In a closed-end pipe of length $105 \, cm$,standing waves are set up corresponding to the third overtone. What distance from the closed end,amongst the following,is a pressure node in $cm$?

The length of an open organ pipe is twice the length of a closed organ pipe. The fundamental frequency of the open pipe is $100 \ Hz$. The frequency of the third harmonic of the closed pipe is: (in $Hz$)

The first resonance length of a resonance tube is $40\, cm$ and the second resonance length is $122\, cm$. The third resonance length of the tube will be... $cm$

Two open organ pipes of fundamental frequencies $n_{1}$ and $n_{2}$ are joined in series. The fundamental frequency of the new pipe so obtained will be

Explain the formation of stationary waves in a closed pipe and derive the equations for natural frequencies (normal modes).

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: