$A$ stationary wave is produced along a stretched string of length $80 \,cm$. The resonant frequencies of the string are $90 \,Hz$, $150 \,Hz$, and $210 \,Hz$. The speed of the transverse wave in the string is: (in $\,m/s$)

The tension in a wire is decreased by $19 \%$. The percentage decrease in frequency will be ......... $\%$

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

$A$ steel wire is used to stretch a spring. An oscillating magnetic field drives the steel wire up and down. $A$ standing wave with three antinodes is created when the spring is stretched by $4.0\, cm$. What stretch of the spring produces a standing wave with two antinodes with the same frequency (in $, cm$)?

The fundamental frequency of a transverse wave of a stretched string subjected to a tension $T_1$ is $300 \ Hz$. If the length of the string is doubled and subjected to a tension of $T_2$, the fundamental frequency of the transverse wave in the string becomes $100 \ Hz$. Then $T_2: T_1=$ (Linear density of the string is constant)

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