The frequency of a sonometer wire is $f$. When the weights producing the tension are completely immersed in water,the frequency becomes $f/2$. When the weights are immersed in a certain liquid,the frequency becomes $f/3$. The specific gravity of the liquid is:

$A$ wire having a linear mass density $9.0 \times 10^{-4} \; \text{kg/m}$ is stretched between two rigid supports with a tension of $900 \; \text{N}$. The wire resonates at a frequency of $500 \; \text{Hz}$. The next higher frequency at which the same wire resonates is $550 \; \text{Hz}$. The length of the wire is $...... \; \text{m}$.

$A$ steel wire of length $1 \,m$, mass $0.1 \,kg$, and uniform area of cross-section $10^{-6} \,m^2$ is rigidly fixed at both ends without any tension. Its temperature is lowered by $20^{\circ} C$ and transverse waves are set up by plucking the wire at the middle. The frequency of the fundamental mode is ($Y = 200 \,GPa$, $\alpha = 1.21 \times 10^{-5} {}^{\circ} C^{-1}$). (in $\,Hz$)

$A$ sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of $9 \,kg$ is suspended from the wire. When this mass is replaced by a mass $M$, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of $M$ is

Two uniform strings $A$ and $B$ made of steel are made to vibrate under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$, the ratio of the lengths of the strings is :

If you set up the $7^{th}$ overtone on a string fixed at both ends,how many nodes and antinodes are set up in it?