AAMC Practice test 3 question help

Started by bretonnia
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bretonnia

Studley Man
10+ Year Member
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I am working on questions in the AAMC practice test #3. Can someone help me figure out this problem? Below is a screen shot of the passage. My questions is: how am I supposed to know which is the first, second or third harmonic (in figure 1a)? Maybe this is a stupid question, but I just can't figure it out.

Thanks in advance

Here is the url if the picture won't show up: http://www.flickr.com/photos/63672153@N02/7611908898/

7611908898
 
Well, the easiest way to think of it is: the harmonic number is directly proportional to the number of nodes (i.e. points of zero displacement) in the wave. So the greater the number of nodes, the greater the harmonic number.

Or you can think of it as the greater the harmonic, the greater the number of waves in a given length. As you can see the wavelength (measured from crest to crest or trough to trough) decreases as you go from the solid black line, to the long-dashed line to the short-dashed line. This shows that the string is "trying to fit" more waves into a given space.

This picture might help.

http://upload.wikimedia.org/wikiped...ingerscale.svg/512px-Moodswingerscale.svg.png
 
I am working on questions in the AAMC practice test #3. Can someone help me figure out this problem? Below is a screen shot of the passage. My questions is: how am I supposed to know which is the first, second or third harmonic (in figure 1a)? Maybe this is a stupid question, but I just can't figure it out.

Thanks in advance

Here is the url if the picture won't show up: http://www.flickr.com/photos/63672153@N02/7611908898/

7611908898

Hey, so this is a standing wave.

Standing waves are basically waves that are either fixed in one end or both end. So first I will take about the waves that are fixed on both ends...when it comes to a standing wave that is fixed on both end, the principle wave length (the largest possible wave length) is always 2L, where L is the length of the string. Look at the solid line, it looks like 1/2 of a complete cycle, which means that 1 length of that string will be 1/2 of a wavelength, and 2 lengths of the string will be a full wavelength (with crest and trough). So this means that the largest wavelength you could possibly get with a string fixed on both sides is 2L (this will always be the first). And from there the second, and third will be L=1 wavelength and L= 3/2 of wavelength.

This same can be said of a string that is fixed only on one end, but instead of 2L being the principle wavelength (or 1st) it will be 4L.