JEE MAIN - Physics (2002 - No. 44)
Tube $$A$$ has bolt ends open while tube $$B$$ has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube $$A$$ and $$B$$ is
$$1:2$$
$$1:4$$
$$2:1$$
$$4:1$$
Explanation
KEY CONCEPT : The fundamental frequency for closed organ pipe is given by $${\upsilon _c} = {v \over {4\ell }}$$ and
For open organ pipe is given by $${\upsilon _0} = {v \over {2\ell }}$$
$$\therefore$$ $${{{\upsilon _0}} \over {{\upsilon _c}}} = {v \over {2\ell }} \times {{4\ell } \over v} = {2 \over 1}$$
For open organ pipe is given by $${\upsilon _0} = {v \over {2\ell }}$$
$$\therefore$$ $${{{\upsilon _0}} \over {{\upsilon _c}}} = {v \over {2\ell }} \times {{4\ell } \over v} = {2 \over 1}$$
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