$A$ 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:

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

$A$ source of frequency $340 \,Hz$ is kept above a vertical cylindrical tube closed at the lower end. The length of the tube is $120 \,cm$. Water is slowly poured in just enough to produce resonance. Then, the minimum height (velocity of sound $= 340 \,m/s$) of the water level in the tube for that resonance is (in $\,m$)

$A$ cylindrical tube,open at both ends,has a vibrating air column of fundamental frequency '$f$' in air. The tube is dipped vertically in water so that half of its length is in water. The fundamental frequency of the vibrating air column is now:

The second overtone of an open organ pipe has the same frequency as the first overtone of a closed organ pipe $L$ metre long. The length of the open pipe will be:

$A$ closed organ pipe has a fundamental frequency $n$. If its length is doubled and its radius is halved,its fundamental frequency nearly becomes-