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 $1:2:3$?

$A$ wire of length $L$, diameter $d$, and density of material $\rho$ is under tension $T$ and has a fundamental frequency of vibration $n_A$. Another wire of length $2L$, diameter $3d$, and density of material $2\rho$ is vibrated under tension $2T$, and its fundamental frequency of vibration becomes $n_B$. The ratio $n_B : n_A$ is

$A$ string is stretched between two rigid supports separated by $75 \,cm$. There are no resonant frequencies between $420 \,Hz$ and $315 \,Hz$. The lowest resonant frequency for the string is (in $\,Hz$)

$L$ length and $M$ mass uniform wire is stretched between two fixed points with tension $F$. $A$ sound of frequency $\mu$ is impressed on it. The maximum vibrational energy exists in the wire when $\mu$ =

$A$ sonometer wire of resonating length $90 \ cm$ has a fundamental frequency of $400 \ Hz$ when kept under some tension. The resonating length of the wire with a fundamental frequency of $600 \ Hz$ under the same tension is . . . . . . $cm$.

Show that when a string fixed at its two ends vibrates in $1$ loop,$2$ loops,$3$ loops and $4$ loops,the frequencies are in the ratio $1 : 2 : 3 : 4$.