1. At the closed ends of a pipe, the air cannot move, since the walls are in the way. Ie, the velocity of the air in the mode must have a node at each of the ends. Alternatively the pressure can be arbitrarily high there, since the wall will push back. The pressure in general will have an antinode at the wall. In between the sound would have the usual sinusoidal dependence that modes in tubes tend to have. The lowest mode would have the velocity be zero at each of the ends, and than have it maximum in the middle. The second mode would have a node for the velocity both at the ends and another in the middle. Thus, to damp out the first mode but not the second one would want to put the gause in a place where the first mode has the air moving, so that as it moves past the gause, the friction damps out the mode. However, it should be at a place where the second mode has a velocity node, so that the air moving in the second mode does not damp itself out by friction against the gauze. Ie you want to be a velocity node of the second mode. The only place where the first does not have a velocity node, but the second mode does is right at the center of the pipe.
[ And on rereading the question, I notice that what I actually asked was where would you put the gauze to damp out the second and not the first. To not damp out the first, it should be a velocity node of the first mode, and the only nodes are at the very ends, where the second also has a node. Thus there is nowhere that you could put the gauze to damp out the second and not the first]
Note also that there is another option to damping out the mode, and that would be to drill a very small hole in the pipe. If the mode had a non-zero pressure fluctuation at that point, that pressure would force air through that small hole, and because it was so small, that air would suffer friction. This would damp out modes which had a non-zero pressure at that point. But if that point was a node for the pressure (not velocity) of that mode, that small hole would not damp out that mode. Now the pressure nodes of the first mode occur right at the middle of the pipe ( the place where the velocity is highest). The nodes for the pressure of the second mode occur 1/4 and 3/4 of the way along the pipe. Thus if you placed a small hole in the pipe just at its midpoint, that would be at a node of the first mode, and would not damp it out, but it would be at a place where the second mode did have large pressure fluctuations and would not damp it out.
2. The lines in the figures are the nodes of the various modes of the drum head. To not damp out a mode by placing your finger lightly, you must place it on a node of that mode. To damp out a mode, you need to put your finger at a place which is NOT a node. The first mode has nodes only at the edge where the drum is attached to its frame. Thus placing your finger anywhere ( but on the edges) will damp out that mode.
To damp out the third but not the second, you would have to put your finger at a point which was a node for the second mode, but was not a mode for the third mode. However the line of nodes of the second mode is also a node for the third. Ie, putting your finger on a node of the second mode, you will also be placing your finger on a mode of the third. There is nowhere that you can place your finger to damp out the third but not the second.
[Note that on the web page, with at least some browsers, a number of the horizontal lines were eliminated, which made this question hard to answer properly.)
3. 150x2=300Hz (one octave). 300x2=600 ( second octave) But 600x2=1200 which is larger than 800Hz. The ratio between 800Hz and 600 Hz is 800/600=4/3=1.333 This is (very near) a perfect fourth ( five semitones). So, 800 Hz is two octaves and five semitones abouve 150Hz.
4. A perfect fourth is very near a frequency ratio of 4/3 ( which is in fact the value of the perfect fourth in Pythagorian tuning). Since the lower is at 220Hz, the higher would be at 220 x4/3= 293.3 Hz.
5. Again only those modes which have a node in the center would be undamped if you held it there. This would be the second, the fourth,...
To lower the bar's frequency, you need to either increase the mass or decrease the stiffness of the bar. Since you are asked to remove material this makes it hard to add mass, so you must decrease the stiffness by more than you decrease the moving mass. Since at the nodes there is no moving mass ( the nodes do not move) this would suggest that removing thickness ( which would decrease the stiffness- make it easier to bend the bar) near the nodes would be most efficient. In fact, for a thick bar, removing thickness always decreases the stiffness by more than it decreases the mass. Thus removing thickness anywhere near the center of the bar ( where it bends and where the bending stiffness actually mades a difference- it does not bend much at the ends) will decrease the stiffness more than it decreases the moving mass, and thus decreases the frequency.
6. Q is a measure, not of how long, but how many cycles the bar oscillates befor it damps out. In particular it is about 4.3 times the number of cycles it takes for the amplitude to decrease to 1/2 its initial value. For a Q of 2000, this would mean that the amplitude would decrease to half in about 465 cylces. Since one cycle takes 1/3000 sec, this would mean it would take about 465/3000= .155 sec for it to decay to half its original amplitude.
In the other case, with a Q of 200, it would only oscillate for about 46 cycles befor it dropped to half its initial amplitude. However in this case one cycle only takes 1/200 sec. Thus it would takes 46/200=.23 sec . This is longer. Thus, even though the Q is smaller, the time taken is longer because the frequency is so much lower.
7. Th Tacoma Narrows bridge is one of prime examples of a false physics explanation. This bridge which collapsed in the fall of 1939 a few months after it opend, shook itself to pieces one morning when the wind increased to about 40 mph. For this to have been a resonance, that resonance must have had a very high Q, for the bridge to begun to vibrate so wildly ( the amplitude of oscillation- about a 45 degree twisting from side to side) was many hundreds of times larger than normal. Thus the Q would have had to be in the hundreds ( since the excess amplitude is proportional to the Q). But this also means that the external oscillating pushing on the bridge must have been tuned exactly to the frequency of the bridge to better than one part in hundreds, since a resonance only responds a lot if the tuning of the external force is very close to the resonant frequency - the higher the Q the closer it has to be.
Since the external force in this case was the wind, the wind would have had to have an oscillating part (eg gusts) which oscillated at exactly the same frequency as the bridge did, to better than one part in hundreds. No wind ever behaves that way. No gusts occur with such regularity. No wind ever maintains its speed to better than one part in hundreds. This explanation is silly, despite the fact that many physics text books say this is the explanation. ( A closer explanation is that the bridge behaves like a clarinet, which also has the feature that a steady blowing by the player causes a huge vibration- we will be looking at why later in the course).
8. The small lump of putty will not do anything to the tension in the string, and thus not to the stiffness of the modes. However if the lump of putty is forced to move while the mode oscillates, it will change the moving the mass. Ie, any mode which moves at the point where the putty is applied will have its frequency decreased (moving mass increases).
At a point a third of the way along the string, the third mode will have a node. But neither the lowerst the second, or the fourth have a node there. Thus all three of these modes will have their frequecies lowered.