The length of the string of a musical instrument is $90 \;cm$ and has a fundamental frequency of $120 \;Hz$. Where (in $cm$) should it be pressed to produce a fundamental frequency of $180 \;Hz$?

$A$ sonometer wire stretched by weight '$w$' is in unison with a tuning fork. The corresponding resonating length is '$L_1$'. If the weight is increased by '$3w$',the corresponding resonating length of the sonometer in unison with the tuning fork becomes '$L_2$'. The ratio $\left(\frac{L_1}{L_2}\right)$ is:

$A$ 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)

$A$ wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $25 \ Hz$. The mass of the wire is $2 \ g$ and its linear mass density is $4 \times 10^{-3} \ kg/m$. What is the tension in the string (in $N$)?

When a vibrating tuning fork is placed on a sound box of a sonometer,$8$ beats per second are heard when the length of the sonometer wire is kept at $101 \,cm$ or $100 \,cm$. Then the frequency of the tuning fork is (Consider that the tension in the wire is kept constant). (in $\,Hz$)

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