It is desired to increase the fundamental resonance frequency in a tube which is closed at one end. This can be achieved by

Select all the correct alternatives. Two identical straight wires are stretched so as to produce $6 \mathrm{beats} \mathrm{per} \mathrm{second}$ when vibrating simultaneously. On changing the tension slightly in one of them, the beat-frequency remains unchanged. Denoting by $T_1, T_2$ the higher and the lower initial tensions in the strings, then it could be said that while making the above changes in tension.

In the figure shown an observer $O_1$ floats (static) on the water surface with ears in air while another observer $O_2$ is moving upwards with constant velocity$\mathit{ }{V}_1\mathit{=}\frac{\mathit{V}}{\mathit{5}}$ in water. The source moves down with constant velocity $V_{\mathit{s}}=\frac{V}{5}$ and emits the sound of frequency $f$. The velocity of sound in air is $V$ and that in water is $4V.$For the situation shown in figure: 

A vibrating string of a certain length $l$ under a tension $T$ resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length $75 \mathrm{cm}$ inside a tube closed at one end. The string also generates $4 \mathrm{beats} \mathrm{per} \mathrm{second}$ when excited along with a tuning fork of frequency $n$. Now when the tension of the string is slightly increased, the number of beats reduces to $2 \mathrm{per} \mathrm{second}$. Assuming the velocity of sound in air to be $340 \mathrm{m} {\mathrm{s}}^{-1}$, the frequency $n$ of the tuning fork in $\mathrm{Hz}$ is