$A$ 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

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:

$A$ sonometer wire resonates with $4$ antinodes between the two bridges for a given tuning fork when a $1 \text{ kg}$ mass is suspended from the wire. Using the same fork, when mass $M$ is suspended, the wire resonates producing $2$ antinodes between the two bridges. (The distance between the bridges remains the same). The value of $M$ is: (in $\text{ kg}$)

The equation of vibration of a stretched string fixed at both ends and vibrating in $5^{\text{th}}$ harmonic is $Y = 3 \sin(0.4x) \cos(200\pi t)$, where '$x$' and '$Y$' are in $cm$ and '$t$' is in seconds. The length of the string is: (in $\pi \text{ cm}$)

Two vibrating strings $A$ and $B$ of the same material but lengths $3 L$ and $2 L$ have radii $2 r$ and $3 r$ respectively. They are stretched under the same tension. String $A$ vibrates in the second overtone and string $B$ in the fundamental mode. The ratio of their frequencies $n_A / n_B$ will be:

$A$ 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?