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:

Question

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: (A)  2f  (B)  3f/2  (C)  f  (D)  f /2

Tardigrade Admin · Accepted Answer

Key Idea: When open end pipe is dipped in water vertically it becomes close end pipe. Fundamental frequency of open pipe, f=(v/2l) ???(i) When half length of tube is dipped vertically in water, then length of the air column becomes half ( l=(l/2) ) and the pipe becomes closed. So, new fundamental frequency of closed pipe f=(v/4l)=(v/4( (l)2 ))=(v/2l) ??(ii) From Eqs. (i) and (ii), we get, f=f Hence, there will be no change in frequency.

Q. 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:

$2 f$

$3 f /2$

$f$

$f$ /2

Q. 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:

2f

3f/2

f

f /2

Q. 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:

$2 f$

$3 f /2$

$f$

$f$ /2