Question

A closed organ pipe has length 'l'. The air in it is vibrating in 3rd overtone with maximum amplitude 'a'. The amplitude at a distance of l/7 from closed end of the pipe is equal to –

• A. a
• B. $\frac{a}{2}$
• C. $\frac{\sqrt{3}a}{2}$
  
• D. zero

Answer 1

The correct option is A a


$l=\frac{3\lambda }{2}+\frac{\lambda }{4}=\frac{7\lambda }{4}\Rightarrow l/7=\frac{\lambda }{4}$

$\therefore$ At distance l/7 from closed end, antinode is formed.