$A$ wire of length $2L$ is made by joining two wires $A$ and $B$ of same lengths but different radii $r$ and $2r$ and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire $A$ is $p$ and that in $B$ is $q$,then the ratio $p : q$ 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 $100 \ Hz$.

The length of a sonometer wire $AB$ is $110\; cm$. Where should the two bridges be placed from $A$ to divide the wire into $3$ segments whose fundamental frequencies are in the ratio of $1:2:3$?

$A$ guitar string of length $90\,cm$ vibrates with a fundamental frequency of $120\,Hz.$ The length of the string producing a fundamental frequency of $180\,Hz$ will be $...........cm$.

In an experiment with a sonometer,when a mass of $180\,g$ is attached to the string,it vibrates with a fundamental frequency of $30\,Hz$. When a mass $m$ is attached,the string vibrates with a fundamental frequency of $50\,Hz$. The value of $m$ is $.........\,g$.

$n$ waves are produced on a string in $1 \ s$. When the radius of the string is doubled,keeping the tension the same,the number of waves produced in $1 \ s$ for the same harmonic will be