Final answer:
Dipping a cylindrical tube that is open at both ends into water so that 3/4 of its length is submerged alters its effective length and increases its fundamental frequency to four times the initial frequency.
Step-by-step explanation:
The question pertains to the fundamental frequency of a cylindrical tube when it is dipped into water, effectively changing its length and, consequently, its fundamental frequency. Considering the cylindrical tube and its acoustical resonance fundamentals, we know that for a tube open at both ends, the fundamental wavelength (the longest wavelength) is twice the length of the tube, implying that fundamental frequency 'f' is associated with a full wavelength of the tube's length. When the tube is dipped into the water so that 3/4 of its length is inside the water, the effective length of the air column becomes 1/4 of the original length, because the water will close off the rest of the tube, leaving only the upper 1/4 as the vibrating air column.
This reduction in length means that the fundamental frequency of the air column (now a closed tube at one end and open at the other) will change. For a closed tube, the fundamental frequency corresponds to 1/4 of the wavelength of the sound wave, effectively making the new fundamental frequency four times the initial frequency if the tube were left open at both ends. Hence, by dipping the tube into water, the air column that resonates at the fundamental frequency will do so at a new frequency that is four times the original frequency 'f'. Therefore, the new fundamental frequency of the air column, now a quarter of its original length, is 4f.