Wave Motion 2 Question 36
36. A cylinder resonance tube open at both ends has fundamental frequency $f$ in air. Half of the length of the tube is dipped vertically in water. The fundamental frequency to the air column now is
(1992, 2M)
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Answer:
Correct Answer: 36. $f$
Solution:
- Fundamental frequency of open pipe $f=\frac{v}{2 l}$ and fundamental frequency of closed pipe,
$$ f^{\prime}=\frac{v}{4(l / 2)}=\frac{v}{2 l} \quad \text { or } \quad f^{\prime}=f $$
