The fundamental frequency of a sonometer wire is $50 \text{ Hz}$ for a given length and tension. If the length is increased by $25 \%$ while keeping the tension constant,what is the percentage change in the frequency of the second harmonic?

$n$ waves are produced on a string in $1 \ s$. When the radius of the string is doubled,keeping the tension the same,the number of waves produced in $1 \ s$ for the same harmonic will be

$A$ 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 $......\,Hz$.

Two wires are fixed in a sonometer. Their tensions are in the ratio $8:1$. The lengths are in the ratio $36:35$. The diameters are in the ratio $4:1$. Densities of the materials are in the ratio $1:2$. If the lower frequency in the setting is $360 \ Hz$,the beat frequency when the two wires are sounded together is: (in $Hz$)

$A$ string $A$ has twice the length,twice the diameter,twice the tension,and twice the density of another string $B$. The overtone of $A$ which will have the same fundamental frequency as that of $B$ is:

$A$ metallic wire of length $L$ is fixed between two rigid supports. If the wire is cooled through a temperature difference $\Delta T$ ($Y =$ Young's modulus,$\rho =$ density,$\alpha =$ coefficient of linear expansion),then the frequency of transverse vibration is proportional to: