$A$ closed organ pipe has length $L$. The air in it is vibrating in the third overtone with a maximum amplitude $a$. The amplitude at a distance $\frac{L}{7}$ from the closed end of the pipe is:

In a resonance pipe, the first and second resonances are obtained at depths $22.7 \,cm$ and $70.2 \,cm$ respectively. What will be the end correction (in $\,cm$)?

In an open organ pipe, $v_3$ and $v_6$ are the $3^{\text{rd}}$ and $6^{\text{th}}$ harmonic frequencies, respectively. If $v_6 - v_3 = 2200 \text{ Hz}$, then the length of the pipe is . . . . . . mm. (Take the velocity of sound in air as $330 \text{ m/s}$.)

$A$ closed pipe and an open pipe have their first overtones identical in frequency. Their lengths are in the ratio:

An open organ pipe is closed at one end such that the third overtone of the closed pipe is found to be higher in frequency by $200 \ Hz$ than the second overtone of the original open pipe. The fundamental frequency of the open pipe is (Neglect end correction). (in $Hz$)

$A$ closed organ pipe and an open pipe of the same length produce $4$ beats when they are set into vibrations simultaneously. If the length of each of them were twice their initial lengths,the number of beats produced will be