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GR9677 #28 |
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Problem
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This problem is still being typed. |
Lab Methods }Oscilloscope
A superposition of two oscillations has the form  \sin(\frac{\omega_1+\omega_2}{2}) ) . This implies that the cosine term is the amplitude of the combined wave.
Similarly, one can see one the lower frequency as the contribution towards the bigger envelope-like wave and the higher-frequency as the zig-zag-gish motion along the envelope.
One oscillation must have a high frequency and the other has a relatively lower frequency. Only choices (D), (A), and (B) show this trait. The high frequency oscillation should have a smaller amplitude than the lower frequency oscillation. Only choices (A) and (D) show this trait. Finally, the amplitude of the lower frequency wave forms the envelope, and the amplitude from that is only about 1 cm; on a 2V scale, this is about 2V---which is closest to choice (D).
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Comments |
ravenseldon666
2008-10-25 00:36:44 | the waves are of different amplitudes so how come they add up to the form product of two sines of a frequency sum and frequency difference? |  | dumbguy
2007-10-16 11:05:44 | Don't forget the scales of the oscilliscope. x is equivilant to freq. while y is equivilant to volts. You'll notice that the period of the the higher freq. wave is about 1 cm.( i used the bottoms of the first two) Since it is given that .5cm/ms, you can deduce that 1 cm is 2 ms. T=1/f, so the period is 1000/2 or 500hz. Choice D is the only one that has this. If not coninved, then check the amplitude. The amplitude is a little more then .5cm, which means it has to be a little more then 1 volt, which fits with D and its freq. as well. If you do the same thing for the wave envolope, you will get a freg. less then 100, around 80 and an amplitude of 2.5 volts. Don't forget when drawing the wave envolope to go thru the center of the higher freq. waves. | | adam
2005-11-09 22:25:22 | From hind sight...
If you count the periods for the high freq that occur during the 1 period of the lower freq you will get a ration about 6:1. This means that 6(low freq)= high freq. This satisfies B and D. But B has the amplitudes wrong (higher freq should have a smaller amplitude than low freq) leaving D as the answer. | |
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