A uniform string of length 0.50\text{ m} and mass 2.0\text{ g} is fixed at both ends under… — JAMB UTME

A uniform string of length $0.50\text{ m}$ and mass $2.0\text{ g}$ is fixed at both ends under a tension of $90\text{ N}$ . When vibrating, the second harmonic of this string resonates with the first overtone of an air column in a pipe closed at one end. Assuming the speed of sound in air is $340\text{ m/s}$ , calculate the length of the pipe in meters.

Answer: 0.85 m

Answer

The length of the pipe is

0.85\text{ m}

The wave speed on the string is computed from tension and linear mass density as $150\text{ m/s}$, yielding a second harmonic frequency of $300\text{ Hz}$. Equating this to the first overtone (third harmonic) frequency formula of a closed air pipe, $f = \frac{3v_{air}}{4L_p}$, yields an exact pipe length of $0.85\text{ m}$.

Step-by-Step Solution

Calculate the linear mass density ($\mu$) of the string

$\mu = \frac{m}{L_s} = \frac{0.0020\text{ kg}}{0.50\text{ m}} = 4.0 \times 10^{-3}\text{ kg/m}$

Mass must be converted to kilograms before determining mass per unit length.

Determine the wave speed ($v_s$) along the stretched string

$v_s = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{90\text{ N}}{4.0 \times 10^{-3}\text{ kg/m}}} = \sqrt{22500} = 150\text{ m/s}$

The velocity of a transverse wave on a string depends on tension and linear mass density.

Calculate the second harmonic frequency of the string ($f_{2,s}$)

$f_{2,s} = \frac{v_s}{L_s} = \frac{150\text{ m/s}}{0.50\text{ m}} = 300\text{ Hz}$

The fundamental frequency is $f_{1,s} = \frac{v_s}{2L_s} = 150\text{ Hz}$, so the second harmonic is twice the fundamental frequency.

Set up the resonance equation for the first overtone of a closed pipe

$f_{3,p} = \frac{3 v_{air}}{4 L_p} = 300\text{ Hz}$

A pipe closed at one end produces only odd harmonics, so the first overtone is the 3rd harmonic.

Solve for the length of the closed pipe ($L_p$)

$L_p = \frac{3 \times 340\text{ m/s}}{4 \times 300\text{ Hz}} = \frac{1020}{1200} = 0.85\text{ m}$

Rearranging the frequency formula gives the required air column length.

Key Concept

Coupled resonance between standing waves on strings and air columns in closed pipes

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