question_answer

An open resonating tube has fundamental frequency of\[x\]. When half of its length is  dipped into water then its fundamental frequency will be:                                                                                                [BHU PMT-2002]

A)                  \[x\]                                    

B)                  \[{{x}^{2}}\]

C)                  \[x\]                                    

D)                  \[m\]

Correct Answer: D

Solution :

                 Key Idea: When half of the tube is dipped in water it behaves as a closed tube.                 For on open pipe of length \[W=\int\limits_{{{V}_{i}}}^{{{V}_{f}}}{PdV}\] antinodes are formed at the two ends and node in between. Hence, fundamental frequency.                                      \[=\int\limits_{{{V}_{i}}}^{{{V}_{f}}}{K{{V}^{\gamma }}dV=\frac{K}{1-\gamma }\left[ {{V}^{1-\gamma }} \right]_{{{V}_{i}}}^{{{V}_{f}}}}\] Fundamental frequency of closed pipe of length \[=\frac{1}{\gamma -1}\left[ \frac{K}{V_{i}^{\gamma -1}}-\frac{K}{V_{f}^{\gamma -1}} \right]\], forming antinodes at open end and node at closed end is \[{{P}_{i}}V_{f}^{\gamma }={{P}_{i}}V_{f}^{\gamma }=K\]frequency of open  pipe.