A student is conducting a resonance column experiment. The tube has a diameter of 4 cm, and the tuning fork vibrates at 512 Hz. The air temperature is 38^oC, where the speed of sound is 336 m/s. The zero mark on the meter stick aligns with the top of the tube. For the first resonance, the water level reads :

During the first resonance :

$$ \frac{\lambda}{4} = l_1 + e $$

where $ l_1 $ is the length of the air column at resonance

Given that :

$$ \text{end correction} = e = 0.6r = 0.6 \times 2 = 1.2 \, \text{cm} $$

where $ r $ is the radius of the tube

The wavelength $ \lambda $ can be expressed as :

$$ \lambda = 4(l_1 + e) $$

The frequency $ f $ is related to the wavelength $ \lambda $ by :

$$ f = \frac{v}{\lambda} = \frac{v}{4(l_1 + e)} $$

Rearranging for $ l_1 $ :

$$ 4(l_1 + e) = \frac{v}{f} $$

Solving for $ l_1 $ :

$$ l_1 = \frac{v}{4f} - e $$

Substitute the given values :

$$ l_1 = \frac{336 \, \text{m/s}}{4 \times 512 \, \text{Hz}} - 0.012 \, \text{m} $$

$$ l_1 = \frac{336}{2048} - 0.012 $$

$$ l_1 = 0.164 \, \text{m} - 0.012 \, \text{m} $$

$$ l_1 = 0.152 \, \text{m} $$

Converting to cm :

$$ l_1 = 15.2 \, \text{cm} $$