to its top, I got an average over the 6 readings of 9.90m/s². The range was 9.04 to 10.32.

The rod is: 0.357m and 0.067kg
The mass at the end is: 0.03kg with a radius of 0.005m

As the point mass is greater than the mass of the rod, is it possible for this pendulum to be a simple pendulum in any respects? Or is definitely a physical pendulum?

 

[haruspex]

The rod is: 0.357m and 0.067kg
The mass at the end is: 0.03kg with a radius of 0.005m

You previously posted
rod mass 0.030 kg
bob mass 0.076 kg
bob radius 5 cm
Now you are saying
rod mass 0.067 kg
bob mass 0.030 kg
bob radius 0.5 cm (which I do not believe)
Please clarify.

 

[Shyanne]

You previously posted
rod mass 0.030 kg
bob mass 0.076 kg
bob radius 5 cm
Now you are saying
rod mass 0.067 kg
bob mass 0.030 kg
bob radius 0.5 cm (which I do not believe)
Please clarify.

Oops, I'm very sorry, my data was all over the place. Here are the final measurements, please ignore previous values:
rod mass: 0.030kg
rod radius: 0.357m = 35.7cm
bob mass: 0.076kg
bob radius: 0.011m = 1.1cm

 

[mfb]

As the point mass is greater than the mass of the rod, is it possible for this pendulum to be a simple pendulum in any respects? Or is definitely a physical pendulum?

1/3 of the total mass is in the rod, it is certainly not negligible.

bob radius: 0.011m = 1.1cm

Radius as in cylinder radius? The length along the rod is still 5 cm?

Edit: by taking the rod as 40cm, the mass as a square block of width 5cm, the measured positions of the mass as being to its top, I got an average over the 6 readings of 9.90m/s2. The range was 9.04 to 10.32.

That looks like the right direction. Is there a systematic trend left? If yes, different bob parameters might improve that even more.

 

[haruspex]

Oops, I'm very sorry, my data was all over the place. Here are the final measurements, please ignore previous values:
rod mass: 0.030kg
rod radius: 0.357m = 35.7cm
bob mass: 0.076kg
bob radius: 0.011m = 1.1cm

Ok. As mfb asks, please clarify what is meant by bob radius here. What shape is the bob?
Also, does the rod project up above the pivot point? If so, by how much, and is that included in the .357m or additional?

 

[Shyanne]

Ok. As mfb asks, please clarify what is meant by bob radius here. What shape is the bob?
Also, does the rod project up above the pivot point? If so, by how much, and is that included in the .357m or additional?

Ok. As mfb asks, please clarify what is meant by bob radius here. What shape is the bob?
Also, does the rod project up above the pivot point? If so, by how much, and is that included in the .357m or additional?

The bob is cylinderical and thus the radius is the 'face' of the cylinder (not the length). The rod does project up above the pivot point, however, I did not measure this but from memory, it is approximately 5cm, so 0.05m which is NOT included in the 0.357m.

So far, this is the formula that I am using: (i've attached it). Is it correct?

 

[mfb]

Instead of l, you'll have to find the center of mass of your system.

The 5 cm above the pivot will contribute to the moment of inertia as well.

 

[Shyanne]

Instead of l, you'll have to find the center of mass of your system.

The 5 cm above the pivot will contribute to the moment of inertia as well.

But if I was investigating the effect of length of the pendulum on its period, where would I include these lengths? Or is it related to the center of mass?
I'm so sorry, I'm pretty hopeless at this, it's far beyond the scope of our curriculum (but it's a design investigation).

 

[mfb]

The distance to the center of mass is some equivalent to the length.

 

[haruspex]

But if I was investigating the effect of length of the pendulum on its period, where would I include these lengths? Or is it related to the center of mass?
I'm so sorry, I'm pretty hopeless at this, it's far beyond the scope of our curriculum (but it's a design investigation).

