Match each vibrating system mode on the left with the correct relationship between its sta… — JAMB UTME

Match each vibrating system mode on the left with the correct relationship between its standing wavelength ( $\lambda$ ) and length ( $L$ ) on the right.

Pipe closed at one end vibrating in its first overtone (third harmonic) $\lambda = \frac{4L}{3}$
Pipe open at both ends vibrating in its first overtone (second harmonic) $\lambda = L$
Stretched string fixed at both ends vibrating in its second overtone (third harmonic) $\lambda = \frac{2L}{3}$
Pipe closed at one end vibrating in its fundamental mode $\lambda = 4L$

Answer

The mode descriptions match their standing wavelength expressions as follows: Pipe closed at one end in its first overtone matches

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

; Pipe open at both ends in its first overtone matches

\lambda = L

; Stretched string in its second overtone matches

\lambda = \frac{2L}{3}

; Pipe closed at one end in its fundamental mode matches

\lambda = 4L

Each pair correctly links the specified boundary condition and mode of vibration to its mathematical relationship between wavelength

\lambda

and physical length

L

Step-by-Step Solution

Identify boundary conditions and available harmonics for each vibrating system.

Closed pipes support odd harmonics only (

n = 1, 3, 5, \dots

) with

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

. Open pipes and fixed strings support all integer harmonics (

n = 1, 2, 3, \dots

) with

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

Boundary conditions constrain node and antinode positions, determining allowed harmonic modes.

Determine the specific harmonic number

n

corresponding to each specified overtone.

First overtone of closed pipe

\rightarrow n = 3

; First overtone of open pipe

\rightarrow n = 2

; Second overtone of fixed string

\rightarrow n = 3

; Fundamental of closed pipe

\rightarrow n = 1

Overtones are higher resonant modes above the fundamental frequency.

Solve for wavelength

\lambda

in terms of system length

L

for each item.

For

n = 3

(closed pipe):

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

. For

n = 2

(open pipe):

L = \lambda \Rightarrow \lambda = L

. For

n = 3

(fixed string):

L = \frac{3\lambda}{2} \Rightarrow \lambda = \frac{2L}{3}

. For

n = 1

(closed pipe):

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

Rearranging each expression establishes the correct matching pair.

Key Concept

Boundary conditions and harmonic wavelength relations in pipes and vibrating strings

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