In the resonance tube experiment,the first resonant length is $l_1$ and the second resonant length is $l_2$. Then,the third resonant length will be:

In the fundamental mode,the time required for a sound wave to reach the closed end of an air-filled pipe is $t$ seconds. What is the frequency of vibration of the air column?

$A$ closed organ pipe of length $L_1$ and an open organ pipe contain diatomic gases of densities $\rho_1$ and $\rho_2$ respectively. The compressibilities of the gases are same in both pipes,which are vibrating in their first overtone with the same frequency. The length of the open organ pipe is (Neglect end correction).

An open organ pipe of length $l$ is sounded together with another organ pipe of length $l + x$ in their fundamental tones $(x << l)$. The beat frequency heard will be (speed of sound is $v$):

Find the fundamental frequency of a closed pipe,if the length of the air column is $42 \, m$. (speed of sound in air $= 332 \, m/s$) (in $, Hz$)

$A$ resonance tube closed at one end is of height $1.5 \ m$. $A$ tuning fork of frequency $340 \ Hz$ is vibrating above the tube. Water is poured in the tube gradually. The minimum height of water column for which resonance is obtained is . (Neglect end correction,speed of sound in air $= 340 \ m/s$) (in $cm$)