Two uniform stretched strings $A$ and $B$,made of steel,are vibrating under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$,the ratio of the lengths of the strings is

$A$ string vibrates with a frequency of $200 \ Hz$. When its length is doubled and tension is altered,it begins to vibrate with a frequency of $300 \ Hz$. The ratio of the new tension to the original tension is

$A$ tuning fork gives $4$ beats with $50 \, cm$ length of a sonometer wire. If the length of the wire is shortened by $1 \, cm$,the number of beats remains the same. The frequency of the fork is ............. $Hz$.

When a sonometer wire vibrates in the third overtone, the number of antinodes and nodes formed on the wire are respectively

$A$ string vibrates in its fundamental mode when a tension $T_1$ is applied to it. If the length of the string is decreased by $25 \%$ and the tension applied is changed to $T_2$, the fundamental frequency of the string increases by $100 \%$, then $\frac{T_2}{T_1} =$ (Linear density of the string is constant)

The fundamental frequency of a sonometer wire is $n$. If the length and diameter of the wire are doubled while keeping the tension constant,what is the new fundamental frequency?