V118. Video Demonstration: Beats

Enter your answers for the following in the corresponding online assessment.

You'll need to have read section 14-9 before doing the following. You'll also need to view the video on beats: Streamed / RealPlayer / Flash / MP4

  1. To the nearest Hz, what beat frequency did you measure when the weight was at the middle position? (diagram C below)

Measuring frequency from a waveform printout

In problem 2, you'll be measuring frequency from printouts of Sound Pressure vs. Time waveforms recorded with Logger Pro. Here's the method to use for getting accurate results.

  1. Print the graph as large as you can for greatest accuracy.

  2. Since the scale divisions aren't marked very fine, here's how to boost precision. With a ruler, measure the full scale distance, call it ΔD, across the time axis to the nearest 0.01 cm. If the corresponding time spanned is ΔT, then the scale factor is SF = ΔT/ΔD. (Note that the time is in the numerator.)

  3. Measure the distance Δd to the nearest 0.01 cm between the two most widely-separated peaks. In order to find the corresponding time interval, Δt, multiply by the scale factor: Δt = (SF)Δd.

  4. Calculate the frequency using f = N/Δt, where N is the number of cycles between the peaks.
  1. Use the method described above to determine the frequency of an unweighted fork, a weighted fork, and both weighted and unweighted forks sounded together. Since the frequencies are close to the same, accurate measurements are particularly important for getting good results. Determine all frequencies to the nearest 0.1 Hz. Links to the graphs are given below.

a.  Unweighted and weighted forks: The two waveforms will print at the same scale on one sheet of paper in portrait orientation. Thus, you need measure the scale factor only once for both waveforms.
b. Weighted and unweighted forks sounded together: Print this in landscape orientation. Note that the time scale is different than for the previous waveforms.

  1. Calculate the beat frequency: fb = f1 - f2, where f1 is the frequency of the unweighted fork.

  2. Find the experimental error between the two values of beat frequency found in items 2b and 3. Use the value from 2b as the accepted value. Give your result to the nearest percent.

  3. The reason the value from 2b is used as the accepted value is because that value has much less uncertainty. In order to see this, you'll carry out what's called a max/min analysis. Let's suppose that the distances you measured had a total uncertainty of 0.04 cm. This would apply both to ΔD and Δd. (One could argue that Δd has a greater uncertainty than ΔD, because it's more difficult to estimate where the peaks are than where the ends of the time scale are. However, we'll stick with 0.04 cm for both for this illustration.) Now consider that when you calculate a Δt, you're actually using the formula, Δt = Δd(ΔT/ΔD) = ΔT(Δd/ΔD). You have to form a ratio of two distances. If you wanted to find how much bigger Δt could be with the uncertainties in play, you'd add 0.04 cm to Δd and subtract the same amount from ΔD. Similarly, if you wanted to find how much smaller Δt could be, you'd subtract 0.04 cm from Δd and add the same amount to ΔD. That would give you a range of values of Δt between which you could be confident that the actual value lay. Here's a table that illustrates the process. While the numbers are made up, they're illustrative of the data in the experiment. We'll assume N = 3.
N = 3 Measurement
or calculated
value

to maximize Δt
or minimize f
(+/- 0.04 cm)

 to minimize Δt
or maximize f
(-/+ 0.04 cm)
Δd (cm) 10.29 10.33 10.25
ΔD (cm) 13.14 13.10 13.18
Δt (s) 0.0157 0.0158 0.0156
f (Hz) 191.5 190.2 192.9

We can write the result for f including the uncertainty as f = 191.5 +/- 1.3 Hz. The relative uncertainty is 100(1.3)/191.5 = 0.7 %.

Now let's construct a similar table for the weighted and unweighted forks sounded together.

N = 3 Measurement
or calculated
value

to maximize Δt
or minimize f
(+/- 0.04 cm)

 to minimize Δt
or maximize f
(-/+ 0.04 cm)
Δd (cm) 15.75 15.79 15.71
ΔD (cm) 18.25 18.21 18.29
Δt (s) 0.3452 0.3468 0.3436
f (Hz) 8.69 8.65 8.73

We can write the result for fb including the uncertainty as fb = 8.69 +/- 0.04 Hz. The relative uncertainty is 100(0.04)/8.69 = 0.5 %.

Now let's examine the results. For both examples above, the relative uncertainty is small, under 1%. Let's look, though, at what happens when finding the beat frequency by taking the difference of f1 and f2. We'll use the value found for f above as f1 = 191.5 +/- 1.3 Hz. Suppose that for f2 we had found f2 = 185.3 +/- 1.5 Hz. For the difference, we'd have f1 - f2 = 6.2 +/- 2.8 Hz. The relative uncertainty of this result is 100(2.8)/6.2 = 45%. What was a small uncertainty for the individual tuning fork frequencies became a huge uncertainty when the frequencies were subtracted. Thus, finding the beat frequency by subtracting the individual frequencies is far more uncertain than the value fb = 8.69 +/- 0.04 Hz found directly from the waveform.

In order to practice the max/min process, carry it out for your measurement of f1 only.

  1. Where do you think the weight was for the measurement made of the frequency of the weighted fork? Choose one of the diagrams from problem 1 and tell why you chose it.

  2. Which note of the chromatic musical scale was the unweighted fork most nearly producing? (If you don't know what chromatic means, it's in the text reading.)

  3. Open this animation and follow the instructions in the description window.  Give your answers to the nearest tenth of a hertz.


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