If the fundamental frequency of a string is $220 \, cps$,the frequency of the fifth harmonic will be ......... $cps$.

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:

$A$ thin wire of length $99 \ cm$ is fixed at both ends as shown in the figure. The wire is kept under a tension and is divided into three segments of lengths $l_1, l_2$ and $l_3$ as shown in the figure. When the wire is made to vibrate,the segments vibrate respectively with their fundamental frequencies in the ratio $1: 2: 3$. Then,the lengths $l_1, l_2$ and $l_3$ of the segments respectively are (in $cm$):

$A$ string on a musical instrument is $50 \ cm$ long and its fundamental frequency is $270 \ Hz$. If the desired frequency of $1000 \ Hz$ is to be produced,the required length of the string is .... $cm$.

$A$ 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 $100 \ Hz$.

The fundamental frequency of a sonometer with a weight of $4\,kg$ is $256\,Hz$. The weight required to produce its octave is .... $kg-wt$