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A cylindrical tube open at both ends, has a vibrating air column of fundamental frequency ' $\mathrm{f}$ ' in air. The tube is dipped vertically in water so that half of its is in water. The fundamental frequency of the vibrating air column is now
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The correct answer is:
f
$v=\lambda f$
Fundamental frequency's $\lambda=\frac{v}{f} \Rightarrow f_0=\frac{v}{2 L}=f$
If the tube is dipped half into liquid then it acts as if closed at $\frac{\mathrm{L}}{2}$.
See figure
The frequency can be written as $\mathrm{f}_{\mathrm{c}}=\frac{\mathrm{v}}{\lambda_1}=\frac{\mathrm{v}}{2 \mathrm{~L}}=\mathrm{f}$
$\therefore$ Fundamental frequency would remain the same ' $\mathrm{f}$ '.
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