Question

A closed organ pipe has length ‘l’ The air in it is vibrating in 3^rd 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. a/2
• C. $\frac{a\sqrt{3}}{2}$
• D. Zero

Answer 1

Answer: (A)

a

Answer 2

The figure shows variation of displacement of particles in a closed organ pipe for 3rd overtone. For third overtone l = $\frac{7\lambda}{4}$or $\lambda = \frac{4\mathcal{l}}{7}$ or $\frac{\lambda}{4} = \frac{\mathcal{l}}{7}$

Hence the amplitude at P at a distance l/7 from closed end is

‘a’ because there is an antinode at that point