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By the end of this lecture, you should:
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A tension in the string is required to have simple wave propagation such as we discuss (in fact we assume the tension is the same at all parts of the string). Whatever "ring" might be used to produce this tension would not only have to be friction-free (or else the end would not be "free") but would have to be of very low mass if not to load the string. In that case, if there was any slope in the string, the tension in the string would apply a transverse component to the ring, which being massless would have infinite acceleration. The only way this can be avoided is if we use the boundary condition that the slope is zero. Another way to look at this is that the massless ring can move very fast and always cancel out any force that would arise if the slope was not zero, and in that way maintain the slope at zero.
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Sound waves arise from overpressure (in fact very slight increases in pressure compared to ambient atmospheric pressure). In the vicinity of an opening and in the absence of unusual conditions like supersonic shock waves, pressure equalizes very rapidly. Thus there can be no overpressure near an opening, and the overpressure variable which represents the sound wave must be zero there, making a node.
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This took a bit of work to describe, but the solution is fairly easy. There will be antinodes at the walls and in the middle where you are moving your hand. The waves will have to be in the second harmonic in both directions (there are two nodes inside the tub in each direction). If the long direction is in x and one corner the origin, we get a water surface displacement \[z(x,y,t) = {A_{2,2}}\cos \left( {\frac{{2\pi x}}{{1.5}}} \right)\cos \left( {\frac{{2\pi y}}{1}} \right)\cos ({\omega _{2,2}}t)\]where \[{\omega _{2,2}} = v\sqrt {{{\left( {\frac{{2\pi }}{{1.5}}} \right)}^2} + {{\left( {\frac{{2\pi }}{1}} \right)}^2}} = 1(7.55) = 7.6rad/s\] or 1.2 Hz. Note: this actually worked quite well in the course developer’s tub! It is a slightly unusual Japanese-style tub with vertical sides and quite rectangular: results may be more complicated in other tubs.
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The general solution would be \[{p_{l,m,n}} = A\cos \left( {\frac{{l\pi x}}{4}} \right)\cos \left( {\frac{{m\pi y}}{3}} \right)\cos \left( {\frac{{n\pi z}}{2}} \right)\cos {\omega _{l,m,n}}t\] if the longest direction is in x, z is vertical, and the origin is on the floor at a corner. The angular frequency is \[{\omega _{l,m,n}} = v\sqrt {{{\left( {\frac{{l\pi }}{4}} \right)}^2} + {{\left( {\frac{{m\pi }}{3}} \right)}^2} + {{\left( {\frac{{n\pi }}{2}} \right)}^2}} \] By making a noise in the center there would have to be an antinode there, much as at the walls, so in all directions there would be second harmonic. So the actual lowest mode solution is \[{p_{2,2,2}} = A\cos \left( {\frac{{2\pi x}}{4}} \right)\cos \left( {\frac{{2\pi y}}{3}} \right)\cos \left( {\frac{{2\pi z}}{2}} \right)\cos {\omega _{2,2,2}}t\] with angular frequency\[{\omega _{2,2,2}} = 340\sqrt {{{\left( {\frac{{2\pi }}{4}} \right)}^2} + {{\left( {\frac{{2\pi }}{3}} \right)}^2} + {{\left( {\frac{{2\pi }}{2}} \right)}^2}} = 1390 rad/s. \] This is a frequency of 221 Hz, a common frequency in speech and near middle A on a piano. This explains why the room would sound "echo-y": part of that effect would be resonance. With a sound generator you would likely be able to explore many node effects in such a room!
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