Waves Ques 48
48. A cylindrical resonance tube open at both ends, has a fundamental frequency, ’ $f$ ’ in air. If half of the length is dipped vertically in water, the fundamental frequency of the air column will be
[1997]
(a) $2 f$
(b) $3 f / 2$
(c) $f$
(d) $f / 2$
Show Answer
Answer:
Correct Answer: 48.(c)
Solution:
- (c) Fundamental frequency of open pipe, $f=\frac{v}{2 l}$
When half of tube is filled with water, then the length of air column becomes half $(l^{\prime}=\frac{l}{2})$ and the pipe becomes closed.
So, new fundamental frequency
$f^{\prime}=\frac{v}{4 l^{\prime}}=\frac{v}{4(\frac{l}{2})}=\frac{v}{2 l}$
Clearly $f^{\prime}=f$.
BETA
