$A$ string is divided into three segments such that the segments possess fundamental frequencies in the ratio $1: 2: 3$. Then,the lengths of the segments are in the ratio:

Two wires of the same material with radii $r$ and $2r$ respectively are welded together end to end. The combination is then used as a sonometer wire under tension $T$. The joint is kept midway between the two bridges. The ratio of the number of loops in the wires,given that the joint is a node,is:

$A$ uniform string is vibrating with a fundamental frequency '$n$'. If the radius and length of the string are both doubled while keeping the tension constant,then the new frequency of vibration 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 ........... $N$.

$A$ string $A$ has twice the length,twice the diameter,twice the tension,and twice the density of another string $B$. The overtone of $A$ which will have the same fundamental frequency as that of $B$ is:

$A$ horizontal stretched string,fixed at two ends,is vibrating in its fifth harmonic according to the equation,$y(x, t) = (0.01 \ m) \sin[(62.8 \ m^{-1}) x] \cos[(628 \ s^{-1}) t]$. Assuming $\pi = 3.14$,the correct statement$(s)$ is (are) :
$(A)$ The number of nodes is $5$.
$(B)$ The length of the string is $0.25 \ m$.
$(C)$ The maximum displacement of the midpoint of the string from its equilibrium position is $0.01 \ m$.
$(D)$ The fundamental frequency is $100 \ Hz$.