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 the closed end of the pipe is equal to

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

Answer 1

The correct option is A $a$

Figure shows variation of displacement of particle in a closed organ pipe for $3^{rd}$ overtone.

For third overtone
$$l=\frac{7\lambda }{4}\text{or}\lambda =\frac{4l}{7}\text{or}\frac{\lambda }{4}=\frac{l}{7}$$ Hence the amplitude at $P$ at a distance $l/7$ from closed end is ${}^{\prime }{a}^{\prime }$ because there is an antinode at that point.