Question

Two narrow organ pipes, are open (length l${}_1$ ) and the other closed (length l${}_2$ ) are sounded in their respective fundamental modes. The beat frequency heard is 5 Hz. If now pipes are sounded in their first overtone then also the beat frequency heard is 5 Hz. Then

• A. $\frac{l_1}{l_2}=\frac{1}{2}$
• B. $\frac{l_1}{l_2}=\frac{1}{1}$
• C. $\frac{l_1}{l_2}=\frac{3}{2}$
• D. $\frac{l_1}{l_2}=\frac{2}{3}$

Answer 1

The correct option is B $\frac{l_1}{l_2}=\frac{1}{1}$

In a closed pipe the $1^{st}$ normal mode and first overtone is $\frac{v}{4l_2}$ and $\frac{3v}{4l_2}$.
In a open pipe the $1^{st}$ normal mode and first overtone is $\frac{v}{2l_1}$ and $\frac{v}{l_1}$.
Now, $\frac{3v}{4l_2}-\frac{v}{4l_2}=5$ and $\frac{v}{l_1}-\frac{v}{2l_1}=5$.

Equating both of them we can see that $\frac{l_1}{l_2}=\frac{1}{1}$.