Four wires of identical length,diameter,and material are stretched on a sonometer. If the ratio of their tensions is $1 : 4 : 9 : 16$,then the ratio of their fundamental frequencies is:

Two uniform stretched steel strings $A$ and $B$ are vibrating under the same tension. The first overtone of $A$ is equal to the second overtone of $B$. If the radius of $A$ is twice that of $B$,then the ratio of the lengths of the strings $l_A/l_B$ is:

$A$ 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).

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

In order to double the frequency of the fundamental note emitted by a stretched string,the length is reduced to $\frac{3}{4}$ of the original length and the tension is changed. The factor by which the tension is to be changed is

$A$ string fixed at both ends resonates at a certain fundamental frequency. Which of the following adjustments would not affect the fundamental frequency?