Calculate the frequency of the second harmonic formed on a string of length $0.5 \ m$ and mass $2 \times 10^{-4} \ kg$ when stretched with a tension of $20 \ N$.

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 $(Hz)$.

$A$ wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45 \;Hz$. The mass of the wire is $3.5 \times 10^{-2} \;kg$ and its linear mass density is $4.0 \times 10^{-2} \;kg \;m^{-1}$. What is
$(a)$ the speed of a transverse wave on the string,and
$(b)$ the tension in the string?

$A$ sonometer wire resonates with a given tuning fork,forming a standing wave with $5$ antinodes between two bridges when a mass of $9 \ kg$ is suspended from the wire. When a mass '$m$' is suspended from the wire,with the same tuning fork and the same length between the two bridges,$3$ antinodes are formed. The value of mass '$m$' is: (in $kg$)

$A$ piano wire with a diameter of $0.90 \ mm$ is replaced by another wire of diameter $0.93 \ mm$ of the same material. If the tension of the wire is kept the same,then the percentage change in the frequency of the fundamental tone is

$A$ stretched string is vibrating in the second overtone. The number of nodes and antinodes between the ends of the string are respectively: