Question

Fill in the blank
A cylindrical resonance tube open at both ends fundamental frequency $F$ in the air. Half-length of the tube is dipped vertically in the water. The fundamental frequency to the air column now is.....................

Answer 1

Fundamental frequency of open organ pipe:- $f=n=\frac{\nu }{2L}$

When tube is half dipped in water, ${\lambda }^{\prime }=\frac{\lambda }{4}$

And length of tube $=\frac{L}{2}$

$\therefore f_0=\frac{\nu }{4l^{\prime }}=\frac{\nu }{4\times \frac{L}{2}}=\frac{\nu }{2L}=f$