At the poles of the earth,a stretched wire of a given length vibrates in unison with a tuning fork. At the equator of the earth,for the same setting,to produce resonance with the same fork,the vibrating length of the wire

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 the tension in a string is increased by $3 \ kg \ wt$,the frequency of the fundamental mode increases in the ratio $2:3$. The initial tension in the string is: (in $kg \ wt$)

$A$ string is clamped at both ends and it is vibrating in its $4^{th}$ harmonic. The equation of the stationary wave is $Y = 0.3 \sin(0.157 x) \cos(200\pi t)$. The length of the string is ..... $m$ (all quantities are in $SI$ units).

$A$ sonometer wire has a length of $114 \ cm$, between two fixed ends. Where should two bridges be placed so as to divide the wire into three segments (in $cm$) whose fundamental frequencies are in the ratio $1:3:4$?

When a string is divided into three segments of length $l_1, l_2$ and $l_3,$ the fundamental frequencies of these three segments are $v_1, v_2$ and $v_3$ respectively. The original fundamental frequency $(v)$ of the string is