© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Recognize the fundamental frequency of the original open tube
For an open cylindrical tube (both ends open) of length $l_0$, the fundamental frequency $f_0$ is given by
$$
f_0 = \frac{v}{2 \, l_0},
$$
where $v$ is the speed of sound in air.
Step 2: Note the change when the tube is dipped in water
When half of the tube is submerged in water, the effective air column length is still $l_0$ from top to bottom. However, the water now closes one end, converting the tube into a “closed pipe” of effective length $l_c = \frac{l_0}{2}$ (since half of the tube is under water and cannot vibrate with air).
Step 3: Write down the fundamental frequency for the new (closed) tube
For a pipe closed at one end, the fundamental frequency $f_c$ is given by
$$
f_c = \frac{v}{4 \, l_c}.
$$
Here, $l_c = \frac{l_0}{2}$. Substituting this into the formula, we get
$$
f_c = \frac{v}{4 \left(\frac{l_0}{2}\right)} = \frac{v}{2 \, l_0}.
$$
Step 4: Compare with the original frequency
We see that
$$
f_c = \frac{v}{2 \, l_0},
$$
which is exactly the same expression as the original open-tube fundamental frequency
$$
f_0 = \frac{v}{2 \, l_0}.
$$
Hence,
$$
f_c = f_0 = f.
$$
Final Conclusion
Therefore, even after half of the tube is submerged in water, the fundamental frequency of the air column remains the same, $f$.
