question_answer

A closed organ pipe has length '\[\ell \]'. The air in it is vibrating in 3rd overtone with maximum amplitude 'a'. The amplitude at a distance off in from closed end of the pipe is equal to -

A) a

B) \[\frac{a}{2}\]

C) \[\frac{a\sqrt{3}}{2}\]

D) zero

Correct Answer: A

Solution :

The figure shows variation of displacement of particles in a closed organ pipe for 3rd overtone. For third overtone \[\ell =\frac{7\lambda }{4}\,\,or\,\,\lambda \,\,=\,\,\frac{4\ell }{7}\,\,or\,\,\frac{\lambda }{4}\,\,=\,\,\frac{\ell }{7}\] Hence the amplitude at P at a distance\[\ell /7\] from closed end is ?a? because there is an antinode at that point