question_answer

A cylindrical tube, open at both ends has a fundamental frequency \[f\] in air. The tube dipped  vertically in water, so that half of it is in water, the fundamental frequency of the air colon is now:

A)  \[2\,\,f\]                            

B)  \[f\]

C)  \[\frac{3f}{4}\]

D)  \[\frac{f}{2}\]

Correct Answer: B

Solution :

Key Idea: When tube is dipped in water its acts like a closed end tube. Frequency of fundamental tone is a cylindrical pipe open at both ends is \[f=\frac{v}{2l}\]                              ?(1) Where v is velocity of sound and \[l\] is length of pipe. When half the length is dipped, it acts like a closed end tube, with fundamental frequency \[f'=\frac{v}{4l'}\] Given,   \[l'=\frac{l}{2}\] \[\therefore \]  \[f'=\frac{v}{4\left( \frac{l}{2} \right)}\]                 ?(2) Comparing Eqs. (1) and (2). We get \[f'=f\] Therefore, the fundamental frequency of air column is same as that for open one.