Match each acoustic or vibrating system operating under boundary conditions on the left wi… — JAMB UTME

Match each acoustic or vibrating system operating under boundary conditions on the left with its corresponding fundamental or harmonic frequency relationship on the right (where $v$ is sound speed in air, $T$ is string tension, $\mu$ is linear mass density, $L$ is length, and $r$ is internal pipe radius).

Pipe closed at one end of length $L$ operating at fundamental frequency (neglecting end correction) $f = \frac{v}{4L}$
Pipe open at both ends of length $L$ operating at fundamental frequency (neglecting end correction) $f = \frac{v}{2L}$
Stretched string of length $L$ fixed at both ends vibrating in its second harmonic mode $f = \frac{1}{L}\sqrt{\frac{T}{\mu}}$
Pipe closed at one end of length $L$ and radius $r$ operating at fundamental frequency with end-correction $f = \frac{v}{4(L + 0.6r)}$

Answer

Pipe closed at one end matches

f = \frac{v}{4L}

; Pipe open at both ends matches

f = \frac{v}{2L}

; Stretched string in second harmonic matches

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

; Pipe closed at one end with end-correction matches

f = \frac{v}{4(L + 0.6r)}

Each system is correctly matched based on wave mechanics boundary conditions: closed pipes produce quarter-wave fundamental modes (

\lambda = 4L

), open pipes produce half-wave fundamental modes (

\lambda = 2L

), the second harmonic of a string doubles the fundamental frequency

f_1 = \frac{1}{2L}\sqrt{T/\mu}

to yield

f_2 = \frac{1}{L}\sqrt{T/\mu}

, and end-correction increases the effective length of a closed pipe to

L + 0.6r

Step-by-Step Solution

Analyze boundary conditions for an ideal closed pipe

Displacement node at closed end, antinode at open end. Length

L = \frac{\lambda}{4} \Rightarrow \lambda = 4L

. Frequency

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

Determines the fundamental mode frequency formula for a closed pipe without end correction.

Analyze boundary conditions for an ideal open pipe

Displacement antinodes at both open ends. Length

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

. Frequency

f = \frac{v}{2L}

Determines the fundamental mode frequency formula for an open pipe.

Calculate the second harmonic frequency of a stretched string

For wave speed

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

, fundamental

f_1 = \frac{c}{2L}

. Second harmonic is

f_2 = 2f_1 = 2\left(\frac{1}{2L}\sqrt{\frac{T}{\mu}}\right) = \frac{1}{L}\sqrt{\frac{T}{\mu}}

Determines the frequency of the first overtone / second harmonic for a vibrating string fixed at both ends.

Apply end correction to a closed pipe

End correction

e = 0.6r

adds to physical length

L

at the open top end, giving

L_{eff} = L + 0.6r

. Fundamental frequency is

f = \frac{v}{4L_{eff}} = \frac{v}{4(L + 0.6r)}

Accounts for the antinode extending slightly beyond the open end of a real tube.

Key Concept

Boundary conditions, standing waves, harmonics in strings and air columns, and end-correction in resonance pipes

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