$A$ wire of density $9 \times 10^3 \,kg/m^3$ is stretched between two clamps $1 \,m$ apart and is subjected to an extension of $4.9 \times 10^{-4} \,m$. What will be the lowest frequency of the transverse vibrations in the wire in $Hz$ $[Y = 9 \times 10^{10} \,N/m^2]$?

One insulated conductor from a household extension cord has a mass per unit length of $\mu$. $A$ section of this conductor is held under tension between two clamps. $A$ subsection is located in a magnetic field of magnitude $B$ directed perpendicular to the length of the cord. When the cord carries an $AC$ current of $i$ at a frequency of $f$,it vibrates in resonance in its simplest standing-wave vibration state. Determine the relationship that must be satisfied between the separation $d$ of the clamps and the tension $T$ in the cord.

The frequency of transverse vibrations in a stretched string is $200 \ Hz$. If the tension is increased four times and the length is reduced to one-fourth the original value,the frequency of vibration will be .... $Hz$.

When the tension in a string is increased by $3 \ kg \ wt$,the frequency of the fundamental mode increases in the ratio $2:3$. The initial tension in the string is: (in $kg \ wt$)

$A$ $1 \, cm$ long string vibrates with a fundamental frequency of $256 \, Hz$. If the length is reduced to $\frac{1}{4} \, cm$ keeping the tension unaltered,the new fundamental frequency will be: (in $, Hz$)

When a vibrating tuning fork is placed on a sound box of a sonometer, $8$ beats per second are heard when the length of the sonometer wire is kept at $101 \,cm$ or $100 \,cm$. Then the frequency of the tuning fork is (Consider that the tension in the wire is kept constant). (in $\,Hz$)