In fundamental mode,the time required for the sound wave to reach up to the closed end of a pipe filled with air is $t$ seconds. The frequency of vibration of the air column is

$A$ string of length $3 \ m$ and linear mass density $0.0025 \ kg/m$ is fixed at both ends. One of its resonance frequencies is $252 \ Hz$. The next higher resonance frequency is $336 \ Hz$. Then the fundamental frequency will be ..... $Hz$.

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

For a particular resonance tube,following are four of the six harmonics below $1000 \,Hz$: $300, 600, 750$ and $900 \,Hz$. The two missing harmonics are:

$A$ 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:

$A$ metal rod of length $125 \ cm$ is clamped at its midpoint. If the speed of sound in the metal is $5000 \ ms^{-1}$, then the fundamental frequency of the longitudinal vibrations of the rod is (in $kHz$)