Q. The lowest resonant length for the resonance tube is $15cm$ . Find the next resonant length for the resonance tube. Consider the end correction of the resonance tube as $1\text{cm}$ .
NTA AbhyasNTA Abhyas 2020
Solution:
In the concept of end correction, we also include the slight air column over the tube that is not a physical part of the tube because the source vibrates in the air and not inside the mouth of the tube.
End correction in a tube is given as $e=0.6r$ where $r$ is the radius of the mouth of tube.
As per data given in question, here
$e=1cm$
Writing the equations for $1^{s t}$ and $3^{r d}$ harmonic repectively, we get
$\frac{\lambda }{4}=l_{1}+e$
and
$\frac{3 \lambda }{4}=l_{2}+e$
Solving these equations , we get
$\frac{1}{3}=\frac{l_{1} + e}{l_{2} + e}=\frac{15 + 1}{l_{2} + 1}$ $48=l_{2}+1\Rightarrow l_{2}=47cm$
End correction in a tube is given as $e=0.6r$ where $r$ is the radius of the mouth of tube.
As per data given in question, here
$e=1cm$
Writing the equations for $1^{s t}$ and $3^{r d}$ harmonic repectively, we get
$\frac{\lambda }{4}=l_{1}+e$
and
$\frac{3 \lambda }{4}=l_{2}+e$
Solving these equations , we get
$\frac{1}{3}=\frac{l_{1} + e}{l_{2} + e}=\frac{15 + 1}{l_{2} + 1}$ $48=l_{2}+1\Rightarrow l_{2}=47cm$