The fundamental frequency of a sonometer wire is $n$. If its radius is doubled and its tension becomes half,while the material of the wire remains the same,the new fundamental frequency will be

$A$ stretched string of $1 \ m$ length and mass $5 \times 10^{-4} \ kg$ is having a tension of $20 \ N$. If it is plucked at $25 \ cm$ from one end,then it will vibrate with a frequency of ... $Hz$.

$A$ tuning fork produces $5 \, Hz$ beats per second when the length of a sonometer wire is either $1 \, m$ or $1.05 \, m$. Calculate the frequency of the tuning fork in $Hz$.

The fundamental frequency of a transverse wave of a stretched string subjected to a tension $T_1$ is $300 \ Hz$. If the length of the string is doubled and subjected to a tension of $T_2$, the fundamental frequency of the transverse wave in the string becomes $100 \ Hz$. Then $T_2: T_1=$ (Linear density of the string is constant)

$A$ 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?

The fundamental frequency of a wire stretched by $2 \ kgwt$ is $100 \ Hz$. The weight required to produce its octave is (in $kgwt$)