The fundamental frequency of a closed organ pipe is $50 \mathrm{Hz}$. The frequency of the second overtone is, 

Find the fundamental frequency of a closed pipe, if the length of the air column is $42 \mathrm{m}$ (speed of sound in air$=332 \mathrm{m} {\mathrm{s}}^{-1}$)

A closed pipe has certain frequency. Now its length is halved. Considering the end correction, its frequency will now become

Two closed pipes produce $10$ beats per second when emitting their fundamental notes. If their lengths are in ratio of $25:26$, then their fundamental frequencies in $\mathrm{Hz}$, are

Two closed organ pipes when sounded together give five beats per second. Their lengths are in the ratio of $100:101$. The fundamental notes (in $\mathrm{Hz}$) produced by them are

The frequency of a stretched uniform wire under tension is in resonance with the fundamental frequency of a closed tube. If the tension in the wire is increased by $8 \mathrm{N}$, it is in resonance with the first overtone of the closed tube. The initial tension in the wire is

A resonance air column of length $20 \mathrm{cm}$, resonates with a tuning fork of frequency $250 \mathrm{Hz}$. The speed of sound in air is

In a long cylindrical tube, the water level is adjusted and the air column is made to vibrate in unison with a vibrating tuning fork kept at the open end. Resonances are obtained when lengths of air column are

In a resonance tube, the first resonance occurs at $16 \mathrm{cm}$ and the second resonance occurs at $49 \mathrm{cm}$. The end corrections will be