Question

In a resonance column experiment the first resonance is obtained when the level of water in tube is 20 cm from the open end resonance will also be obtained when the water level is at a distance of

Answer 1

Answer:

Given:

1st resonance is obtained at 20 cm in open organ tube.

To find:

2nd resonance at what depth.

Concept:

In an open organ tube , the open end forms anti-node and closed end forms node.

Let depth at which fundamental frequency happens be l , wavelength of standing wave be λ , depth of 2nd resonance be l2

For fundamental frequency , we can say that :

$\boxed{ \sf{ \huge{ \red{l = \dfrac{ \lambda}{4} }}}}$

For 2nd frequency , we can say that :

$\boxed{{ \sf{ \large{ \green{ \therefore l_{2} = \dfrac{ \lambda}{4} + \dfrac{ \lambda}{2} }}}}}$

${ \sf{ \large{ \green{ \therefore l_{2} = \dfrac{ 3\lambda}{4} }}}}$

So we can understand that

$\boxed{ \boxed{ \sf{l_{2} = 3 \times l}}}$

$\sf{ \implies \: l_{2} = 3 \times 20}$

$\sf{ \implies \: l_{2} = 60 \: cm}$

So final answer :

$\boxed{ \red{ \bold{ \sf{2nd \: resonance \: at \: 60 \: cm}}}}$