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 $180 \;Hz$?

$A$ stretched string is divided into three segments of lengths $50\,cm$,$40\,cm$,and $10\,cm$ with the help of bridges. Their vibrations will have frequencies in the ratio:

$A$ wire of density $8 \times 10^3\,kg/m^3$ is stretched between two clamps $0.5\,m$ apart. The extension developed in the wire is $3.2 \times 10^{-4}\,m$. If Young's modulus $Y = 8 \times 10^{10}\,N/m^2$,the fundamental frequency of vibration in the wire will be $......\,Hz$.

$A$ uniform string is vibrating with a fundamental frequency '$n$'. If the radius and length of the string are both doubled while keeping the tension constant,then the new frequency of vibration is:

$A$ violin emits sound waves of frequency $n_1$ under tension $T$. When tension is increased by $44\%$,keeping the length and mass per unit length constant,the frequency of sound waves becomes $n_2$. The ratio of frequency $n_2$ to frequency $n_1$ is:

$A$ string of length $1 \,m$ and mass $2 \times 10^{-5} \,kg$ is under tension $T$. When the string vibrates, two successive harmonics are found to occur at frequencies $750 \,Hz$ and $1000 \,Hz$. The value of tension $T$ is . . . . . . Newton.