Sound Waves Legacy Problem #16 Guided Solution
Problem*
A 1.65-meter length string is forced to vibrate in its fifth harmonic. Determine the locations of the nodal positions. Express the locations as a distance measured from one of the ends of the string.
Audio Guided Solution
Like a wire or a rope, a string has a set of natural frequencies at which it vibrates with. And when a string vibrates with one of its natural frequencies, it begins to establish a standing wave pattern, a distinct pattern with so-called nodes and antinodes. Now these nodes are points along the string which appear to not even be moving at all. They're nodes, or points of no displacement. The antinodes, on the other hand, are points which undergo vigorous displacement up and down from its rest position. They're points of maximum displacement. Now the various frequencies which cause these standing wave patterns to be established in the string are referred to as harmonic frequencies. There's a number of them, and they're all mathematically related. We call the lowest frequency the first harmonic, or the fundamental frequency. That's when there's one section within the length of the string that's vibrating up and down. Of course, the second harmonic frequency is simply two times the frequency, and it has two sections which are vibrating vigorously up and down. And then, of course, the fifth harmonic would have five sections equal length within the length of the string that are vibrating up and down with an antinode. Now in this question, I wish to find the location of the nodal positions within the fifth harmonic standing wave pattern for a 1.65 meter length string. So what I need to first do is find out how long the equal length sections are that are vibrating up and down. So I simply take the 1.65 meters, and I divide by five, since there's actually five sections vibrating up and down for the fifth harmonic. That would give me 0.33 meters, which means that every 0.33 meters there's going to be a new node that distinguishes one section vibrating up and down from the next section vibrating up and down. And, of course, there's nodes at the two ends as well. So the positions of the nodes along the string are at 0 meters, 0.33 meters, twice 0.33, or 0.66 meters, 0.99, 1.32 meters, and finally at the very opposite end of the string, 1.65 meters.
Solution
0 m, 0.33 m, 0.66 m, 0.99 m, 1.32 m, and 1.65 m
Habbits of an Effective Problem Solver
- Read the problem carefully and develop a mental picture of the physical situation. If necessary, sketch a simple diagram of the physical situation to help you visualize it.
- Identify the known and unknown quantities and record in an organized manner, often times they can be recorded on the diagram itself. Equate given values to the symbols used to represent the corresponding quantity (e.g., \(\descriptive{v}{v,velocity} = 345\unit{\meter\per\second}\), \(\descriptive{λ}{λ,wavelength} = 1.28 \unit{m}\), \(\descriptive{f}{f,frequency} = \colorbox{gray}{Unknown}\)).
- Use physics formulas and conceptual reasoning to plot a strategy for solving for the unknown quantity.
- Identify the appropriate formula(s) to use.
- Perform substitutions and algebraic manipulations in order to solve for the unknown quantity.
Read About It!
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