$A$ tuning fork vibrating with a sonometer having $20 \ cm$ wire produces $5$ beats per second. The beat frequency does not change if the length of the wire is changed to $21 \ cm$. The frequency of the tuning fork (in Hertz) must be

$A$ uniform wire of length $L$,diameter $D$,and density $\rho$ is stretched by a tension $T$. The frequency $f$ of the wire is proportional to:

$A$ $1 \, cm$ long string vibrates with a fundamental frequency of $256 \, Hz$. If the length is reduced to $\frac{1}{4} \, cm$ keeping the tension unaltered,the new fundamental frequency will be: (in $, Hz$)

In order to double the frequency of the fundamental note emitted by a stretched string,the length is reduced to $\frac{3}{4}$ of the original length and the tension is changed. The factor by which the tension is to be changed 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 ........... $N$.

$A$ uniform string of length $L$ and mass $M$ is fixed at both ends while it is subject to a tension $T$. It can vibrate at frequencies $(v)$ given by the formula (where $n=1, 2, 3, \ldots$):