Saved Bookmarks
| 1. |
A closed organ pipe of radiusr_(1) and an open organ pipe of radius r_(2) and having same length 'L' resonate when exited with a given tuning fork. Closed organ pipe and open organ pipe resonates in fundamental mode, then : |
|
Answer» `1.2 (r_(2)-r_(1)) =L` `implies (lambda)/4 = L+e` and for open organ pipe `(lambda)/2 = L + 2e_(2)` SO `4(L+e_(1))=2(L+2e_(2))` Now, `e_(1) = 0.6 r_(1)` and `e_(2) = 0.6 r_(2)` `2L = 4(e_(2)-e_(1))` `L = 2(e_(2)-e_(1))` `=2 xx 0.6 (r_(2)-r_(1))` `L = 1.2 (r_(2)-r_(1))`. |
|