$A$ steel wire of length $1.25 \ m$ is stretched between two rigid supports. The tension in the wire produces an elastic strain of $0.14 \%$. The fundamental frequency of the wire is (Density and Young's modulus of steel are $7.7 \times 10^3 \ kg \ m^{-3}$ and $2.2 \times 10^{11} \ N \ m^{-2}$ respectively). (in $Hz$)

$A$ string with a mass density of $4 \times 10^{-3} \, kg/m$ is under a tension of $360 \, N$ and is fixed at both ends. One of its resonance frequencies is $375 \, Hz$. The next higher resonance frequency is $450 \, Hz$. The mass of the string is:

$A$ sonometer wire has a length of $114 \ cm$, between two fixed ends. Where should two bridges be placed so as to divide the wire into three segments (in $cm$) whose fundamental frequencies are in the ratio $1:3:4$?

Four wires of identical length,diameter,and material are stretched on a sonometer. If the ratio of their tensions is $1 : 4 : 9 : 16$,then the ratio of their fundamental frequencies is:

In an experiment to study standing waves,you use a string whose mass per unit length is $\mu = (1.0 \pm 0.1) \times 10^{-4} \ kg/m$. You look at the fundamental mode,whose frequency $f$ is related to the length $L$ and tension $T$ of the string by the equation $L = \frac{1}{2f} \sqrt{\frac{T}{\mu}}$. You make a plot with $L$ on the $y$-axis and $\sqrt{T}$ on the $x$-axis,and find that the best-fitting line is $y = (8.0 \pm 0.3) \times 10^{-3}x + (0.2 \pm 0.04)$ in $SI$ units. What is the value of the frequency of the wave (including the error)? Express your result in $SI$ units $(Hz)$.

$A$ stationary wave is produced along a stretched string of length $80 \,cm$. The resonant frequencies of the string are $90 \,Hz$, $150 \,Hz$, and $210 \,Hz$. The speed of the transverse wave in the string is: (in $\,m/s$)