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

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

An open pipe is in resonance in its $2^{nd}$ harmonic with a tuning fork of frequency ${f_1}$. Now,it is closed at one end. If the frequency of the tuning fork is increased slowly from ${f_1}$,then again a resonance is obtained with a frequency ${f_2}$. If in this case the pipe vibrates in its $n^{th}$ harmonic,then:

If $v$ is the speed of sound in air,then what is the shortest length of a closed pipe that resonates at a frequency $n$?

The fundamental frequency of a closed pipe is $400 \text{ Hz}$. If $\left(\frac{1}{3}\right)^{rd}$ length of the pipe is filled with water, the frequency of the $2^{\text{nd}}$ harmonic of the pipe will be (Neglect end correction). (in $\text{ Hz}$)

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