Yes, it does seem to be beyond what you expect.
I will lay out the (almost) full formula based on the information you have provided. Let:
L = total length of rod
D = distance from top of rod to pivot
m = mass of rod
M = mass of cylindrical bob
R = radius of bob
H = length of bob
X = distance from pivot to centre of bob
I will ignore the radius of the rod.
Moment of inertia of bob about a horizontal axis through its centre = (M/4)(R²+H²/3)
To get its MOI about the axis, add MX². (Parallel axis theorem.)
Moment of inertia of the rod about its centre = mL²/12
To get its MOI about the pivot, add m(L/2-D)²
Add the two MOIs about the pivot to get the total MOI. Call this I.

The restoring torque is τ=g sin(θ) (Mx+m(L/2-D)).
The period will be 2π√(I/τ)

 

[Shyanne]

Yes, it does seem to be beyond what you expect.
I will lay out the (almost) full formula based on the information you have provided. Let:
L = total length of rod
D = distance from top of rod to pivot
m = mass of rod
M = mass of cylindrical bob
R = radius of bob
H = length of bob
X = distance from pivot to centre of bob
I will ignore the radius of the rod.
Moment of inertia of bob about a horizontal axis through its centre = (M/4)(R²+H²/3)
To get its MOI about the axis, add MX². (Parallel axis theorem.)
Moment of inertia of the rod about its centre = mL²/12
To get its MOI about the pivot, add m(L/2-D)²
Add the two MOIs about the pivot to get the total MOI. Call this I.

The restoring torque is τ=g sin(θ) (Mx+m(L/2-D)).
The period will be 2π√(I/τ)

Haruspex, thank you very much for this! I'm truly grateful!
If you don't mind me asking, for the formula that you have provided for the bob; is that for a cylindrical bob?

Thank you so much once again!

 

[haruspex]

is that for a cylindrical bob?

Yes.
It's unusual for me/us to supply such a complete solution, but I feel it is appropriate in this case. However, I'd like to believe you understand how these formulae come to be . Please ask me to explain any parts you don't get.

 

[Shyanne]

Yes.
It's unusual for me/us to supply such a complete solution, but I feel it is appropriate in this case. However, I'd like to believe you understand how these formulae come to be . Please ask me to explain any parts you don't get.

I think I understand all the components, however I do have one last question (sorry): I understand that the torque is τ=mghsin(θ) but I do not understand why we are multiplying the masses by both X and (L/2)-D instead of only X.

 

[haruspex]

I think I understand all the components, however I do have one last question (sorry): I understand that the torque is τ=mghsin(θ) but I do not understand why we are multiplying the masses by both X and (L/2)-D instead of only X.

The two mass centres are different distances from the pivot. The bob is at distance X, the rod's centre is at L/2 from the end above the pivot, so at distance L/2-D from the pivot.
(I have assumed the rod stays fixed and the bob slides on the rod.)

 

[mfb]

With the approach of post #29 and values from the thread, I get values between 9.86 and 10.33 m/s^2. That looks reasonable.
If the bob length is reduced to 3 cm, the values fit even better with a range from 9.82 to 9.99.

Who designed this experiment and who wrote instructions for it?
Looks like a horrible experiment design to me.

 

[Shyanne]

The two mass centres are different distances from the pivot. The bob is at distance X, the rod's centre is at L/2 from the end above the pivot, so at distance L/2-D from the pivot.
(I have assumed the rod stays fixed and the bob slides on the rod.)

I see, thank you! I've plotted the results and they seem MUCH better than before! Is there are any particular reason that the 0.05cm pendulum length yielded the most in

The two mass centres are different distances from the pivot. The bob is at distance X, the rod's centre is at L/2 from the end above the pivot, so at distance L/2-D from the pivot.
(I have assumed the rod stays fixed and the bob slides on the rod.)

Ah I see, thank you! Also, according to this site (http://hyperphysics.phy-astr.gsu.edu/hbase/icyl.html), shouldn't the mass of interia of the bob be I= M((R^2/4)+(H^2/4))?