Testing Can You Solve Physics GRE Problem #28 Using an Oscilloscope?

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The discussion revolves around a GRE physics question involving an oscilloscope and the analysis of a signal composed of a low-frequency sinusoid and a higher-frequency sinusoid. The original poster expresses difficulty in solving the problem and seeks assistance. A participant explains the signal characteristics, noting that the low-frequency component has a peak-to-peak voltage of 5V and an amplitude of 2.5V, while the high-frequency component has an amplitude of approximately 1.25V. They also discuss the frequency ratio of the components, suggesting a low-frequency component of about 400 Hz and a high-frequency component of 500 Hz, which aligns with the answer choice D. The original poster appreciates the clarification, particularly the distinction between the low and high-frequency sinusoids, which had been challenging due to their blending in the graph.
quantumworld
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GRE GRE GRE... I need some help :blushing:

this question is kinda of experimental question, dealing with oscilloscope, I tried to work it by calculating the beat, but it didn't work out...
may I provide the link to the question, because there is figure that follows it,
it is page 26, number 28.
http://phys.columbia.edu/~hbar/Physics-GRE.pdf
I am really stuck at it... thought about it for a long time.

Many thanks for all efforts
 
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You can tell the signal is one low-frequency sinusoid with another, smaller, higher-frequency sinusoid added to it. The minimum of the low-frequency component is right around 2 cm, at about 1.5 divisions (3V), while the maximum is right around 5 cm, at about 4 divisions (8V). That means the signal is 5V peak-to-peak, or has an amplitude of 5/2 = 2.5V. There is only one answer which fits this, choice D.

You can double check this answer easily. The second component certainly does have an amplitude of about 1.25V, 2.5V peak-to-peak, or a little over one division peak-to-peak. Also, there seem to be about six cycles of the high frequency component for every one cycle of the low-frequency component, so 83:500 Hz seems like the correct ratio.

You could go a step further and calculate the frequency of the low-frequency component. It appears to be about 5 cm per cycle, or about 2.5 milliseconds per cycle, or about 1/0.0025 = 400 Hz. The frequency given in choice D, 500 Hz, is probably close enough for comfort. (I wouldn't do all these checks on the test to avoid wasting time, but, if you're really not confident in your first answer, they will help.)

- Warren
 
Thank you Warren for your reply,
I could see your explanation, if I thought of the low frequency sinusoid as the average of the given graph, and the high frequency sinuoid as the wiggling of it. My problem was trying to distinguish which is which, because they are already blended.

THanks again, and please correct me if I was wrong.
 
That's a reasonable way to think of it.

- Warren
 
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