Show that the velocity of travelling wave produced in a string is `v= sqrt((T)/(mu))`

Explanation

Let us consider an elemental segment in the string as shown in the figure. Let A and B be two points on the string at an instant of time. Ket dl and dm be thelength and mass of the elemental string, respectively. By definition, linear mass density `mu` is

`mu=(dm)/(dl) " "...(1)`
`dm=mu dl " "...(2)`
Consider an element string AB having a curvature which looks like an arc of a circle with centre at O, radius R and the are subtending an angle `theta` at the origin O as shown in figure. The angle `thete` can be written in terms of arc length and radius as `(dl)/(R)= theta`. The centripetal acceleration supplied by the tension in the string is
`a_(cp)=(v^(2))/(R) " "....(3)`
The, centripetla force can be obtained when mass of the string (dm) is included in equation (3)
`F_(cp)=((dvm)v^(2))/(R) " "...(4)`
Experienced by elemental string can be calculated by substituing equation (2) in equation (4) we get the centripetal force as
`((dm)v^(2))/(R)=(muv^(2) dl)/(R) " "....(5)`
The tention T acts along the tengent of the elemental segment of the string at A and B. since the arc length is very small, variation in the tension force can be ignored. T can be resolved into horizontal component `T cos ((theta)/(2))` and vertical component `T sin ((theta)/(2))`. The horizontal components at A and B are equal in magnitude but opposite in direction. Hence they cencel each other. Since the elemental are length AB is taken to be very small, the verticaly components at A and B appears to acts vertical towards the centre of the arc and hence, they add up. The net radial fore `F_(r)` is
`F_(r)=2T sin ((theta)/(2)) " "...(6)`
Since the amplitude of the wae is very small when it is compared with the length of the string, the since of small angle is approximated as `sin ((theta)/(2))~~ (theta)/(2)`. Hence equation (6) can be written as,
`F_(r)=2Rxx(theta)/(2)=T theta " "...(7)`
But `theta=(dl)/(R)` therefore substituting is equation (7) we get,
`F_(r)=T(dl)/(R)" "...(8)`
Applying Newton's second law to the elemental string in the radial direction, under equilibrium, the radial component of the force is equal to the centripetal force. By equating equation (5) and equation (8), we get,
`T=(dl)/(R)= muv^(2)(dl)/(R)`
`v=sqrt((T)/(mu))` measured in `ms^(-1)`