Final answer:
The second antinode for a standing wave formed by two opposing waves on a string is located at x = 0.75 meters. This is derived by setting the spatial part of the wave functions to equal nπ, where n is an odd integer, and solving for x when n=3.
Step-by-step explanation:
To find the location of the antinode having the second smallest value of x, we need to consider the superposition of two waves. The waves are represented by y=(6.0cm)cos π/2[(2.00m⁻¹)x+(8.00s⁻¹)t] and y=(6.0cm)cos π/2[(2.00m⁻¹)x−(8.00s⁻¹)t] for x ≥0. When these two waves interfere, they create a standing wave pattern with nodes and antinodes. The nodes occur where the two waves are out of phase and cancel each other out, while the antinodes occur at points where the waves are in phase and their amplitudes add constructively.
To solve this, we can note that the expression inside the cosine function for both waves involves π/2, 2.00 m⁻¹, and 8.00 s⁻¹. Since the waves are moving in opposite directions, when added, the time-dependent parts will cancel, leaving only the spatial parts. So, we have π/2(2.00m⁻¹)*2x = nπ for antinodes, where n is an odd integer. Here, the smallest value of x that satisfies the antinode condition would be when n=1. The second smallest value of x for which an antinode occurs is when n=3. Therefore, we have π/2(2.00m⁻¹)*2x = 3π, which simplifies to x = 0.75 m. Hence, the second antinode is located at x = 0.75 meters.