The fundamental frequency of a closed organ pipe is 400 Hz. What will be the fundamental frequency of oscillation of an organ pipe of the same length?

Hint: A transverse wave is the one which travels through a medium, making crests and troughs. Wave on a string is an example of a transverse wave. A longitudinal wave is the one which travels through the medium making compressions and rarefactions. Sound waves are an example of longitudinal waves. When two waves approach each other in the same place, on the same line, standing waves are formed.

Formula used: $\nu = \dfrac{v}{4L}$ (for closed organ pipe) and $ \dfrac{v}{2L}$ (for open organ pipe)

Complete step by step answer:
Organ pipe is related to the experiments on sound waves. Open organ pipe is the one in which both ends are opened and then sound is passed through it. Closed organ pipe is the one in which only one end is open and the other is closed and then sound is passed.
Now, for a closed organ pipe, the fundamental frequency is given$\nu = \dfrac{v}{4L}$, where ‘v’ is the velocity of sound in the medium of organ pipe and ‘L’ being the length of pipe. For an open organ pipe, the fundamental frequency is given $\nu = \dfrac{v}{2L}$, where ‘v’ is the velocity of sound in the medium of organ pipe and ‘L’ being the length of pipe.
Given, $\nu = \dfrac{v}{4L} = 400 Hz$
Hence, $\nu_{open} = \dfrac{v}{2L} = 2\times\dfrac{v}{4L} = 2\times\nu = 2\times 400 = 800 Hz$

Hence the fundamental frequency of the corresponding length open organ pipe will be 800 Hz.

Note: Fundamental frequency is the frequency corresponding to the $1^{st}$harmonic of the frequency of the pipe. One must know that sound waves create standing waves inside the tube so that we can analyze the situation and know the parameters like frequency and conduct the experiment. Students are advised to derive these expressions once by their own.