The first overtone of a stretched wire of given length is $320 \ Hz$. The first harmonic is .... $Hz$.

$A$ 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:

If the length of a stretched string is shortened by $40 \%$ and the tension is increased by $44 \%$,then the ratio of the final and initial fundamental frequencies is :

$A$ uniform string resonates with a tuning fork at a maximum tension of $32 \,N$. If it is divided into two segments by placing a wedge at a distance one-fourth of the length from one end,then to resonate with the same frequency,the maximum value of tension for the string will be ........... $N$.

$A$ segment of wire vibrates with a fundamental frequency of $450 \,Hz$ under a tension of $9 \,kg-wt$. The tension at which the fundamental frequency of the same wire becomes $900 \,Hz$ is:

$A$ string is under tension of $180 \ N$ and mass per unit length $2 \times 10^{-3} \ kg/m$. It produces two consecutive resonant frequencies with a tuning fork,which are $375 \ Hz$ and $450 \ Hz$. The mass of the string is: (in $g$)