Match each vibrating acoustic system setup on the left with the correct mathematical expre… — JAMB UTME

Match each vibrating acoustic system setup on the left with the correct mathematical expression for its resonant frequency ( $f$ ) on the right, where $v$ is the speed of sound in air, $L$ is the physical length of the pipe or string, $e$ is the end correction per open end, $T$ is tension, and $\mu$ is linear mass density.

Fundamental mode of a pipe closed at one end, taking into account end correction $f = \frac{v}{4(L + e)}$
Fundamental mode of a uniform stretched string fixed at both ends $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$
Fundamental mode of a pipe open at both ends, taking into account end corrections at both open ends $f = \frac{v}{2(L + 2e)}$
First overtone of a pipe closed at one end, neglecting end correction $f = \frac{3v}{4L}$

Answer

The fundamental mode of a pipe closed at one end with end correction matches

f = \frac{v}{4(L + e)}

; the fundamental mode of a stretched string matches

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

; the fundamental mode of a pipe open at both ends with end correction at both ends matches

f = \frac{v}{2(L + 2e)}

; and the first overtone of a closed pipe without end correction matches

f = \frac{3v}{4L}

Each setup corresponds directly to its derived wave equation: closed pipes produce fundamental frequency

f = \frac{v}{4(L+e)}

for one open end, open pipes produce

f = \frac{v}{2(L+2e)}

for two open ends, stretched strings depend on tension and mass per unit length as

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

, and the first overtone of a closed pipe is its third harmonic

f = \frac{3v}{4L}

Step-by-Step Solution

Analyze boundary conditions and effective acoustic length for closed and open pipes.

A closed pipe has one displacement antinode at the open end and one node at the closed end, adding an effective end correction

e

to its physical length

L

(

L_{\text{eff}} = L + e

). An open pipe has two open ends, giving an effective length

L_{\text{eff}} = L + 2e

Air displacement antinodes occur slightly outside open pipe boundaries by a distance

e

per open end.

Derive the frequency formula for the fundamental mode of a closed pipe with end correction.

For the fundamental mode,

L + e = \frac{\lambda}{4} \implies \lambda = 4(L + e)

. Frequency

f = \frac{v}{\lambda} = \frac{v}{4(L + e)}

The distance between a node and an adjacent antinode is one-quarter of a wavelength.

Derive the fundamental frequency for a stretched string fixed at both ends.

L = \frac{\lambda}{2} \implies \lambda = 2L

. Using wave velocity

v = \sqrt{\frac{T}{\mu}}

f = \frac{v}{2L} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

Nodes exist at both fixed ends in a vibrating string, making the fundamental wavelength twice the length.

Derive the fundamental frequency of an open pipe considering both end corrections.

L + 2e = \frac{\lambda}{2} \implies \lambda = 2(L + 2e)

, so

f = \frac{v}{2(L + 2e)}

Antinodes occur at both open ends, placing half a wavelength within the effective acoustic length.

Determine the first overtone frequency for a closed pipe without end correction.

The first overtone is the third harmonic (

n = 3

), so

L = \frac{3\lambda}{4} \implies \lambda = \frac{4L}{3}

, which gives

f = \frac{3v}{4L}

Closed pipes support only odd integer multiples of the fundamental frequency.

Key Concept

Standing Waves and Resonance in Air Columns and Strings

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