$L$ length and $M$ mass uniform wire is stretched between two fixed points with tension $F$. $A$ sound of frequency $\mu$ is impressed on it. The maximum vibrational energy exists in the wire when $\mu$ =

An iron load of $2 \ kg$ is suspended in air from the free end of a sonometer wire of length $1 \ m$. $A$ tuning fork of frequency $256 \ Hz$ is in resonance with $\frac{1}{\sqrt{7}}$ times the length of the sonometer wire. If the load is immersed in water, the length of the wire in metres that will be in resonance with the same tuning fork is (Specific gravity of iron $= 8$).

The correct graph between the frequency $n$ and the square root of density $(\rho)$ of a wire,keeping its length,radius,and tension constant,is:

$A$ second harmonic has to be generated in a string of length $l$ stretched between two rigid supports. The points where the string has to be plucked and touched are

Two vibrating strings of the same material but lengths $L$ and $2L$ have radii $2r$ and $r$ respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes,the one of length $L$ with frequency $n_1$ and the other with frequency $n_2$. The ratio $n_1/n_2$ is given by

$A$ wire of density $9 \times 10^3 \text{ kg/m}^3$ is stretched between two clamps $1 \text{ m}$ apart and is subjected to an extension of $4.9 \times 10^{-4} \text{ m}$. The lowest frequency of transverse vibration in the wire is ..... $\text{Hz}$ $(Y = 9 \times 10^{10} \text{ N/m}^2)$.