$A$ 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 .... $cm$.

Two wires $W_1$ and $W_2$ have the same radius $r$ and respective densities $\rho_1$ and $\rho_2$ such that $\rho_2 = 4\rho_1$. They are joined together at the point $O$,as shown in the figure. The combination is used as a sonometer wire and kept under tension $T$. The point $O$ is midway between the two bridges. When a stationary wave is set up in the composite wire,the joint is found to be a node. The ratio of the number of antinodes formed in $W_1$ to $W_2$ is

When a stretched wire of fundamental frequency $f$ is divided into three segments,the fundamental frequencies of these three segments are $f_1, f_2$ and $f_3$ respectively. Then the relation among $f_1, f_2, f_3$ and $f$ is (Assume tension is constant).

$A$ stretched wire of length $260 \ cm$ is set into vibrations. It is divided into three segments whose frequencies are in the ratio $2:3:4$. Their lengths must be

When tension $T$ is applied to a sonometer wire of length $\ell$,it vibrates with the fundamental frequency $n$. Keeping the experimental setup same,when the tension is increased by $8 \ N$,the fundamental frequency becomes three times the earlier fundamental frequency $(n)$. The initial tension applied to the wire in newton was:

Two wires of the same material are vibrating under the same tension. If the first overtone of the first wire is equal to the second overtone of the second wire and the radius of the first wire is twice the radius of the second,then the ratio of the length of the first wire to the second wire is: