The closed and open organ pipes have the same length $L$. When they are vibrating simultaneously in their first overtone,they produce $4$ beats per second. The length of the open pipe is made half $(L/2)$ and that of the closed pipe is made two times $(2L)$ the original. Now,the number of beats produced if the two pipes are vibrating in their fundamental modes simultaneously is:

An open pipe of length $l$ is vibrating in the $3^{rd}$ overtone with a maximum amplitude $A$. The amplitude at a distance of $\frac{l}{16}$ from any open end is

The lengths of the two organ pipes open at both ends are $L$ and $(L+L_1)$. If they are sounded together,the beat frequency will be ($v=$ velocity of sound in air).

$A$ pipe's lower end is immersed in water such that the length of the air column from the top open end has a length of $25 \, cm$. The speed of sound in air is $350 \, m/s$. The air column is found to resonate with a tuning fork of frequency $1750 \, Hz$. By what minimum distance should the pipe be raised in order to make the air column resonate again with the same tuning fork?

The fundamental frequency of an open pipe of length $0.5 \ m$ is equal to the frequency of the first overtone of a closed pipe of length $l_c$. The value of $l_c$ in meters is:

An organ pipe $P_1$, closed at one end and containing a gas of density $\rho_1$ is vibrating in its first harmonic. Another organ pipe $P_2$, open at both ends and containing a gas of density $\rho_2$ is vibrating in its third harmonic. Both the pipes are in resonance with a given tuning fork. If the compressibility of gases is equal in both pipes, the ratio of the lengths of $P_1$ and $P_2$ is (assume the given gases to be monoatomic)