$A$ stretched string is vibrating in the second overtone. The number of nodes and antinodes between the ends of the string are respectively:

Four wires of identical length,diameter,and material are stretched on a sonometer. If the ratio of their tensions is $1 : 4 : 9 : 16$,then the ratio of their fundamental frequencies is:

The frequency of a sonometer wire is $f$. When the weights producing the tension are completely immersed in water,the frequency becomes $f/2$. When the weights are immersed in a certain liquid,the frequency becomes $f/3$. The specific gravity of the liquid is:

$A$ sonometer wire is in unison with a tuning fork of frequency '$n$' when it is stretched by a weight of specific gravity '$d$'. When the weight is completely immersed in water,'$x$' beats are produced per second,then

$A$ 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 $100 \ Hz$.

Two uniform wires of the same material are vibrating under the same tension. If the first overtone of the first wire is equal to the $2^{\text{nd}}$ overtone of the $2^{\text{nd}}$ wire and the radius of the first wire is twice the radius of the $2^{\text{nd}}$ wire,then the ratio of the length of the first wire to the $2^{\text{nd}}$ wire is: