In a resonance tube experiment using a tuning fork of constant frequency, the first two co… — JAMB UTME

In a resonance tube experiment using a tuning fork of constant frequency, the first two consecutive resonant lengths of the air column above the water level are measured to be $23.5\text{ cm}$ and $73.5\text{ cm}$ respectively. What is the end correction of the tube in centimeters?

Cevap: 1.5 cm

Cevap

The end correction of the tube is

1.5\text{ cm}

In a resonance tube closed at one end by water, consecutive resonances occur when the air column length increases by half a wavelength. Subtracting the first resonant length from the second gives

\frac{\lambda}{2} = 73.5\text{ cm} - 23.5\text{ cm} = 50.0\text{ cm}

, which yields

\lambda = 100.0\text{ cm}

and

\frac{\lambda}{4} = 25.0\text{ cm}

. The first resonance condition accounts for end correction through

L_1 + e = \frac{\lambda}{4}

. Substituting

L_1 = 23.5\text{ cm}

gives

e = 25.0\text{ cm} - 23.5\text{ cm} = 1.5\text{ cm}

Adım Adım Çözüm

Determine the wavelength using consecutive resonant positions

$\frac{\lambda}{2} = L_2 - L_1 = 73.5\text{ cm} - 23.5\text{ cm} = 50.0\text{ cm}$, so $\lambda = 100.0\text{ cm}$

For a column closed at one end, consecutive resonances occur at intervals of half a wavelength.

Calculate the quarter-wavelength value

$\frac{\lambda}{4} = \frac{100.0\text{ cm}}{4} = 25.0\text{ cm}$

The fundamental mode position of the displacement antinode corresponds to a distance of one quarter-wavelength from the closed end.

Calculate the end correction

$e = \frac{\lambda}{4} - L_1 = 25.0\text{ cm} - 23.5\text{ cm} = 1.5\text{ cm}$

The effective length for the first resonance includes the physical length plus the end correction.

Anahtar Kavram

End Correction in Resonance Air Columns

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