If the length of a stretched string is reduced by $40 \%$ and the tension is increased by $44 \%$,then the ratio of the final to the initial frequencies of the stretched string is:

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:

The total length of a sonometer wire between fixed ends is $110 \ cm$. Two bridges are placed to divide the length of the wire in the ratio $6 : 3 : 2$. The tension in the wire is $400 \ N$ and the mass per unit length is $0.01 \ kg/m$. What is the minimum common frequency with which the three parts can vibrate in $Hz$?

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:

If the length of a stretched string is shortened by $40 \%$ and the tension is increased by $44 \%$,then the ratio of the final and initial fundamental frequencies is :