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

Explanation

Velocity of transverse waves in a stretched string: Let us compute the velocity of transverse travelling waves on a string . When a jerk is given at one end (left end ) of the rope , the wave pulses move towards right end with a velocity v . This means that the pulses move with a velocity v and respect to an observer who is at rest frame . Suppose an observer also moves with same velocity v in the direction of motion of the wave pulse , then that observer will notice that the wave pulse is stationary and the rope is moving with pulse with the same velocity v . Consider an elemental segment in the string . Let A and B be two points on the string at an instant of time . Let dl and dm be the length and mass of the elemental string , respectively . By definition , linear mass density `mu` is
`mu = (dm)/(dl) " " ... (1)`
`dm = mu dl " " ... (2)`
The elemental string AB has a curvature which looks like an arc of a circle with centre at 0 , radius Rand the arc subtending an angle `theta` at the origin O. The angle `theta` can be written in terms of arc length and radius as `theta = (dl)/(R)` . The centripetal acceleration supplied by the tension in the string is
`a_(cp) = (v^(2))/(R) " " ... (3)`
Then , centripetal force can be obtained when mass of the string (dm) is included in equation (3) .
`F_(cp) =((dm) v^(2))/(R) " " ... (4)`
The centripetal force experienced by elemental string can be calculated by substituting equation (2) in equation ( 4) we get
`((dm) v^(2))/(R) = (mu v^(2) dl)/(R) " " ... (5) `

The tension T acts along the tangent 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. We can resolve T into horizontal component T cos `((theta)/(2))`and vertical component T sin `((theta)/(2))` The horizontal components at A and Bare equal in magnitude but opposite in direction, therefore, they cancel each other. Since the elemental arc length AB is taken to be very small, the vertical components at A and B appears to acts vertical towards the centre of the arc and hence, they add up. The net radial force F is
`F_(r)= 2 T sin ((theta)/(2)) " " ... (6)`
Since the amplitude of the wave is very small when it is compared with the length of the string , the sine of small angle is approximated as sin `((theta)/(2)) = (theta)/(2)` . Hence equation (6) can be written as `F_(r) = 2 T xx (theta)/(2) = T theta " " .... (7)`
But `theta = (dl)/(R)` ,, therefore substituting in 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. Hence equating equation (5) and equation (8) , we have `T(dl)/(R) = mu v^(2) (dl)/(R)`
`v = sqrt((T)/(mu))` measured in `ms^(-1) " " ... (9)`