Match each vibrating acoustic system setup on the left with its corresponding displacement… — JAMB UTME

Match each vibrating acoustic system setup on the left with its corresponding displacement standing wave characteristics (number of nodes, antinodes, and wavelength $\lambda$ in terms of pipe or string length $L$ ) on the right.

Pipe closed at one end vibrating at its fundamental frequency1 node and 1 antinode; $\lambda = 4L$
Pipe open at both ends vibrating at its fundamental frequency1 node and 2 antinodes; $\lambda = 2L$
String fixed at both ends vibrating in its second harmonic3 nodes and 2 antinodes; $\lambda = L$
Pipe closed at one end vibrating in its first overtone2 nodes and 2 antinodes; $\lambda = \frac{4}{3}L$

Answer

Pipe closed at fundamental matches 1 node, 1 antinode ($\lambda = 4L$); Pipe open at fundamental matches 1 node, 2 antinodes ($\lambda = 2L$); String fixed at second harmonic matches 3 nodes, 2 antinodes ($\lambda = L$); Pipe closed at first overtone matches 2 nodes, 2 antinodes ($\lambda = \frac{4}{3}L$).

Each standing wave profile is uniquely determined by boundary constraints: fixed ends and closed pipe ends form displacement nodes, whereas open pipe ends form displacement antinodes. Counting the number of quarter-wavelength segments yields the exact relationship between wavelength $\lambda$ and system length $L$.

Step-by-Step Solution

Identify boundary conditions for displacement standing waves

Fixed ends of strings and closed ends of pipes are displacement nodes. Open ends of pipes are displacement antinodes.

Physical constraints prevent particle displacement at rigid boundaries while allowing maximum oscillation amplitude at open boundaries.

Calculate node/antinode count and wavelength for fundamental modes

For a closed pipe fundamental, $L = \frac{\lambda}{4}$, so $\lambda = 4L$ (1 node, 1 antinode). For an open pipe fundamental, $L = \frac{\lambda}{2}$, so $\lambda = 2L$ (1 node, 2 antinodes).

The distance between a consecutive node and antinode is $\frac{\lambda}{4}$, while the distance between two consecutive antinodes is $\frac{\lambda}{2}$.

Calculate node/antinode count and wavelength for higher harmonics

For a string fixed at both ends in the 2nd harmonic, two complete half-wavelength loops exist ($L = \lambda$), giving 3 nodes and 2 antinodes. For a closed pipe in its first overtone (3rd harmonic), $L = \frac{3\lambda}{4}$, so $\lambda = \frac{4}{3}L$, giving 2 nodes and 2 antinodes.

The harmonic number dictates how many quarter-wavelength or half-wavelength segments fit within length $L$.

Key Concept

Boundary conditions, node-antinode distributions, and wavelength formulas for standing waves in strings and pipes

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic