The fundamental frequency of a sonometer wire is $n$. If the length,tension,and diameter of the wire are all tripled,what is the new fundamental frequency?

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 $\,N$)

$A$ sonometer wire is in unison with a tuning fork when it is stretched by weight $w$ and the corresponding resonating length is $L_1$. If the weight is reduced to $\frac{w}{4}$,the corresponding resonating length becomes $L_2$. The ratio $\frac{L_1}{L_2}$ is:

If the length of a stretched string is shortened by $x \%$ and the tension is increased by $44 \%$,then the ratio of the initial and final fundamental frequencies is $1:2$. The value of '$x$' is:

Two wires of the same material with radii $r$ and $2r$ respectively are welded together end to end. The combination is then used as a sonometer wire under tension $T$. The joint is kept midway between the two bridges. The ratio of the number of loops in the wires,given that the joint is a node,is:

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)