Calculate the frequency of the second harmonic formed on a string of length $0.5 \ m$ and mass $2 \times 10^{-4} \ kg$ when stretched with a tension of $20 \ N$.

$A$ sonometer wire has a length of $114 \ cm$,between two fixed ends. Where should two bridges be placed so as to divide the wire into three segments (in $cm$) whose fundamental frequencies are in the ratio $1:3:4$?

The transverse displacement in a stretched string is given by $y = 0.06 \sin \left( \frac{2\pi}{3}x \right) \cos (120\pi t)$,where $x$ and $y$ are in $m$ and $t$ is in $s$. The length of the string is $1.5 \, m$ and its mass is $3.0 \times 10^{-2} \, kg$. The tension in the string is ..... $N$.

If the tension and diameter of a sonometer wire of fundamental frequency $n$ are doubled and density is halved,then its fundamental frequency will become

The fundamental frequency of a sonometer wire of length $l$ is $n_0$. $A$ bridge is now introduced at a distance of $\Delta l$ (where $\Delta l \ll l$) from the center of the wire. The lengths of wire on the two sides of the bridge are now vibrated in their fundamental modes. Then,the beat frequency is nearly

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: